Thursday 26 September 2013

படித்ததில் பிடித்தது-நாய்க்குட்டிகள் விற்பனைக்கு


நாய்க் குட்டிகள் விற்பனைக்கு’ என்று எழுதிய பலகையை தனது கடைக் கதவுக்கு மேல் மாட்டிக் கொண்டிருந்தார்
அதன் உரிமையாளர்.

அந்தப் பலகை
குழந்தைகளை ஈர்க்கும் என்று நினைத்தார் அவர்.

அதன்படியே ஒரு சிறுவன், கடையின் முன் வந்து நின்றான்.
"நாய்க்குட்டிகளை நீங்கள் என்ன விலைக்கு விற்கப் போகிறீர்கள்?" என்று கேட்டான்.

"முப்பது டாலரிலிருந்து ஐம்பது டாலர் வரை" - கடைக்காரர் பதில் சொன்னார்.

அந்தக் குட்டிப் பையன் தனது பேண்ட் பைக்குள் கைவிட்டுக் கொஞ்சம் சில்லறைகளை எடுத்தான்.
"எங்கிட்ட 2.37 டாலர் இருக்கு. நான் நாய்க்குட்டிகளைப் பார்க்கலாமா?" என்று கேட்டான்.

கடை உரிமையாளர் புன்னகைத்து, உள் பக்கம் திரும்பி விசிலடித்தார். நாய்க் கூண்டிலிருந்து ஒரு பெண் இறங்கி நடைபாதை வழியாக ஓடி வந்தாள்.
அவளுக்குப் பின்னால், முடியாலான பந்துகளைப் போல ஐந்து குட்டியூண்டு நாய்க்குட்டிகள் ஓடிவந்தன.
ஒரு குட்டி மட்டும் மிகவும் பின்தங்கி மெதுவாக வந்தது.
பின் தங்கி, நொண்டி நொண்டி வந்த அந்தக் குட்டியை உடனே கவனித்த சிறுவன்,
"என்னாச்சு அதுக்கு?" என்று கேட்டான்.

அந்தக் குட்டி நாயைப் பரிசோதித்த கால்நடை மருத்துவர், அதற்குப் பிற்பகுதி சரியாக வளர்ச்சி யடையவில்லை. எனவே எப்போதும் நொண்டித்தான் நடக்கும், முடமாகத் தான் இருக்கும் என்று கூறிவிட்டதாக விளக்கினார் கடைக்காரர்.

சிறுவனின் முகத்தில் ஆர்வம்.
"இந்தக் குட்டிதான் எனக்கு வேணும்."
"அப்படின்னா நீ அதுக்குக் காசு கொடுக்க வேணாம். நான் அதை உனக்கு இலவசமாகவே தர்றேன்" என்றார் கடைக்காரர்.

அந்தக் குட்டிப் பையனின் முகத்தில் இப்போது சிறு வருத்தம்.
கடைக்காரரின் கண்களை நேருக்கு நேராகப் பார்த்து விரல் நீட்டிச் சொன்னான்.
"நீங்க ஒண்ணும் எனக்கு இலவசமாகக் கொடுக்க வேணாம். மற்ற நாய்க் குட்டிகளைப் போலவே இதுவும் விலை கொடுத்து வாங்கத் தகுதியானது தான்.
நான் இந்தக் குட்டிக்கு உரிய முழுத் தொகையையும் கொடுக்கிறேன்.
ஆனா, இப்போ எங்கிட்ட 2.37 டாலர்தான் இருக்கு. பாக்கித் தொகையை மாசாமாசம் 50 சென்ட்டா கொடுத்துக் கழிச்சிடறேன்."

ஆனாலும் கடைக்காரர் விடவில்லை.
"பையா... இந்த நாய்க் குட்டியால உனக்கு எந்தப் பிரயோஜனமும் இல்லை.
இதால மற்ற நாய்க்குட்டிகளைப் போல ஓடமுடியாது...
குதிக்க முடியாது... உன்னோட விளையாட முடியாது."

