在 C 中传递 char 指针 [英] Passing char pointer in C

问题描述

好的，我正在尝试将 char 指针传递给另一个函数.我可以用一个 char 数组来做到这一点，但不能用一个 char 指针来做到这一点.问题是我不知道它的大小，所以我无法在 main() 函数中声明任何关于大小的内容.

Okay so I am trying to pass a char pointer to another function. I can do this with an array of a char but cannot do with a char pointer. Problem is I don't know the size of it so I cannot declare anything about the size within the main() function.

#include <stdio.h>

void ptrch ( char * point) {
    point = "asd";
}

int main() {
    char * point;
    ptrch(point);
    printf("%s
", point);
    return 0;
}

但这不起作用，这两个起作用:

This does not work however, these two works:

1)

#include <stdio.h>

int main() {
    char * point;
    point = "asd";
    printf("%s
", point);
    return 0;
}

2)

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

void ptrch ( char * point) {
    strcpy(point, "asd");
}

int main() {
    char point[10];
    ptrch(point);
    printf("%s
", point);
    return 0;
}

所以我试图了解问题的原因和可能的解决方案

So I am trying to understand the reason and a possible solution for my problem

推荐答案

void ptrch ( char * point) {
    point = "asd";
}

您的指针按值传递，此代码复制，然后覆盖副本.所以原来的指针是不变的.

Your pointer is passed by value, and this code copies, then overwrites the copy. So the original pointer is untouched.

附:需要注意的是，当您执行 point = "blah" 时，您正在创建一个字符串文字，任何 modify 的尝试都是 未定义的行为，所以它真的应该是 const char *

P.S. Point to be noted that when you do point = "blah" you are creating a string literal, and any attempt to modify is Undefined behaviour, so it should really be const char *

修复 - 像@Hassan TM 那样传递一个指向指针的指针，或者如下所示返回指针.

To Fix - pass a pointer to a pointer as @Hassan TM does, or return the pointer as below.

const char *ptrch () {
    return "asd";
}

...
const char* point = ptrch();

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