Home Interview Questions and AnswersTechnical Interview Questions and AnswersC Programming C Language Pointer Interview Questions and Answers For Freshers Part-6

c-program-pointers13. When would you use a pointer to a function?

Pointers to functions are interesting when you pass them to other functions. A function that takes function pointers says, in effect, “Part of what I do can be customized. Give me a pointer to a function, and I’ll call it when that part of the job needs to be done. That function can do its part for me.” This is known as a “callback.” It’s used a lot in graphical user interface libraries, in which the style of a display is built into the library but the contents of the display are part of the application.

As a simpler example, say you have an array of character pointers (char*s), and you want to sort it by the value of the strings the character pointers point to. The standard qsort() function uses function pointers to perform that task. qsort() takes four arguments

1. a pointer to the beginning of the array,

2. the number of elements in the array,

3. the size of each array element, and

4. a comparison function.

and returns an int.

The comparison function takes two arguments, each a pointer to an element. The function returns 0 if the pointed-to elements compare equal, some negative value if the first element is less than the second, and some positive value if the first element is greater than the second. A comparison function for integers might look like this:

int icmp( const int *p1, const int *p2 )


return *p1 – *p2;


The sorting algorithm is part of qsort(). So is the exchange algorithm; it just copies bytes, possibly by callingmemcpy() or memmove(). qsort() doesn’t know what it’s sorting, so it can’t know how to compare them. That part is provided by the function pointer.

You can’t use strcmp() as the comparison function for this example, for two reasons. The first reason is thatstrcmp()’s type is wrong; more on that a little later. The second reason is that it won’t work. strcmp() takes two pointers to char and treats them as the first characters of two strings. The example deals with an array of character pointers (char*s), so the comparison function must take two pointers to character pointers(char*s). In this case, the following code might be an example of a good comparison function:

int strpcmp( const void *p1, const void *p2 )


char * const *sp1 = (char * const *) p1;

char * const *sp2 = (char * const *) p2;

return strcmp( *sp1, *sp2 );


The call to qsort() might look something like this:

qsort( array, numElements, sizeof( char * ), pf2 );

qsort() will call strpcmp() every time it needs to compare two character pointers (char*s).

Why can’t strcmp() be passed to qsort(), and why were the arguments of strpcmp() what they were?

A function pointer’s type depends on the return type of the pointed-to function, as well as the number and types of all its arguments. qsort() expects a function that takes two constant void pointers:

void qsort( void *base,

size_t numElements,

size_t sizeOfElement,

int (*compFunct)( const void *, const void *) );

Because qsort() doesn’t really know what it’s sorting, it uses a void pointer in its argument (base) and in the arguments to the comparison function. qsort()’s void* argument is easy; any pointer can be converted to avoid* without even needing a cast. The function pointer is harder.

For an array of character arrays, strcmp() would have the right algorithm but the wrong argument types. The simplest, safest way to handle this situation is to pass a function that takes the right argument types forqsort() and then casts them to the right argument types. That’s what strpcmp() does.

If you have a function that takes a char*, and you know that a char* and a void* are the same in every environment your program might ever work in, you might cast the function pointer, rather than the pointed- to function’s arguments, in this way:

char     table[ NUM_ELEMENTS ][ ELEMENT_SIZE ];

/* … */

/* passing strcmp() to qsort for array of array of char */


( int (*)( const void *, const void * ) ) strcmp );

14. Can the size of an array be declared at runtime?

No. In an array declaration, the size must be known at compile time. You can’t specify a size that’s known only at runtime. For example, if i is a variable, you can’t write code like this:

char array[i]; /* not valid C */

Some languages provide this latitude. C doesn’t. If it did, the stack would be more complicated, function calls would be more expensive, and programs would run a lot slower.

If you know that you have an array but you won’t know until runtime how big it will be, declare a pointer to it and use malloc() or calloc() to allocate the array from the heap.

If you know at compile time how big an array is, you can declare its size at compile time. Even if the size is some complicated expression, as long as it can be evaluated at compile time, it can be used.

/* A program that copies the argv array and all the pointed-to

strings. It also deallocates all the copies. */

#include <stdlib.h>

#include <string.h>

int main(int argc, char** argv)


char** new_argv;

int i;

/* Since argv[0] through argv[argc] are all valid, the

program needs to allocate room for argc+1 pointers. */

new_argv = (char**) calloc(argc+1, sizeof (char*));

/* or malloc((argc+1) * sizeof (char*)) */

printf(“allocated room for %d pointers starting at %P\n”, argc+1, new_argv);

/* now copy all the strings themselves

(argv[0] through argv[argc-1]) */

for (i = 0; i < argc; ++i) {

/* make room for ‘\0’ at end, too */

new_argv[i] = (char*) malloc(strlen(argv[i]) + 1);

strcpy(new_argv[i], argv[i]);

printf(“allocated %d bytes for new_argv[%d] at %P, “”copied \”%s\”\n”,

strlen(argv[i]) + 1, i, new_argv[i], new_argv[i]);


new_argv[argc] = NULL;

/* To deallocate everything, get rid of the strings (in any

order), then the array of pointers. If you free the array

of pointers first, you lose all reference to the copied

strings. */

for (i = 0; i < argc; ++i)



printf(“freed new_argv[%d] at %P\n”, i, new_argv[i]);

argv[i] = NULL;



printf(“freed new_argv itself at %P\n”, new_argv);

return 0;


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