C++ std::string 和格式字符串
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原文地址: http://stackoverflow.com/questions/1575698/
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std::string and format string
提问by Scott
I found the below code through Google. It almost does what I want it to do, except it doesn't provide a way to indicate the precision like '%.*f' does in C-type format strings. Also, it doesn't provide anything further than 5 decimal places. Am I going to have to stick with C strings and snprintf?
我通过谷歌找到了下面的代码。它几乎可以完成我想要它做的事情，只是它没有提供一种方法来指示像 '%.*f' 在 C 类型格式字符串中所做的精度。此外，它不提供超过 5 个小数位的任何内容。我将不得不坚持使用 C 字符串和 snprintf 吗？
#include <string>
#include <sstream>
#include <iostream>
template <class T>
std::string to_string(T t, std::ios_base & (*f)(std::ios_base&))
{
std::ostringstream oss;
oss << f << t;
return oss.str();
}
int main()
{
std::cout<<to_string<double>(3.1415926535897931, std::dec)<<std::endl;
return 0;
}
采纳答案by 1800 INFORMATION
You want to use the std::setprecisionmanipulator:
你想使用std::setprecision操纵器：
int main()
{
std::cout << std::setprecision(9) << to_string<long>(3.1415926535897931, std::dec)
<< '\n';
return 0;
}
回答by GManNickG
C++ wouldn't be successful if it couldn't do something C could.
如果 C++ 不能做 C 可以做的事情，它就不会成功。
You need to check out manipulators.
你需要检查操纵器。
If you want C-style formatting (which I do prefer, it's more terse), check out Boost.Format.
如果你想要 C 风格的格式（我更喜欢，它更简洁），请查看Boost.Format。
回答by MSalters
Taking the almost-correct answer (note that std::dec is redundant in this simple case):
采取几乎正确的答案（注意 std::dec 在这个简单的情况下是多余的）：
int main()
{
std::cout << std::setprecision(9) << std::dec << 3.1415926535897931 << std::endl;
return 0;
}
However, if you wanted the to_string function to behave as desired, that's a bit more difficult. You'd need to pass setprecision(9)to the to_string<T>function, and it doesn't accept arguments of that type. You'd want a templated version:
但是，如果您希望 to_string 函数按预期运行，那就有点困难了。您需要传递setprecision(9)给该to_string<T>函数，并且它不接受该类型的参数。你想要一个模板版本：
template <class T, class F>
std::string to_string(T t, F f)
{
std::ostringstream oss;
oss << f << t;
return oss.str();
}
int main()
{
std::cout << to_string<double>(3.1415926535897931, std::setprecision(9)) << std::endl;
return 0;
}
This works because you really didn't need std::dec in to_string. But if you needed to pass more manipulators, the simple solution is to add template <class T, class F1, class F2> std::string to_string(T t, F1 f1, F2 f2)etcetera. Technically this doesn't scale very well, but it's going to be so rare that you probably don't need it at all.
这是有效的，因为您确实不需要在 to_string 中使用 std::dec。但是如果你需要传递更多的操纵器，简单的解决方案是添加template <class T, class F1, class F2> std::string to_string(T t, F1 f1, F2 f2)等等。从技术上讲，这不能很好地扩展，但它会非常罕见，以至于您可能根本不需要它。
回答by Jerry Coffin
Have you looked at Boost::format?
你看过Boost::format吗？
Edit: It's not entirely clear what you want. If you just want to write to a string, with formatting, you can use normal manipulators on a stringstream. If you want to use printf-style formatting strings, but retain type-safety, Boost::format can/will do that.
编辑：目前还不完全清楚你想要什么。如果您只想写入带有格式的字符串，则可以在字符串流上使用普通操纵器。如果你想使用 printf 风格的格式化字符串，但保留类型安全性，Boost::format 可以/将会这样做。
