错误:没有上下文类型信息的重载函数的地址 [英] error: address of overloaded function with no contextual type information

问题描述

代码:

class que {
public:
    que operator++(int) {}  // 1
    que &operator++() {}
    que &operator+=(int n) {
        que& (que::*go)();
        go = 0; if(n > 0) go = &que::operator++ ; // 2
        //go = (n > 0) ?    (&que::operator++) : 0 ;    // 3
    }
};

int main() {
    que iter;
    iter += 3;
    return 0;
}

我想将第2行替换为第3行("if"语句替换为?:").
如果我取消注释3，则编译器会给我一个错误.
如果我删除第1行，则第3行有效.
问题是:编译器向我要什么?
错误:错误:没有上下文类型信息的重载函数的地址
编译器:gcc-4.5.2

I want to replace line 2 by line 3("if" statement for "?:").
If I uncomment 3, compiler gives me an error.
If I delete line 1, then line 3 works.
Question is: what does compiler want from me?
Error: error: address of overloaded function with no contextual type information
Compiler: gcc-4.5.2

推荐答案

错误:没有上下文类型信息的重载函数的地址

error: address of overloaded function with no contextual type information

有两个具有operator++名称的函数(即消息的重载函数"位)，您需要指定所需的函数(即上下文类型信息"之一):

There are two functions with the operator++ name (that's the 'overloaded function' bit of the message), you need to specify which one you want (that's the 'contextual type information' one):

n > 0 ? (que& (que::*)())&que::operator++ : 0

您必须考虑到上面的子表达式独立于封闭的完整表达式，即对go的赋值.因此，它必须本身是正确的，即它不能使用go的类型来选择正确的重载，因为它不是此特定子表达式的一部分.

You have to consider that the above subexpression is independent from the enclosing full expression, the assignment to go. So it must be correct on its own, i.e. it can't use the type of go to pick the correct overload because it's not part of this particular subexpression.

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