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C Aptitude Questions and Answers



 

Predict the output or error(s) for the following:
53.    main()
{
                        char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}

Answer:
A

Explanation:
Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”. Since this string becomes the format string for printf and ASCII value of 65 is ‘A’, the same gets printed.

54.     void ( * abc( int, void ( *def) () ) ) ();
 
Answer::
 abc is a  ptr to a  function which takes 2 parameters .(a). an integer variable.(b).        a ptrto a funtion which returns void. the return type of the function is  void.
Explanation:
                        Apply the clock-wise rule to find the result.

55.     main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
            }

Answer:
No output

Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.

56.     main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}

Answer:
“Strings are not equal”
“Strings are not equal”
….

Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.

57.     main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}
 
Answer:
Compiler Error: Lvalue required.

Explanation:
As we know that increment operators return rvalues and  hence it cannot appear on the left hand side of an assignment operation.

58.     void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}

Answer:
garbage-value 0

Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.

59.      void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
            printf(“%d”, i);
}

Answer:
32767

Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.

60.      main()
{
                        int i=10,j=20;
            j = i, j?(i,j)?i:j:j;
                        printf("%d %d",i,j);
}
 
Answer:
10 10

Explanation:
                        The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
                        if(i,j)
                             {
if(i,j)
                             j = i;
                        else
                            j = j;                        
                        }
               else
                        j = j;      

61.     1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.

 
Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
            *a='F'       : illegal
                                    a="Hi"       : legal
 
2. 'const' applies to 'a'  rather than to the value of a (constant pointer to char )
            *a='F'       : legal
            a="Hi"       : illegal
 

3. Same as 1.

62.     main()
{
                        int i=5,j=10;
            i=i&=j&&10;
                        printf("%d %d",i,j);
}
 
Answer:
1 10

Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.

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