在统一初始化期间使用尾随逗号时是否存在任何潜在的语义差异?
std::vector< std::size_t > v1{5, }; // allowed syntax
std::vector< std::size_t > v2{10};
我可以使用尾随逗号使编译器选择std::vector::vector(std::initializer_list< std::size_t >)
吗?构造函数而不是 std::vector::vector(std::size_t, const std::size_t &)
或者还有其他提到语法的技巧吗?
我可以用它来检测是否有std::initializer_list
- 构造函数重载?
考虑以下代码,必须选择哪个构造函数?
struct A { A(int) { ; } A(double, int = 3) { ; } };
A a{1};
A b{2, };
此代码 is accepted by gcc 8
和 A(int)
在这两种情况下都被选中。
最佳答案
首先,C++ 语法规则使尾随 ,
braced-init-list 是可选的。报价 dcl.init/1
A declarator can specify an initial value for the identifier being declared. The identifier designates a variable being initialized. The process of initialization described in the remainder of [dcl.init] applies also to initializations specified by other syntactic contexts, such as the initialization of function parameters ([expr.call]) or the initialization of return values ([stmt.return]).
initializer: brace-or-equal-initializer ( expression-list ) brace-or-equal-initializer: = initializer-clause braced-init-list initializer-clause: assignment-expression braced-init-list braced-init-list: { initializer-list ,opt } { designated-initializer-list ,opt } { }
其次,您几乎无法覆盖重载解决系统。它将始终使用 std::initializer_list
构造函数,如果你使用这样的语法和这样的std::initializer_list
构造函数可用。
A constructor is an initializer-list constructor if its first parameter is of type std::initializer_list or reference to possibly cv-qualified std::initializer_list for some type E, and either there are no other parameters or else all other parameters have default arguments. [ Note: Initializer-list constructors are favored over other constructors in list-initialization ([over.match.list])....
下面的程序打印 Using InitList
:
#include <iostream>
#include <initializer_list>
struct X{
X(std::initializer_list<double>){ std::cout << "Using InitList\n"; }
X(int){ std::cout << "Using Single Arg ctor\n"; }
};
int main(){
X x{5};
}
尽管 5
是 int
类型的文字,选择单参数构造函数应该是有意义的,因为它是完美匹配的;和 std::initializer_list<double>
构造函数需要 double
的列表.然而,规则有利于 std::initializer_list<double>
因为它是一个 initializer-list 构造函数。
结果,即使是下面的程序也因为缩小转换而失败:
#include <iostream>
#include <initializer_list>
struct Y{
Y(std::initializer_list<char>){ std::cout << "Y Using InitList\n"; }
Y(int, int=4){ std::cout << "Y Using Double Arg ctor\n"; }
};
int main(){
Y y1{4777};
Y y2{577,};
Y y3{57,7777};
}
针对您在下面的评论,“如果 std::initializer_list 没有重载,或者它不是第一个构造函数的参数怎么办?” - 然后重载决议不会选择它。演示:
#include <iostream>
#include <initializer_list>
struct Y{
Y(int, std::initializer_list<double>){ std::cout << "Y Using InitList\n"; }
Y(int, int=4){ std::cout << "Y Using Double Arg ctor\n"; }
};
int main(){
Y y1{4};
Y y2{5,};
Y y3{5,7};
}
打印:
Y Using Double Arg ctor
Y Using Double Arg ctor
Y Using Double Arg ctor
如果没有 initializer-list 构造函数 可用,那么 {initializer-list...,}
根据 dcl.init/16,初始化程序几乎可以退回到直接初始化。 ,其语义由 dcl.init/16 的前一段涵盖
关于c++ - 统一初始化中的尾随逗号,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46070793/