C PUZZLE

•    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.

      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.

•    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.

•   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.

•    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.

•   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.

•    #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.

•   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".

      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.

•   #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.

•       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.

•    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.