【发布时间】:2016-05-14 01:10:43
【问题描述】:
我知道我可以使用字符数组和初始化列表来填充字符串。
看起来编译器做了一些从 int 到 initializer_list 或 allocator 的隐式提升。但我不知道为什么它没有给我任何警告以及为什么它使它隐含。
你能解释一下字符串 s4 和 s5 会发生什么吗?
#include <iostream>
#include <string>
using namespace std;
class A{
};
int main() {
// string::string(charT const* s)
string s1("12345");
// 5 - because constructor takes into account null-terminated character
cout << s1.size() << endl;
// string(std::initializer_list<charT> ilist)
string s2({'1','2','3','4','5'});
// 5 - because string is built from the contents of the initializer list init.
cout << s2.size()<<endl;
// string::string(charT const* s, size_type count)
string s3("12345",3);
// 3 - Constructs the string with the first count characters of character string pointed to by s
cout << s3.size() << endl;
// basic_string( std::initializer_list<CharT> init,const Allocator& alloc = Allocator() ); - ?
string s4({'1','2','3','4','5'},3);
// 2 - why this compiles (with no warning) and what this result means?
cout << s4.size() << endl;
string s5({'1','2','3','4','5'},5);
// 0 - why this compiles (with no warning) and what this result means?
cout << s5.size() << endl;
// basic_string( std::initializer_list<CharT> init,const Allocator& alloc = Allocator() );
// doesn't compile, no known conversion for argument 2 from 'A' to 'const std::allocator<char>&'
//string s6({'1','2','3','4','5'},A());
//cout << s6.size() << endl;
return 0;
}
【问题讨论】:
-
您能向我解释一下您的期望,而不是会发生什么吗?我在这段代码中看不到任何无意的东西。
-
是的,当然。我期望两个构造函数具有相同的行为,一个也采用字符数组和初始化列表。
标签: c++ string c++11 constructor initializer-list