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！