C++字符串 - 使用初始化列表构造函数时的奇怪行为

Question

我知道我可以使用字符数组和初始化列表来填充字符串.

它看起来编译器从int到initializer_list或allocator进行了一些隐式的提升.但我不知道为什么它没有给我任何警告,为什么它暗示它.

你能解释一下字符串s4和s5会发生什么吗？

http://ideone.com/5Agc2T

#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;
}

Answer 1

string s6({'1','2','3','4','5'},3);
string s7({'1','2','3','4','5'},5);

实际上,那些初始化不只是调用std::initializer_list构造函数.第二个参数不能隐式转换为std::allocator,因此考虑其他构造函数.调用的构造函数是具有此签名的构造函数:

basic_string( const basic_string& other, 
              size_type pos, 
              size_type count = std::basic_string::npos,
              const Allocator& alloc = Allocator() );

该std::initializer_list构造函数用于创建临时std::string从支撑-INIT列表传递作为other参数传递给上述构造.临时可以绑定到它,因为它是const的引用.因此,第二个参数是pos参数,它用作子串复制结构的起点.

所以s6在区间的字符[3, 5)(即,"45")和s7在区间中的字符[5,5)(即,"").