C：接收指针的指针的函数，因此可以分配外部指针

Question

How can I make the following work? The idea is for the function to allocate an external pointer so I can use this concept in another program, but I can't do that because gcc keeps telling me that the argument is from an incompatible pointer type... It should be simple, but I'm not seeing it.

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

int allocMyPtr(char *textToCopy, char **returnPtr) {
    char *ptr=NULL;
    int size=strlen(textToCopy)+1;
    int count;

    ptr=malloc(sizeof(char)*(size));
    if(NULL!=ptr) {
        for(count=0;count<size;count++) {
            ptr[count]=textToCopy[count];
        }
        *returnPtr = ptr;
        return 1;
    } else {
        return 0;
    }
}

int main(void) {
    char text[]="Hello World\n";
    char *string;

    if(allocMyPtr(text,string)) {
        strcpy(string,text);
        printf(string);
    } else {
        printf("out of memory\n");
        return EXIT_FAILURE;
   }
   free(string);
   return EXIT_SUCCESS;
}

Answer 1

几乎是正确的，但是由于您的函数需要一个指向指针的指针，因此必须使用address-of运算符将指针的地址传递给该函数：

allocMyPtr(text, &string)

Answer 2

use &string instead to fix your problem the type related to this input parameter is char ** and not char *

if(allocMyPtr(text,&string)) {

Just a remark concerning your source code:

The allocMyPtr() function already do a copy from text to string.

so why you make copy agian with strcpy. it's useless

strcpy(string,text); // this useless

Answer 3

You are passing string using pass by value in allocMyPtr() you should use pass by adress so that pointer should match otherwise compiler keep tellin you about ,

incompatible type char * to char **

do this :

if(allocMyPtr(text,&string)) { }