打印后出现奇怪的分割错误

问题描述：

编写了一个简单的交换程序，效果很好；但是在打印所有内容后给出分段错误.

Wrote a simple swap program, works well; But gives a Segmentation Fault after printing everything.

#include <stdio.h>

void swap(int* p1,int* p2){

    int* temp;
    *temp = *p1;
    *p1 = *p2;
    *p2 = *temp;
}

int main(){ 

    int a,b;
    a = 9; b = 8;
    printf("%d %d \n",a,b);
    swap(&a,&b);    
    printf("%d %d \n",a,b);

    return 0;
}

输出:

9 8  
8 9  
Segmentation fault

我应该忽略这一点并继续前进吗，还是发生了真正奇怪的事情?

Should I simply ignore this and move forward or is there something really strange going on ?

答

int * temp;* temp = * p1;

在C和C ++中是未定义的行为.(在使用时，指针必须始终指向您拥有的内存，而指针则不是).

is undefined behaviour in C and C++ as you are using an uninitialised pointer. (At the point of use, a pointer must always point to memory that you own, and your pointer isn't).

使用 int temp;temp = * p1; ，或者更好的是， int temp = * p1;

Use int temp; temp = *p1; instead, or better still, int temp = *p1;

打印后出现奇怪的分割错误

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