c++ - 函数参数字符串按引用传递

Question

void print(const string& str){
    cout << str <<endl;;
}

int main(){
    print(string("asdf"));
}

我想我可以这样理解创建了一个临时字符串对象，并通过引用将其传递给函数。参数 str 实际上是与那个临时字符串对象完全相同的对象。

但是如果我用 print("asdf"); 替换调用呢？在这种情况下到底发生了什么？我知道涉及隐式转换。但我不确定它是何时以及如何完成的？我的意思是是否也创建了一个临时字符串对象？

Answer 1

你的理解是正确的。

What is exactly happening in this case? I know implicit conversion is involved.

发生完全相同的事情，但涉及隐式转换，而不是显式调用构造函数。

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But I am not sure when and how it is done?

引自 cppreference :

Implicit conversions are performed whenever an expression of some type T1 is used in context that does not accept that type, but accepts some other type T2

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I mean is it there is a temporary string object created as well?

是的。