函数返回另一个函数的返回 [英] Function returning the return of another function

问题描述

如果我想调用 Bar()而不是 Foo()， Bar()是否会给我返回副本(其他开销)是什么Foo()返回，还是返回 Foo()放在临时堆栈上的同一对象?

If I want to call Bar() instead of Foo(), does Bar() return me a copy (additional overhead) of what Foo() returns, or it returns the same object which Foo() places on the temporary stack?

vector<int> Foo(){  
    vector<int> result;  
    result.push_back(1);  
    return result;  
}  
vector<int> Bar(){  
    return Foo();  
}

推荐答案

两者都可能发生.但是，大多数编译器在进行优化后就不会立即进行复制.

Both may happen. However, most compiler will not do copy as soon as you optimize.

您的代码表明应该有副本.但是，允许编译器删除所有不会更改语义和程序的副本.

Your code indicate there should be a copy. However, the compiler is allowed to remove any copy that do not change the semantic and the program.

注意:这就是为什么您永远都不能拥有一个复制构造函数，该复制构造函数除了正确复制外什么也不做，因为您永远无法确定是否会真正完成复制.

Note: This is why you should NEVER have a copy constructor that does anything but copying correctly as you can never be sure if a copy will be actually done or not.

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