உடனே, அந்தப் பையன் குனிந்து தனது இடது கால் பேண்டை உயர்த்தினான்.
வளைந்து, முடமாகிப் போயிருந்த அக்காலில் ஓர் உலோகப் பட்டை மாட்டப்பட்டிருந்தது.
இப்போது அவன் கடைக்காரரை நிமிர்ந்து பார்த்துச் கொன்னான்.
"என்னாலும் தான் ஓட முடியாது... குதிக்க முடியாது. இந்தக் குட்டி நாயின் கஷ்டத்தைப் புரிஞ்சிக்கிறவங்க தான் இதுக்குத் தேவை!"

வாசிக்க கண் கலங்குது !!! பிடிச்சா பகிருங்க !

Saturday 14 September 2013

You think you are the best programmer out there? Well here are some challenges!


 
Are you a C,C++ or C# programmer who is looking forward to get some coding adventure and test your skills in some coding contests?  Then this is for you.
Annual contests:

1. International Conference on Functional Programming (ICFP) This has been running for a decade and happens in June or July each year. Though it's based in Germany, anyone can enter using any programming language, from any location. It's free to enter and your team isn't limited by size. 

2. The BME International: The BME International is an intense free to enter contest that takes place in Europe once a year for teams of three, and you have to bring your own computers and software. This year, the 7th edition took place in Budapest. This contest has had some interesting challenges in the past including driving a car over a virtual terrain? Other past tasks included controlling an oil-company, driving an assembly line robot and programming for secret communication. All programs were written in a 24-hour intense period!

3. International Collegiate Programming Contest: One of the longest running- this contest started in 1970 at Texas A&M and has been run by the ACM since 1989 and has IBM's involvement since 1997. One of the bigger contests, it has thousands of teams from universities and colleges competing locally, regionally and ultimately in the a world final. The contest pits teams of three university students against eight or more complex, real-world problems, with a gruelling five-hour deadline.

4. The Obfuscated C contest: The Obfuscated C contest has been running for nearly 20 years. This is done on the internet, with email submissions. All you have to do is write the most obscure or obfuscated Ansi C program in under 4096 characters length according to the rules. The 19th contest took place back in January/February 2007.

5. The Loebner Prize: The Loebner Prize is not a general programming contest but an AI challenge to enter a computer program that can do the Turing test, ie talk to a human sufficiently well to make the judges believe they are talking to a human. The Judge program, written in Perl will ask questions like "What time is it?", or "What is a hammer?" as well as comparisons and memory. The prize for the best entrant is $2,000 and a Gold Medal.

6. Chatterbox Challenge: It is similar to the Loebner Prize is the Chatterbox Challenge. This is to write the best chatter bot- a web based (or downloadable) application written in any language that can carry on text conversations. If it has an animated display that syncs with text then that is even better- you get more points!

7. International Problem Solving Contest (IPSC): This is more for fun, with teams of three entering via the web. There are 6 programming problems over a 5 hour period. Any programming language is allowed.

8. The Rad Race: Competitors in teams of two have to complete a working business program using any language over two days. This is another contest where you have to bring along equipment, including a router, computer(s), cables, a printer etc. The next one will be in Hasselt, Belgium in October 2007.

9. The Imagine Cup: Students at school or college compete by writing software applicable to the set theme which for 2008 is "Imagine a world where technology enables a sustainable environment." Entries started August 25th 2007.

10. ORTS Competition: ORTS (Open Real Time Strategy game) is a programming environment for studying real-time AI problems such as path-finding, dealing with imperfect information, scheduling, and planning in the domain of RTS games. These games are fast-paced and very popular. Using the ORTS software once every year there is a series of battles to see whose AI is best.

11. The International Obfuscated C Code Contest: Abbreviated IOCCC is a programming contest for the most creatively obfuscated C code. It started in 1984 and the 20th competition started in 2011. Entries are evaluated anonymously by a panel of judges. The judging process is documented in the competition guidelines and consists of elimination rounds. By tradition, no information is given about the total number of entries for each competition. Winning entries are awarded with a category, such as "Worst Abuse of the C preprocessor" or "Most Erratic Behavior", and then announced on the official IOCCC website. There's no prize except if your program is featured on the site then you won!

12. Google Code Jam: Running since 2008, it's open to anyone aged 13 or other, and you or a close relative don't work for Google or a subsidiary country and you don't live in a banned country: Quebec, Saudi Arabia, Cuba, Syria, Burma (Myanmar). (The contest is prohibited by law). There's a qualification round and three other rounds and the top 25 travel to a Google office for the Grand Final.

