关于c ++转换:参数1从'[some_class]'到'[some_class]&'的未知转换 [英] about c++ conversion : no known conversion for argument 1 from ‘[some_class]' to ‘[some_class]&’

问题描述

我正在使用C ++，但出现了一个我不知道确切原因的错误.我已经找到了解决方案，但仍然想知道为什么.

I'm working on C++, and had an error that I didn't know the exact reason. I've found the solution, but still want to know why.

    class Base
    {
        public:
               void something(Base& b){}
    };

    int main()
    {
        Base b;
        b.something(Base());
        return 0;           
    }

编译代码时，出现以下错误:

when I compile the code, I got this following error :

abc.cpp:12:20: error: no matching function for call to ‘Base::something(Base)’
abc.cpp:12:20: note: candidate is:
abc.cpp:6:7: note: void Base::something(Base&)
abc.cpp:6:7: note:   no known conversion for argument 1 from ‘Base’ to ‘Base&’

但是当我将b.something(Base())替换为

but when I replaced b.something(Base()) into

Base c;
b.something(c);

错误消失了，我想知道为什么吗???他们不是同一类型吗?只是我写的方式很重要，但含义应该是相同的??

the error is gone, I'm wondering why??? aren't they have the same type? It only matters how I write it, but the meaning should be the same???

谢谢你们！

推荐答案

您要在此处传递一个临时 Base 对象:

You are passing a temporary Base object here:

b.something(Base());

但是您尝试在此处将其绑定到非常量左值引用:

but you try to bind that to a non-const lvalue reference here:

void something(Base& b){}

这在标准C ++中是不允许的.您需要一个 const 参考.

This is not allowed in standard C++. You need a const reference.

void something(const Base& b){}

执行此操作时:

Base c;
b.something(c);

您没有通过临时工.非常量引用绑定到 c .

you are not passing a temporary. The non-const reference binds to c.

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