Thread: C programmers help please...........help with pointers

int StrPrint(char *str)
main()
char str[24]="pointing to a function.";
int (*ptr)(char *str);

ptr=StrPrint;
if (!(*ptr)(str))
printf("Done!\n");

return 0;
}
int StrPrint (char *str)
{
printf("&s\n", str);
return 0;
}
/* I don't really know why there are pointers declared inside of a function,
don't understand int (*ptr)(char *str); and the if statement. I mean this book just jumped into declaring pointers in a function without really explaining them. I was like you know understanding pointer a little bit, moving up in the address declaring and all that, but that just blew me off. Please any help will be greatlly appreciated */

Aren't you a little new to be messing around with function pointers?

Code:

#include <stdio.h>

// a function which accepts a char pointer and returns an int
// in this case, it prints the string and returns 0
int StrPrint(char *str);

int main() {
    char str[]="pointing to a function.";

    // a pointer to a function
    // the function being pointed to accepts a char * and returns an int
    // just like StrPrint
    int (*ptr)(char *str);

    // a function name without () is just a pointer to that function
    // so this is just copying a pointer value into a variable
    // just like any other assignment would do
    ptr=StrPrint;

    // this is how you would call the function normally
    if( !StrPrint(str) )
        printf( "Done!\n" );

    // this is an old way - where you have to specify that ptr
    // is actually a pointer
    if( !(*ptr)(str) )
        printf( "Done!\n" );

    // this is the ANSI-C way of referring to the function via a pointer
    // you can see the similarity to calling the actual function by name
    if( !ptr(str) )
        printf( "Done!\n" );

    return 0;
}

int StrPrint(char *str) {
    printf( "%s\n", str );
    return 0;
}