我对将指针传递给 char 数组有什么误解?
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;
}
最佳答案
在函数 get_response
中,您将 request
的地址存储在临时变量 response
中。你想把它存储在response
指向的地方。
*response = request;
关于c - 将指向 C 数组的指针传递给函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9414661/