[英]Difference between a+b and operator+(a,b)
Consider the following program: 考虑以下程序:
#include<functional>
typedef std::function< int( int ) > F;
F operator+( F, F )
{
return F();
}
int f( int x ) { return x; }
int main()
{
operator+(f,f); // ok
f+f; // error: invalid operands to binary expression
}
Why does the last line f+f;
为什么最后一行
f+f;
not compile? 无法编译? Why is it not identical to
operator+(f,f);
为什么它与
operator+(f,f);
? ? A reference to the standard would be appreciated.
对标准的引用将不胜感激。
The type of f
is a built-in type. f
的类型是内置类型。 Operations on objects of built-in types never consider user-define operators. 内置类型的对象上的操作永远不会考虑用户定义的运算符。 Calling
operator+(f, f)
explicitly force two conversions which will not happen unless they are forced. 调用
operator+(f, f)
显式强制执行两次转换,除非将其强制执行,否则不会发生。 The relevant clause is 13.3.1.2 [over.match.oper] paragraph 1: 相关条款是13.3.1.2 [over.match.oper]第1段:
If no operand of an operator in an expression has a type that is a class or an enumeration, the operator is assumed to be a built-in operator and interpreted according to Clause 5. ...
如果表达式中没有运算符的操作数具有类或枚举的类型,则假定该运算符是内置运算符,并根据第5章进行解释。
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