在函数中更改指针的值 [英] Change pointer's value in function

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);
}

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