在c ++ 0x中的用户定义文字的重载规则

本文介绍了在c ++ 0x中的用户定义文字的重载规则的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我对重载规则有些困惑，I am a little confused about overloading rules,我们说有以下文字运算符：let's say there are following literal operators,unsigned long long operator "" _xx(unsigned long long cooked_literal_int); //1unsigned long long operator "" _xx(const char * raw_literal_string); // 2unsigned long long operator "" _xx(long double cooked_literal_double); // 3 3，则重载很明显，13_xx //call 113.5_xx //call 3如果1& 2，if 1 & 2 are defined, 13_xx //call 113.5_xx //call 2如果2& 3 if 2 & 3 are defined13_xx // call 2 or 3??13.5_xx // call 3混乱来自最新的c ++ 0x标准n3225 2.14.8 / 3，The confusion comes from latest c++0x standard n3225 2.14.8/3, operatorX（n ULL）operator "" X (n ULL)否则，S应包含一个原始字面量运算符或文字运算符模板（13.5.8），但不能同时包含两者。如果S包含一个原始的字面量运算符，字面量L被视为一个形式的调用Otherwise, S shall contain a raw literal operator or a literal operator template (13.5.8) but not both. If S contains a raw literal operator, the literal L is treated as a call of the form operatorX（n）operator "" X ("n")否则（S包含文字运算符模板），L被视为Otherwise (S contains a literal operator template), L is treated as a call of the form operatorX （）operator "" X <’c1’, ’c2’, ... ’ck’>()其中n是源字符序列c1c2 ... ck。where n is the source character sequence c1c2...ck.这说明，如果1存在（无符号长整数参数），13_xx将调用1，否则13_xx将调用2.从13.5.8， This says that, if 1 is present (an unsigned long long parameter), 13_xx shall call 1, otherwise, 13_xx shall call 2. And from 13.5.8,根据我的理解，如果1不存在，13_xx可隐式转换为double并调用3. From my understanding, if 1 is not present, 13_xx can be implicitly converted to double and call 3.因此，如果1不存在，则2& 3从标准描述中以某种方式有效。Therefore if 1 is not present, both 2 & 3 are somehow valid from the standard description.我希望有人能帮助我清除我的疑虑。非常感谢。I hope someone can help me clear my doubts. Many thanks.推荐答案我相信13.5.8 / 7说明了这个问题：I believe that 13.5.8/7 clarifies this issue :根据我的理解，只有当通过用户定义的字面调用 隐式调用之后，常规的重载解析规则才会用于文字运算符。From my understanding, regular overload resolution rules are only implied for literal operators when called outside an implicit invocation through user-defined literals.所以我认为如果2& 3， 13_xx 调用2（原始字面量操作符）。So I think that if 2 & 3 are defined, 13_xx calls 2 (the raw literal operator). 这篇关于在c ++ 0x中的用户定义文字的重载规则的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！