C++ 将 int 附加到 std::string
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Append int to std::string
提问by navylover
I tried two different ways to append an intto a std::string, and to my surprise, I got different results:
我尝试了两种不同的方法将 an 附加int到 a std::string，令我惊讶的是，我得到了不同的结果：
#include <string>
int main()
{
std::string s;
s += 2; // compiles correctly
s = s + 2; // compiler error
return 0;
}
Why does it compile and work correctly when I use the +=operator, but fail when I use the +operator?
为什么当我使用+=运算符时它可以编译并正常工作，但使用+运算符时却失败？
I don't think the question is like How to concatenate a std::string and an int?
我不认为问题类似于如何连接 std::string 和 int？
In that question,no answer uses +=operator.And the difference between +=and +operator of std::stringis the key to solve my doubt.
在那个问题中，没有答案使用+=运算符。+=和+运算符之间的区别std::string是解决我疑问的关键。
Frankly,the question is a good example for explaining why c++ is so difficult to master.
坦率地说，这个问题是解释为什么 c++ 如此难以掌握的一个很好的例子。
回答by iBug
TL;DRoperator+=is a class member function in class string, while operator+is a template function.
TL;DRoperator+=是 中的类成员函数class string，而operator+是模板函数。
The standard class template<typename CharT> basic_string<CharT>has overloaded function basic_string& operator+=(CharT), and string is just basic_string<char>.
标准类template<typename CharT> basic_string<CharT>具有重载函数basic_string& operator+=(CharT)，而字符串只是basic_string<char>.
As values that fits in a lower type can be automatically cast into that type, in expression s += 2, the 2 is nottreated as int, but charinstead. It has exactlythe same effect as s += '\x02'. A char with ASCII code 2 (STX) is appended, notthe character '2' (with ASCII value 50, or 0x32).
由于适合较低类型的值可以自动转换为该类型，因此在 expression 中s += 2， 2不被视为int，char而是。它正好相同的效果s += '\x02'。附加了 ASCII 代码 2 (STX)的字符，而不是字符“2”（ASCII 值为 50 或 0x32）。
However, string does not have an overloaded member function like string operator+(int), s + 2is not a valid expression, thus throws an error during compilation. (More below)
但是，string 没有像 那样的重载成员函数string operator+(int)，s + 2它不是有效的表达式，因此在编译期间会引发错误。（更多见下文）
You can use operator+ function in string in these ways:
您可以通过以下方式在字符串中使用 operator+ 函数：
s = s + char(2); // or (char)2
s = s + std::string(2);
s = s + std::to_string(2); // C++11 and above only
For people concerned about why 2 isn't automatically cast to charwith operator+,
对于担心为什么 2 不会自动转换为charwith 的人operator+，
template <typename CharT>
basic_string<CharT>
operator+(const basic_string<CharT>& lhs, CharT rhs);
The above is the prototype[note]for the plus operator in s + 2, and because it's a template function, it is requiring an implementation of both operator+<char>and operator+<int>, which is conflicting. For details, see Why isn't automatic downcasting applied to template functions?
上面是加号操作符 in的原型[注]s + 2，因为它是一个模板函数，所以需要同时实现operator+<char>and operator+<int>，这是冲突的。有关详细信息，请参阅为什么不将自动向下转换应用于模板函数？
Meanwhile, the prototype of operator+=is:
同时，原型operator+=为：
template <typename CharT>
class basic_string{
basic_string&
operator+=(CharT _c);
};
You see, no template here (it's a class member function), so the compiler deduces that type CharT is charfrom class implementation, and int(2)is automatically cast into char(2).
你看，这里没有模板（它是一个类成员函数），所以编译器推断出 CharT 类型char来自类实现，并int(2)自动转换为char(2).
Note:Unnecessary code is stripped when copying from C++ standard include source. That includes typename 2 and 3 (Traits and Allocator) for template class "basic_string", and unnecessary underscores, in order to improve readability.
注意：从 C++ 标准包含源复制时，会删除不必要的代码。这包括模板类“basic_string”的 typename 2 和 3（特征和分配器），以及不必要的下划线，以提高可读性。
回答by Bathsheba
s += 2;is not doing what you think it's doing. It calls the overloaded +=operator to a char. It does notappend the character'2', but rather the character with value2, and the result will depend on the encodingused on your platform.
s += 2;不是在做你认为它在做的事情。它将重载+=运算符调用到 a char。它并没有追加字符'2'用，而是字符值2，结果将取决于编码的平台上使用。
There is no operator overload defined to allow s + 2to compile1. Hence the error.
没有定义运算符重载来允许s + 2编译1。因此错误。
The solution in both cases is to use std::to_string(2)rather than the intliteral 2.
在这两种情况下的解决方案是使用std::to_string(2)而不是int文字 2。
1Essentially the reason is because operator+=is not a template function, but std::operator+is, and overload resolution will favour a non-template function over a template one.
1本质上的原因是因为operator+=不是模板函数，而是模板函数，std::operator+并且重载决议将有利于非模板函数而不是模板函数。
回答by Cory Kramer
The correct way to add to your stringwould be
添加到您的正确方法string是
std::string s;
s += std::to_string(2);
s = s + std::to_string(2);
回答by Xatyrian
While @CoryKramer answer gives you the correct way to add an integer to a string, it doesn't explain why the instruction s = s + 2does not compile.
虽然@CoryKramer 的答案为您提供了向字符串添加整数的正确方法，但它并没有解释为什么指令s = s + 2无法编译。
The difference between the two instruction is that in the first one you use the std::string's += operatorwhile in the second instruction, the compiler tries to cast 2to a string.
两条指令之间的区别在于，在第一条指令中使用std::string's += 运算符，而在第二条指令中，编译器尝试转换2为字符串。
There is no implicit conversion between intand std::string. however, you can cast an intto char, so this is why s += 2works.
int和之间没有隐式转换std::string。但是，您可以强制转换为intto char，所以这就是有效的原因s += 2。
