按分配给const引用的值返回 [英] return by value assigned to const reference

问题描述

我正在修复某些代码中的另一个错误，并遇到了一些我认为是错误的代码；但是，此代码在gcc 4.4、4.5和4.6下进行编译，并且看起来像预期的功能。谁能告诉我这是否是有效的c ++？

I was fixing another bug in some code and came across some code that I would have thought was a bug; however, this code compiles under gcc 4.4, 4.5, and 4.6 and appears to function as "expected". Can anyone tell me if this is valid c++?

struct foo {
     int bar;
};

foo myfunction(foo const &orig) {
    foo fooOnStack = orig;
    fooOnStack.bar *= 100;
    return fooOnStack;
}

void myOtherFunction(foo const &orig) {
    foo const &retFoo = myfunction();
    // perhaps do some tests on retFoo.bar ...
}

如果这是有效的c ++，是否有人知道这是合法的？

If this is valid c++, does anyone know the rationale behind this being legal?

推荐答案

是的，这是合法的C ++。对临时对象形成对const的引用可将临时对象的生存期延长至引用的生存期。

Yes, this is legal C++. Forming a reference-to-const to a temporary extends the lifetime of the temporary to the lifetime of the reference.

这篇关于按分配给const引用的值返回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！