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取消引用字符串迭代器会产生int

[英]Dereferencing string iterator yields int

I get this error 我收到这个错误

comparison between pointer and integer ('int' and 'const char *')

For the following code 对于以下代码

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

int main()
{
    std::string s("test string");
    for(auto i = s.begin(); i != s.end(); ++i)
    {
        cout << ((*i) != "s") << endl;
    }
}

Why does dereferencing the string iterator yield an int and not std::string ? 为什么解除引用字符串迭代器会产生一个int而不是std::string

Actually, it does not yield an int , it yields a char (because a string iterator iterates over the characters in the string). 实际上,它不会产生一个int ,它会产生一个char (因为字符串迭代器迭代字符串中的字符)。 Since the other operand of != is not a char (it's a const char[2] ), standard promotions and conversions are applied to the arguments: 由于!=的另一个操作数不是char (它是一个const char[2] ),标准的促销和转换将应用于参数:

  • char is promoted to int via integral promotion char通过整体推广升级为int
  • const char[2] is converted to const char* via array-to-pointer conversion, const char[2]通过数组到指针转换转换为const char*

This is how you arrive at the int and const char* operands the compiler is complaining about. 这是你到达编译器抱怨的intconst char*操作数的方法。

You should compare the dereferenced iterator to a character, not to a string: 您应该将解除引用的迭代器与字符进行比较而不是与字符串进行比较:

cout << ((*i) != 's') << endl;

"" encloses a string literal (type const char[N] ), '' encloses a character literal (type char ). ""包含一个字符串文字(类型const char[N] ), ''包含一个字符文字( char类型)。

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