取消引用地图迭代器时，返回对临时对象的引用 [英] return reference to temporary when dereferencing map iterator

问题描述

考虑此代码

#include <iterator>
#include <vector>

const int& foo(const std::vector<int>& x,unsigned i) {
    auto it = x.begin();
    std::advance(it,i);
    return *it;
}

clang和gcc都不会发出错误/警告，但这是

Both clang and gcc emit no errors/warnings, but this:

#include <iterator>
#include <map>

const std::pair<int,int>& bar(const std::map<int,int>& x,unsigned i){
    auto it = x.begin();
    std::advance(it,i);
    return *it;
}

使用 clang 编译，并使用-Werror结果:

compiled with clang and using -Werror results in :

<source>:14:12: error: returning reference to local temporary object [-Werror,-Wreturn-stack-address]
    return *it;
           ^~~

以及 gcc :

<source>: In function 'const std::pair<int, int>& bar(const std::map<int, int>&, unsigned int)':
<source>:14:13: error: returning reference to temporary [-Werror=return-local-addr]
     return *it;
             ^~

是什么使gcc和clang拒绝bar?为什么foo还可以?

What makes gcc and clang reject bar and why is foo fine?

推荐答案

问题是value_type相对="nofollow noreferrer"> std::map<int,int> 不是std::pair<int,int>，而是std::pair<const int,int>.然后对于return *it;，必须创建并返回一个临时std::pair<int,int>. (std::pair<int,int>可能是std::pair<const int,int>中的转换.)临时文件将被立即销毁，并将返回的参考物悬挂起来.

The problem is that the value_type of std::map<int,int> is not std::pair<int,int>, but std::pair<const int,int>. Then for return *it;, a temporary std::pair<int,int> has to be created and returned. (std::pair<int,int> could be converted from std::pair<const int,int>.) The temporary will be destroyed immediately and left the returned reference dangled.

要解决问题，您可以将返回类型更改为const std::pair<const int,int>&或const std::map<int,int>::value_type &.

To sovle the issue you can change the return type to const std::pair<const int,int>& or const std::map<int,int>::value_type &.

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