什么使这种指针的使用不可预测？ [英] What makes this usage of pointers unpredictable?

问题描述

我目前正在学习指针，我的教授提供这段代码作为示例：

I'm currently learning pointers and my professor provided this piece of code as an example:

//We cannot predict the behavior of this program!

#include <iostream>
using namespace std;

int main()
{
    char * s = "My String";
    char s2[] = {'a', 'b', 'c', '\0'};

    cout << s2 << endl;

    return 0;
}

他在评论中写道我们无法预测程序的行为。究竟是什么让它不可预测呢？我看不出有什么错误。

He wrote in the comments that we can't predict the behavior of the program. What exactly makes it unpredictable though? I see nothing wrong with it.

推荐答案

程序的行为是不存在的，因为它是不成形的。 / p>

The behaviour of the program is non-existent, because it is ill-formed.

char* s = "My String";

这是非法的。

正确的行是：

const char* s = "My String";

除此之外，程序很好。你的教授应该喝更少的威士忌！

Other than that, the program is fine. Your professor should drink less whiskey!

这篇关于什么使这种指针的使用不可预测？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！