元组作为函数参数中的默认值

Question

如何将元组作为函数的默认值？

这些原型引发编译错误：

void f(std::tuple<int, int>=std::tuple<int, int>(1, 1));
void f(std::tuple<int, int>=std::make_tuple(1, 1));
void f(std::tuple<int, int>=(1, 1));

Answer 1

void f(std::tuple<int, int> t = std::make_tuple(1, 1));

或

void f(std::tuple<int, int> = std::make_tuple(1, 1));

或

void f(std::tuple<int, int> = std::tuple<int, int>(1, 1));

或

void f(std::tuple<int, int> = std::tuple<int, int>{1, 1});

Answer 2

整个事情是你错过了名字 :) 你忘了给你的变量命名。

void f(std::tuple<int, int>x=std::make_tuple(1., 1.))

应该可以

查看此 docs 了解更多信息。

attr(optional) decl-specifier-seq declarator = initializer

您的定义中缺少 "declarator" 部分。

Each declarator introduces a single variable or function into the program. The most simple form of declarator is just the name you're introducing - the declarator-id.

Answer 3

你只缺一个 space:

void f(std::tuple<int, int> =std::make_tuple(1, 1));

照原样，>= 被解析为单个标记。

添加一个变量名也可以方便地回避这个问题。