在函数中更改指针的值

本文介绍了在函数中更改指针的值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我想在函数中更改变量的值. 我的代码是这样的:

I want to change the variable's value in the function. my code is like this:

void change(char *buf){ char str = "xxxxxxx"; *buf = &str; } int main(){ char *xxx = NULL; change(xxx); }

当我用valgrind调试时，它说:

when I debug with valgrind, it says:

==3709== Invalid write of size 1 ==3709== at 0x80483CA: change (test.c:5) ==3709== by 0x80483E5: main (test.c:10) ==3709== Address 0x0 is not stack'd, malloc'd or (recently) free'd ==3709== ==3709== ==3709== Process terminating with default action of signal 11 (SIGSEGV) ==3709== Access not within mapped region at address 0x0 ==3709== at 0x80483CA: change (test.c:5) ==3709== by 0x80483E5: main (test.c:10)

有人可以帮助我吗?我是C的新手.

Can anyone help me? I'm new in C....

推荐答案

使用指向指针的指针:

void change(char **buf) { *buf = "xxxxxxx"; } int main(void) { char *xxx = NULL; change(&xxx); }