将指向C数组的指针传递给函数

Question

What do I misunderstand about passing pointers to char arrays?

Request pointer in fun: 0x7fffde9aec80
Response pointer in fun: 0x7fffde9aec80
Response pointer: (nil), expected: 0x7fffde9aec80
Response itself: (null), expected: Yadda

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

int get_response(char *request, char **response) {
    response = &request;
    printf("Request pointer in fun: %p\n", request);
    printf("Response pointer in fun: %p\n", *response);
    return 0;
}

int main() {
    char *response = NULL, request[] = "Yadda";

    get_response(request, &response);

    printf("Response pointer: %p, expected: %p\n", response, request);
    printf("Response itself: %s, expected: %s\n", response, request);

    return 0;
}

Answer 1

in the function get_response you store the address of request in the temporary variable response . You want to store it where response points to.

*response = request;

Answer 2

尝试*response = request ：您想要将响应指针的内容设置为请求内容。

Answer 3

You want *response = request; instead of response = &request; in get_response(...)

Answer 4

Firstly, with the currect declaration and your usage of get_response , the parameter response is declared as char** , which is a pointer to a char* , eg a pointer to a pointer. This would be useful if you somehow needed to modify the pointer actually pointing to the memory containing your response , but in this case this is not needed.