构造std :: string使用<<运算符 [英] Constructing std::string using << operator

问题描述

如果我们要构造一个复杂的字符串，像这样：
我有10个朋友和20个关系（其中10和20是一些变量的值）它像这样：

If we want to construct a complex string, say like this: "I have 10 friends and 20 relations" (where 10 and 20 are values of some variables) we can do it like this:

std::ostringstream os;
os << "I have " << num_of_friends << " friends and " << num_of_relations << " relations";

std::string s = os.str();

但它有点太长了。如果在你的代码中的不同方法中你需要多次构造复合字符串，你必须总是在其他地方定义一个std :: ostringstream的实例。

But it is a bit too long. If in different methods in your code you need to construct compound strings many times you will have to always define an instance of std::ostringstream elsewhere.

我创建了一些额外的代码来实现这一点：

I created some extra code for being able to do this:

struct OstringstreamWrapper
{
     std::ostringstream os;
};

std::string ostream2string(std::basic_ostream<char> &b)
{
     std::ostringstream os;
     os << b;
     return os.str();
}

#define CreateString(x) ostream2string(OstringstreamWrapper().os << x)

// Usage:
void foo(int num_of_friends, int num_of_relations)
{
     const std::string s = CreateString("I have " << num_of_friends << " and " << num_of_relations << " relations");
}

但是也许在C ++ 11或Boost中有更简单的方法？

But maybe there is simpler way in C++ 11 or in Boost?

推荐答案

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

template<typename T, typename... Ts>
std::string CreateString(T const& t, Ts const&... ts)
{
    using expand = char[];

    std::ostringstream oss;
    oss << std::boolalpha << t;
    (void)expand{'\0', (oss << ts, '\0')...};
    return oss.str();
}

void foo(int num_of_friends, int num_of_relations)
{
    std::string const s =
        CreateString("I have ", num_of_friends, " and ", num_of_relations, " relations");
    std::cout << s << std::endl;
}

int main()
{
    foo(10, 20);
}

在线演示

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