如何解决C中的指针错误？

Question

I am starting to learn about pointers in C.

How can I fix the error that I have in function x() ?

This is the error:

Error: a value of type "char" cannot be assigned to an entity of type "char *".

This is the full source:

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

void x(char **d, char s[]) {
    d = s[0]; // here i have the problem
}

void main() {
    char *p = NULL;
    x(&p, "abc");
}

Answer 1

In function x() you pass d which is a char ** (a pointer to a string pointer), and char s[] (an array of char , passed as similarly to a pointer to char ).

So in the line:

d = s[0];

s[0] is a char , whereas char **d is a pointer to a pointer to char . These are different, and the compiler is saying you cannot assign from one to the other.

However, did your compiler really warn you as follows?

Error: a value of type "char" cannot be assigned to an entity of type "char *"

Given the code sample, it should have said char ** at the end.

I think what you are trying to make x do is copy the address of the string passed as the second argument into the first pointer. That would be:

void x(char **d, char *s)
{
    *d = s;
}

That makes p in the caller point to the constant xyz string but does not copy the contents.

If the idea was to copy the string's contents:

void x(char **d, char *s)
{
    *d = strdup(s);
}

and ensure you remember to free() the return value in main() , as well as adding #include <string.h> at the top.

Answer 2

Here's what you can do, so it would compile in two versions.

Version 1.

void x(char **d, char s[]) {
    d = (char**)s[0];
}

or Version 2.

void x(char **d, char *s) {
    *d = s; 
}

Hope this helps.

Answer 3

A more proper way would be to use strcpy :

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

void x(char **d) {
    *d = malloc(4 * sizeof(char));        
    strcpy(*d, "abc");
}

int main() {
    char *p;
    x(&p);
    printf("%s", p);
    free(p);
    return 0;
}

Outputs: abc