vector< bool> :: operator []行为不当? [英] vector<bool>::operator[] misbehavior?

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问题描述

可能重复:
为什么vector< bool> :: reference不返回对bool的引用?

Possible Duplicate:
Why vector<bool>::reference doesn’t return reference to bool?

我曾经认为,使用std::vector::operator[]可以获取所访问项目的副本,但似乎并非总是如此.至少,对于vector<bool>,以下测试代码将给出不同的结果:

I used to think that with std::vector::operator[] we get deep copies of the accessed item, but it seems that it is not always true. At least, with vector<bool> the following test code gives a different result:

#include <iostream>
#include <vector>
using namespace std;

template <typename T>
void Test(const T& oldValue, const T& newValue, const char* message)
{
    cout << message << '\n';

    vector<T> v;
    v.push_back(oldValue);
    cout << " before:  v[0] = " << v[0] << '\n';

    // Should be a deep-copy (?)       
    auto x = v[0];   
    x = newValue;

    cout << " after:   v[0] = " << v[0] << '\n';
    cout << "-------------------------------\n";
}

int main()
{
    Test<int>(10, 20, "Testing vector<int>");
    Test<double>(3.14, 6.28, "Testing vector<double>");
    Test<bool>(true, false, "Testing vector<bool>");
}

输出(使用VC10/VS2010 SP1编译的源代码):

Output (source code compiled with VC10/VS2010 SP1):

Testing vector<int>
 before:  v[0] = 10
 after:   v[0] = 10
-------------------------------
Testing vector<double>
 before:  v[0] = 3.14
 after:   v[0] = 3.14
-------------------------------
Testing vector<bool>
 before:  v[0] = 1
 after:   v[0] = 0
-------------------------------

我希望x = newValue赋值之后的v[0] 仍等于其先前的值,但这似乎不正确. 这是为什么? 为什么vector<bool>很特别?

I would have expected that v[0] after the x = newValue assignment would still be equal to its previous value, but this seems not true. Why is that? Why is vector<bool> special?

推荐答案

vector<bool>::operator[]既不产生bool也不引用bool.它仅返回一个有点像参考对象的代理对象.这是因为没有对单个位的引用,并且vector<bool>实际上以压缩方式存储了bool.因此,通过使用auto,您仅创建了该引用类对象的副本.问题是C ++不知道该对象充当引用.您必须在此处强制将衰减到值",方法是将auto替换为T.

vector<bool>::operator[] neither yields a bool nor a reference to a bool. It just returns a little proxy object that acts like a reference. This is because there are no references to single bits and vector<bool> actually stores the bools in a compressed way. So by using auto you just created a copy of that reference-like object. The problem is that C++ does not know that this object acts as a reference. You have to force the "decay to a value" here by replacing auto with T.

这篇关于vector&lt; bool&gt; :: operator []行为不当?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!

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