C ++中的构造函数调用或函数样式转换

Question

If I have the following c++ code:

class foo{
public:
    explicit foo(int i){};
};
void f(const foo &o){
}

And then I call

f(foo(1));

Is foo(1) constructor call or function-style cast?

Answer 1

他们是一样的东西。

Answer 2

它是一个函数式转换，导致构造函数调用，所以两者都是。

Answer 3

5.2.3 Explicit type conversion (functional notation)

1 A simple-type-specifier (7.1.6.2) or typename-specifier (14.6) followed by a parenthesized expression-list constructs a value of the specified type given the expression list. If the expression list is a single expression, the type conversion expression is equivalent (in definedness, and if defined in meaning) to the corresponding cast expression (5.4). ...

Your code creates a temporary, using the constructor you have with the argument's value 1, and binds it to a const reference. The temporary's lifetime ends at the end of the statement where it was created.