在C语言中：指向函数时使用void和int关键字

Question

I am trying to do a program where the function is pointed by a pointer. It goes as follows:-

This is the first program which uses "void" return type.

#include<stdio.h>
#include<conio.h>
void CharPrint(char *ptr);
main()
{
    char *str="Hello World";
    void (*ptr1)(char *ptr);
    ptr1=CharPrint;
    if((*ptr1)(str))
        printf("Done");
    return 0;
}
void CharPrint(char *ptr)
{
    printf("%s\n",ptr);
}

It throws many errors. They are:-

The second program is as follows:-

#include<stdio.h>
#include<conio.h>
int CharPrint(char *ptr);
main()
{
    char *str="Hello World";
    int (*ptr1)(char *ptr);
    ptr1=CharPrint;
    if((*ptr1)(str))
        printf("Done");
    return 0;
}
int CharPrint(char *ptr)
{
    printf("%s\n",ptr);
    return 0;
}

This program runs without any hiccup.

The output is:-

My problem is that in the first output, why is it showing " Not an allowed type in function main " on line 9. The other lines are also arising doubts but this line is bugging me the most. Any help?

Answer 1

Your first function does not return anything. Thus, you cannot test if((*ptr1)(str)) .

Answer 2

void (*ptr1)(char *ptr);
ptr1=CharPrint;
if((*ptr1)(str))
    printf("Done");

What is the if testing if the return value is void ? Just change the last two lines to:

((*ptr1)(str));