Question

我有以下代码：

#include <iostream> 

bool function(int a, int b, int &foo) {
    std::cout << "I have been called and I";
    if (a > b)// some magic that maybe changes 'foo'
    {
        foo++;
        std::cout << " did change the variable" << std::endl;
        return true;//inform that i have changed the value
    }
    std::cout << " did NOT change the variable" << std::endl;

    return false;
};

int main()
{
    bool changed = false;
    int bar = 0;
    for (size_t i = 0; i < 10; i++)
    {
        changed = changed || function(i,4,bar);
    }
    std::cout << "Bar:" << bar;
    std::cin.get();
    std::cin.get();
    return 0;
}

我有一个函数，它对两个变量做了一些魔术，这取决于它可能会改变foo变量。无论是否改变它，它都会返回bool。

假设我把这个函数放在一个循环中。我称它为10次，我想知道这些调用是否改变了变量。所以我对上面代码的期望是：

I have been called and I did NOT change the variable
I have been called and I did NOT change the variable
I have been called and I did NOT change the variable
I have been called and I did NOT change the variable
I have been called and I did NOT change the variable
I have been called and I did change the variable// 'changed' is now true
I have been called and I did change the variable
I have been called and I did change the variable
I have been called and I did change the variable
I have been called and I did change the variable
Bar:5

但是，没有。相反，我得到：

I have been called and I did NOT change the variable
I have been called and I did NOT change the variable
I have been called and I did NOT change the variable
I have been called and I did NOT change the variable
I have been called and I did NOT change the variable
I have been called and I did change the variable// 'changed' is now true
Bar:1

在第五次调用中，变量发生了变化，这是正确的。但剩下的四个电话甚至没有发生。我明白了，因为第五次调用返回true，然后'changed'变量将始终为true，无论剩余的调用返回什么。我不在乎，我希望它能保持'真实'，毕竟这是正确的。但我的观点是，剩下的四个调用可能会将'bar'变量更改为完全不同的值，并且我需要在循环后使用此“正确更改”值。

所以有人可以解释一下，为什么函数不被调用只是因为它的返回值无关紧要？因为我没有看到这意味着函数内部的代码也是无关紧要的，特别是当一些参数通过非const引用传递时。（据我所知，函数甚至可能完全终止程序。）

我不是在寻找解决方案，更像解释为什么会发生这种情况。

我使用VS 2015，标准DEBUG和RELEASE模式编译了这个错误的结果。

如果我消除'changed'变量并且只是在循环中调用函数，显然我得到了正确的输出。

Answer 1

这是由于short-circuit evaluation。

标准声明（强调我的）：

5.15逻辑OR运算符
...
与|不同，||保证从左到右评价;此外，如果第一个操作数的计算结果为true ，则不会计算第二个操作数。

尝试交换参数：

changed = function(i,4,bar) || changed;

如果输出无关紧要，则不会调用函数

1 个答案: