我可以通过函数更改已初始化的char指针吗? [英] Can I change an initialized char pointer via function?

问题描述

这是我的main.c:

#include <stdio.h>

void changeStuff(char *stuff){
    stuff=  "Hello everyone";
}

int main(int argc, char **argv) {
    char *stuff;
    changeStuff(stuff);
    printf(stuff);
    return 0;
}

构建此文件时，会收到以下警告:

When I build this, I get this warning:

warning: ‘stuff’ is used uninitialized in this function [-Wuninitialized]

当我运行该程序时，什么都没有打印.

When I run this program, nothing is printed.

由于似乎不可能在声明后定义没有值的char*，我该如何更改传递给函数的char*的值?

Since it does not seem possible to define a char* with no value after it has been declared, how do I change the value of a char* passed to a function?

推荐答案

在C语言中，函数参数按值传递.为了修改指针值，您需要将指针传递给指针，例如:

In C, function arguments are passed-by-value. In order to modify a pointer value, you need to pass a pointer to the pointer, like:

#include <stdio.h>

void changeStuff(char **stuff){
    *stuff=  "Hello everyone";
}

int main(int argc, char **argv) {
    char *stuff;
    changeStuff(&stuff);
    printf("%s\n", stuff);
    return 0;
}

请注意，直接将用户定义的值传递给printf()不是一个好主意.请参阅:如何利用格式字符串漏洞?

Note that it's not a good idea to directly pass user-defined values to printf(). See: How can a Format-String vulnerability be exploited?

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