Ongoing contests:

13. Hutter Prize: If you can improve on the compression of 100 MB of Wikipedia data by 3% or better then you can win cash prizes. Currently the smallest compression is 15,949,688. For every 1% reduction (minimum 3%) you win €500.

14. Project Euler: This is an ongoing series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. computationally the problems should be solvable in less than a minute. A typical problem is "Find the first ten digits of the sum of one-hundred 50-digit numbers."

15. Sphere Online Judge. Run at Gdansk University of Technology in Poland, they have regular programming contests - with over 125 completed. Solutions are submitted to an automatic online judge that can deal with C, C++ and C# 1.0 and many other languages.

16. Intel's Threading Programming Problems: Running from September 2007 until the end of September 2008 Intel have their own Programming Challenge with 12 programming tasks, one per month that can be solved by threading. You get awarded points for solving a problem, coding elegance, code execution timing, use of the Intel Threading Building Blocks and bonus points for posting in their problem set discussion forum. Any language but C++ is probably the preferred language.

17. Codechef: Codechef is India's first, non-commercial, multi-platform online coding competition, with monthly contests in more than 35 different programming languages including C, C++ and C#. Winners of each contest get prizes, peer recognition and an invitation to compete at the CodeChef Cup, an annual live event.
 

Friday 13 September 2013

ஹைக்கூ(Hikoo) கவிதை- உனக்காக அங்கு நான்..

கவலைகள் உன்னை 
நோகடிக்கும் பொழுது 
உன் விழியோரம் 
வழியும் நீர்த்துளி 
துடைக்க 
உனக்காக அங்கு நான்...!

ஹைக்கூ(Hikoo) கவிதை ~ நான்

உன் கண்ணீருக்கு 
காரணம் 
நானாயிருக்க கூடாது...
உன் கண்ணில் 
நீர் வருமென்றால்... 
அன்று 
நானேயிருக்க கூடாது...!

Wednesday 4 September 2013

C Aptitude Questions and Answers(Part-2)


Predict the output or error(s) for the following:
36.    main()
{
int i;
printf("%d",scanf("%d",&i));  // value 10 is given as input here
}

Answer:
1

Explanation:
Scanf returns number of items successfully read and not 1/0.  Here 10 is given as input which should have been scanned successfully. So number of items read is 1.
37.       #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
            }


Answer:
100
38.      main()
{
int i=0;
 
for(;i++;printf("%d",i)) ;
printf("%d",i);
}

Answer:
            1

Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).
39.       #include
main()
{
  char s[]={'a','b','c','\n','c','\0'};
  char *p,*str,*str1;
  p=&s[3];
  str=p;
  str1=s;
  printf("%d",++*p + ++*str1-32);
}

Answer:
M

Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77("M");
40.       #include
main()
{
  struct xx
   {
      int x=3;
      char name[]="hello";
   };
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
Initialization should not be done for structure members inside the structure declaration
41.       #include
main()
{
struct xx
 {
              int x;
              struct yy
               {
                 char s;
                 struct xx *p;
               };
                         struct yy *q;
            };
            }


Answer:
Compiler Error

Explanation:
in the end of nested structure yy a member have to be declared
42.       main()
{
 extern int i;
 i=20;
 printf("%d",sizeof(i));
}

Answer:
Linker error: undefined symbol '_i'.

Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.
43.       main()
{
printf("%d", out);
}

int out=100;
Answer:
Compiler error: undefined symbol out in function main.

Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.
predict the output or error(s) for the following:
44.    main()
{
 extern out;
 printf("%d", out);
}
 int out=100;

Answer:
100     

            Explanation:  
This is the correct way of writing the previous program.
45.       main()
{
 show();
}
void show()
{
 printf("I'm the greatest");
}

Answer:
Compier error: Type mismatch in redeclaration of show.

Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
46.  main( )
{
  int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
  printf(“%u %u %u %d \n”,a,*a,**a,***a);
        printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
       }

Answer:
100, 100, 100, 2
114, 104, 102, 3
47.   main( )
{
  int a[ ] = {10,20,30,40,50},j,*p;
  for(j=0; j<5 j="" strong="">
    {
printf(“%d” ,*a);
a++;
    }
    p = a;
   for(j=0; j<5 j="" strong="">
      {
printf(“%d ” ,*p);
p++;
      }
 }

Answer:
Compiler error: lvalue required.
                       
