C function prototype: \\char *strinv(const char *s);

Question

char *strinv(const char *s); //that's the given prototype

I'm a bit insecure about the *strinv part. Does it mean that the function is automatically dereferenced when called? Or that the function is defined as a pointer?

Thanks in advance for clarification.

Answer 1

This function declaration

char * strinv(const char *s);

declares a function that has the return type char *. For example the function can allocate dynamically memory for a string and return pointer to that string.

Here is a demonstrative program that shows how the function for example can be defined.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char * strinv(const char *s)
{
    size_t n = strlen( s );

    char *t = malloc( n + 1 );

    if ( t != NULL )
    {
        size_t i = 0;

        for ( ; i != n; i++ ) t[i] = s[n-i-1];

        t[i] = '\0';
    }

    return t;
}

int main(void) 
{
    const char *s = "Hello Worlds!";

    char *t = strinv( s );

    puts( t );

    free( t );

    return 0;
}

The program output is

!sdlroW olleH

A declaration of a pointer to the function can look the foolowing way

char * ( *fp )( const char * ) = strinv;

To dereference the pointer and call the pointed function you can write

( *fp )( s );

though it is enough to write

fp( s );