如果输出无关紧要，则不会调用函数

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，取决于它可能会更改foo变量。无论是否更改，它都会返回bool。

假设我把这个函数放在一个循环中。我调用它10次，我想知道这些调用中是否有任何一次更改了变量。因此，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，我得到了：

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，变量发生了更改，这是正确的。但剩下的四个电话都没有发生。我明白了，因为第五个调用返回true，那么无论其余调用返回什么，"changed"变量都将始终为true。我不在乎，我希望它保持"真实"，这毕竟是正确的。但我的观点是，剩下的四个调用可能已经将"bar"变量更改为完全不同的值，我需要在循环后使用这个"正确更改"的值。

有人能解释一下吗，为什么仅仅因为函数的返回值不相关就不调用它？因为我看不出这意味着函数中的代码也不相关，尤其是当一些参数通过非常数引用传递时。（据我所知，该功能甚至可能完全终止程序。）

我并不是在寻找解决方案，更像是解释为什么会发生这种情况。

我用VS2015编译了这个，标准的DEBUG和RELEASE模式有同样错误的结果。

显然，如果我去掉"changed"变量，只调用循环中的函数，我会得到正确的输出。