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
48.       main( )
{
 static int  a[ ]   = {0,1,2,3,4};
 int  *p[ ] = {a,a+1,a+2,a+3,a+4};
 int  **ptr =  p;
 ptr++;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 *ptr++;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 *++ptr;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 ++*ptr;
       printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
}

Answer:
            111
            222
            333
            344
49.       main( )
{
 void *vp;
 char ch = ‘g’, *cp = “goofy”;
 int j = 20;
 vp = &ch;
 printf(“%c”, *(char *)vp);
 vp = &j;
 printf(“%d”,*(int *)vp);
 vp = cp;
 printf(“%s”,(char *)vp + 3);
}

Answer:
            g20fy

Explanation:
Since a void pointer is used it can be type casted to any  other type pointer. vp = &ch  stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly  the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.
50.    main ( )
{
 static char *s[ ]  = {“black”, “white”, “yellow”, “violet”};
 char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
 p = ptr;
 **++p;
 printf(“%s”,*--*++p + 3);
}

Answer:
            ck

Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p =  s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.
51.    main()
{
 int  i, n;
 char *x = “girl”;
 n = strlen(x);
 *x = x[n];
 for(i=0; i

   {
printf(“%s\n”,x);
x++;
   }
 }

Answer:
(blank space)
irl
rl
l

Explanation:
Here a string (a pointer to char) is initialized with a value “girl”.  The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position.  Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.
Predict the output or error(s) for the following:
52.     int i,j;
            for(i=0;i<=10;i++)
            {
            j+=5;
            assert(i<5 strong="">
            }

Answer:
Runtime error: Abnormal program termination.
                                    assert failed (i<5 br="">
Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,
            #undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of. 
53.       main()
            {
            int i=-1;
            +i;
            printf("i = %d, +i = %d \n",i,+i);
            }


Answer:
 i = -1, +i = -1

Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).
54. What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).
55.  what will be the position of the file marker?
            a: fseek(ptr,0,SEEK_SET);
            b: fseek(ptr,0,SEEK_CUR);

Answer :
            a: The SEEK_SET sets the file position marker to the starting of the file.
                        b: The SEEK_CUR sets the file position marker to the current position
            of the file.
56.       main()
            {
            char name[10],s[12];
            scanf(" \"%[^\"]\"",s);
            }
            How scanf will execute?

Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it  reads all character upto another quotation mark.
57.       What is the problem with the following code segment?
            while ((fgets(receiving array,50,file_ptr)) != EOF)

                                    ;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.
58.   main()
            {
            main();
            }

Answer:
 Runtime error : Stack overflow.

Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.
59.      main()
            {
            char *cptr,c;
            void *vptr,v;
            c=10;  v=0;
            cptr=&c; vptr=&v;
            printf("%c%v",c,v);
            }

Answer:
Compiler error (at line number 4): size of v is Unknown.

Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.
60.       main()
            {
            char *str1="abcd";
            char str2[]="abcd";
            printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
            }

Answer:
2 5 5

Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.
61.      main()
            {
            char not;
            not=!2;
            printf("%d",not);
            }

Answer:
0

Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.
62.       #define FALSE -1
            #define TRUE   1
            #define NULL   0
            main() {
               if(NULL)
                        puts("NULL");
               else if(FALSE)
                        puts("TRUE");
               else
                        puts("FALSE");
               }

Answer:
TRUE

Explanation:
The input program to the compiler after processing by the preprocessor is,
            main(){
                        if(0)
                                    puts("NULL");
            else if(-1)
                                    puts("TRUE");
            else
                                    puts("FALSE");
                        }
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.
Predict the output or error(s) for the following:
63.     main()
            {
            int k=1;
            printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
            }

Answer:
1==1 is TRUE

Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".
64.       main()
            {
            int y;
            scanf("%d",&y); // input given is 2000
            if( (y%4==0 && y%100 != 0) || y%100 == 0 )
                 printf("%d is a leap year");
            else
                 printf("%d is not a leap year");
            }

Answer:
2000 is a leap year

Explanation:
An ordinary program to check if leap year or not.
65.       #define max 5
            #define int arr1[max]
            main()
            {
            typedef char arr2[max];
            arr1 list={0,1,2,3,4};
            arr2 name="name";
            printf("%d %s",list[0],name);
            }

Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.
66.       int i=10;
            main()
            {
             extern int i;
              {
                 int i=20;
                        {
                         const volatile unsigned i=30;
                         printf("%d",i);
                        }
                  printf("%d",i);
               }
            printf("%d",i);
            }

Answer:
30,20,10

Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
            const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.
67.       main()
            {
                int *j;
                {
                 int i=10;
                 j=&i;
                 }
                 printf("%d",*j);
}

Answer:
10

Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.
68.      main()
            {
            int i=-1;
            -i;
            printf("i = %d, -i = %d \n",i,-i);
            }

Answer:
i = -1, -i = 1

Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.
69.       #include
main()
 {
   const int i=4;
   float j;
   j = ++i;
   printf("%d  %f", i,++j);
 }

Answer:
Compiler error

Explanation:
i is a constant. you cannot change the value of constant
70.       #include
main()
{
  int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };
  int *p,*q;
  p=&a[2][2][2];
  *q=***a;
  printf("%d..%d",*p,*q);
}

Answer:
garbagevalue..1

Explanation:
p=&a[2][2][2]  you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.
71.      #include
main()
  {
    register i=5;
    char j[]= "hello";                    
     printf("%s  %d",j,i);
}

Answer:
hello 5

Explanation:
if you declare i as register  compiler will treat it as ordinary integer and it will take integer value. i value may be  stored  either in register  or in memory.
72.      main()
{
              int i=5,j=6,z;
              printf("%d",i+++j);
             }

Answer:
11

Explanation:
the expression i+++j is treated as (i++ + j) 

Tuesday 3 September 2013

C Aptitude Questions and Answers(Part-1)



Predict the output or error(s) for the following:
1.    void main()
{
            int  const * p=5;
            printf("%d",++(*p));
}

Answer:
Compiler error: Cannot modify a constant value.
Explanation:   
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".
2.    main()
{
            char s[ ]="man";
            int i;
            for(i=0;s[ i ];i++)
            printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}

Answer:
                        mmmm
                       aaaa
                       nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally  array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the  case of  C  it is same as s[i].
3.      main()
{
            float me = 1.1;
            double you = 1.1;
            if(me==you)
printf("I love U");
else
                        printf("I hate U");
}

Answer:
I hate U

Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value  represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) . 
4.      main()
            {
            static int var = 5;
            printf("%d ",var--);
            if(var)
                        main();
            }

Answer:
5 4 3 2 1

Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively. 
5.      main()
{
             int c[ ]={2.8,3.4,4,6.7,5};
             int j,*p=c,*q=c;
             for(j=0;j<5 j="" strong="">
                        printf(" %d ",*c);
                        ++q;     }
             for(j=0;j<5 j="" strong="">
printf(" %d ",*p);
++p;     }
}

Answer:
                        2 2 2 2 2 2 3 4 6 5

Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.
6.      main()
{
            extern int i;
            i=20;
printf("%d",i);
}

Answer: 
Linker Error : Undefined symbol '_i'
Explanation:
                        extern storage class in the following declaration,
                                    extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .
Predict the output or error(s) for the following:
7.      main()
{
            int i=-1,j=-1,k=0,l=2,m;
            m=i++&&j++&&k++||l++;
            printf("%d %d %d %d %d",i,j,k,l,m);
}

Answer:
                        0 0 1 3 1

Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression  ‘i++ && j++ && k++’ is executed first. The result of this expression is 0    (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.
8.      main()
{
            char *p;
            printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
                        1 2

Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.
9.      main()
{
            int i=3;
            switch(i)
             {
                default:printf("zero");
                case 1: printf("one");
                           break;
               case 2:printf("two");
                          break;
              case 3: printf("three");
                          break;
              } 
}

Answer :
three

Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.
10.      main()
{
              printf("%x",-1<<4 strong="">
}

Answer:
fff0

Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.
11.      main()
{
            char string[]="Hello World";
            display(string);
}
void display(char *string)
{
            printf("%s",string);
}

Answer:
Compiler Error : Type mismatch in redeclaration of function display

Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.
12.      main()
{
            int c=- -2;
            printf("c=%d",c);
}

