問題描述

我需要創建一個 Bar 對象，它有一個私有對象 Foo f.

I need to create a Bar object, which has a private object Foo f.

但是Foo對象參數的值應該通過特定的方法int genValue()來傳遞.

However, the value of Foo object parameter should be passed by the specific method int genValue().

如果我在構造函數作用域 Bar(){...} 中初始化 f，編譯器會大喊錯誤，就像沒有構造函數 Foo().

If I initialize f in the constructor scope Bar(){...}, the compiler yell error, something like there is no constructor Foo().

如果我像這樣構造Bar(): f(genValue())，編譯器會報錯:

If I construct like this Bar(): f(genValue()), the compiler yells the error:

test.cpp: In constructor ‘Bar::Bar()’:
test.cpp:16:19: error: cannot bind non-const lvalue reference of type ‘int&’ to an rvalue of type ‘int’
 Bar(): f(genValue()){    
            ~~~~~~~~^~
test.cpp:7:2: note:   initializing argument 1 of ‘Foo::Foo(int&)’    
 Foo(int &x) {    
 ^~~

示例代碼:

class Foo {
public:
    Foo(int &x) {
        this->x = x;
    }
private:
    int x;
};

class Bar {
public:
    Bar(): f(genValue()){
    }
private:
    Foo f;

    int genValue(){
        int x;
        // do something ...
        x = 1;
        return x;
    }
};

int main() {

    Bar bar ();

    return 0;
}

如果我不想修改 Foo 類并且它的參數值應該從 genValue() 傳遞，我該如何解決這個問題?而且，我不想使用純指針 (*)，但使用智能指針的解決方案是可以的！

How can I fix the problem, if I don't want to modify Foo class and its argument value should be passed from genValue()? And, I don't want to use pure pointer (*), but a solution with smart pointer is okay!

推薦答案

非const 引用參數，例如int&，只能引用一個左值"，這是一個命名變量.

A non-const reference parameter, such as an int&, can only refer to an "lvalue," which is a named variable.

auto takes_nonconst_reference = [](int&){};
auto takes_const_reference = [](const int&){};
auto takes_value = [](int){};
auto returns_int = []{return 42;};

int foo = 1;

// OK
takes_nonconst_reference(foo);
takes_const_reference(foo);
takes_const_reference(returns_int());
takes_value(foo);
takes_value(returns_int());

// compilation error, value returned from a function is not a named variable
takes_nonconst_reference(returns_int());

在這種特殊情況下，由于您的類存儲了構造函數參數的副本，您應該按值傳遞它(int，而不是 int& 或 const int&).

In this particular case, since your class is storing a copy of the constructor parameter, you should pass it by value (int, not int& nor const int&).

這篇關于錯誤:無法將“int&"類型的非常量左值引用綁定到“int"類型的右值的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！