int *change(int*a,int b){int c=30;*a=~(1<<2);b=20;return &c;} int main(void) {int a=1;b=2;int *c;c=change(&a,b);printf("0x%X,%d\n",a,b);printf("%d\n",*c);return 0;}输出结果
时间: 2024-03-20 21:38:35 浏览: 12
这段代码的输出结果为:
```
0xFFFFFFFB,2
未定义行为
```
原因是:
与上一个问题一样,`change` 函数中的变量 `c` 是一个局部变量,它的地址被返回给了 `main` 函数中的指针 `c`。但是,当 `change` 函数执行完毕后,变量 `c` 所在的栈帧被销毁,`c` 指向的地址变成了未定义行为。所以在 `printf("%d\n",*c)` 语句中,尝试访问已经被销毁的变量 `c`,造成了未定义行为。
此外,在 `main` 函数中,变量 `b` 没有被定义,所以编译器会报错。需要将 `int a=1; b=2;` 修改为 `int a=1, b=2;`。
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