Answer:
                                    c=2;

Explanation:
Here unary minus (or negation) operator is used twice. Same maths  rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can  only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.
13.      #define int char
main()
{
            int i=65;
            printf("sizeof(i)=%d",sizeof(i));
}

Answer:
                        sizeof(i)=1

Explanation:
Since the #define replaces the string  int by the macro char
14.      main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}

Answer:
i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol.  ! is a unary logical operator. !i (!10) is 0 (not of true is false).  0>14 is false (zero).
Predict the output or error(s) for the following:
15.      #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:
77       

Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
 Now performing (11 + 98 – 32), we get 77("M");
 So we get the output 77 :: "M" (Ascii is 77).
16.      #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}

Answer:
SomeGarbageValue---1

Explanation:
p=&a[2][2][2]  you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.
17.      #include
main()
{
struct xx
{
      int x=3;
      char name[]="hello";
 };
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
You should not initialize variables in declaration
18.      #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
            struct xx *p;
};
struct yy *q;
};
}

Answer:
Compiler Error

Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.
19.      main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}

Answer:
hai

Explanation:
\n  - newline
\b  - backspace
\r  - linefeed
20.      main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

Answer:
45545

Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the  evaluation is from right to left, hence the result.
21.      #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer:
64

Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64
22.      main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s   %s",p,p1);
}

Answer:
ibj!gsjfoet

Explanation:
                        ++*p++ will be parse in the given order
Ø  *p that is value at the location currently pointed by p will be taken
Ø  ++*p the retrieved value will be incremented
Ø  when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.
23.      #include
#define a 10
main()
{
#define a 50
printf("%d",a);
}

Answer:
50

Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.
24.      #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}

Answer:
100

Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input  program to compiler looks like this :
                        main()
                        {
                             100;
                             printf("%d\n",100);
                        }
            Note:  
100; is an executable statement but with no action. So it doesn't give any problem
Predict the output or error(s) for the following:
25.   main()
{
printf("%p",main);
}

Answer:
                        Some address will be printed.

Explanation:
            Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.
26.       main()
{
clrscr();
}
clrscr();

           
Answer:
No output/error

Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).
27.       enum colors {BLACK,BLUE,GREEN}
 main()
{
 
 printf("%d..%d..%d",BLACK,BLUE,GREEN);
  
 return(1);
}

Answer:
0..1..2

Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
28.       void main()
{
 char far *farther,*farthest;
 
 printf("%d..%d",sizeof(farther),sizeof(farthest));
  
 }

Answer:
4..2 

Explanation:
            the second pointer is of char type and not a far pointer
29.       main()
{
 int i=400,j=300;
 printf("%d..%d");
}

Answer:
400..300

Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program, then printf will take garbage values.
30.       main()
{
 char *p;
 p="Hello";
 printf("%c\n",*&*p);
}

Answer:
H

Explanation:
* is a dereference operator & is a reference  operator. They can be    applied any number of times provided it is meaningful. Here  p points to  the first character in the string "Hello". *p dereferences it and so its value is H. Again  & references it to an address and * dereferences it to the value H.
31.       main()
{
    int i=1;
    while (i<=5)
    {
       printf("%d",i);
       if (i>2)
              goto here;
       i++;
    }
}
fun()
{
   here:
     printf("PP");
}

Answer:
Compiler error: Undefined label 'here' in function main

Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.
32.       main()
{
   static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
    int i;
    char *t;
    t=names[3];
    names[3]=names[4];
    names[4]=t;
    for (i=0;i<=4;i++)
           
printf("%s",names[i]);
}

Answer:
Compiler error: Lvalue required in function main

Explanation:
Array names are pointer constants. So it cannot be modified.
33.     void main()
{
            int i=5;
            printf("%d",i++ + ++i);
}

Answer:
Output Cannot be predicted  exactly.

Explanation:
Side effects are involved in the evaluation of   i
34.       void main()
{
            int i=5;
            printf("%d",i+++++i);
}

Answer:
Compiler Error

Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.
35.       #include
main()
{
int i=1,j=2;
switch(i)
 {
 case 1:  printf("GOOD");
                break;
 case j:  printf("BAD");
               break;
 }
}

Answer:
Compiler Error: Constant expression required in function main.

Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
            Note:
Enumerated types can be used in case statements.

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