How to get a formatted std::string in c++ without length liminations

Question

What is the best way (short, using standard libraries and easy to understand) to do this in c++:

std::string s = magic_command("%4.2f", 123.456f)

without length limitations (char s[1000] = ...)
where "%4.2f" is any c format string (which would be given to printf for example)

I am aware of the snprintf malloc combo suggested for pure c in

writing formatted data of unknown length to a string (C programming)

but is there a better, less verbose, way to do this with c++?

I am also aware of the std::ostringstream method suggested in

Convert float to std::string in C++

but I want to pass a c format string such as "%4.2f", and I could not find a way to do this with ostringstream.

I can recommend Boost.Format, especially because it's templated. If you, for whatever reason, don't want to use it, there's also Poco::PatternFormatter: appinf.com/docs/poco/Poco.PatternFormatter.html — Robin
– Robin, Commented Aug 20, 2012 at 15:43

score 8 · Accepted Answer · 2012-08-21 14:23:51Z

8

You can try Boost.Format:

std::string s = boost::str(boost::format("%4.2f") % 123.456f);

It's not included in the standard, but Boost is about as standard as a non-standard library can get.

edited Aug 21, 2012 at 14:23

answered Aug 20, 2012 at 14:35

user784668

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1 Comment

Rostyslav Dzinko Over a year ago

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Luchian Grigore · Accepted Answer · 2012-08-20 14:35:01Z

I would use a std::stringstream (in combination with setprecision) instead and use .str() afterwards to get the std::string.

zneak · Accepted Answer · 2013-08-28 16:04:47Z

C++ completely abandons the concept of format string, so there will be no standard way to do it. You could implement magic_command yourself using asprintf (its vasprintf variant actually), though.

Note that *asprintf are GNU/BSD extensions. As such, they don't exist on Windows. Also, this solution is not type-safe, and will only accept POD types (so no classes, structs or unions).

std::string magic_command(const std::string& format, ...) { char* ptr; va_list args; va_start(args, format); vasprintf(&ptr, format.c_str(), args); va_end(args); std::unique_ptr<char, decltype(free)&> free_chars(ptr, free); return std::string(ptr); }

Ciro Santilli OurBigBook.com · Accepted Answer · 2022-09-04 08:18:26Z

New answer 10 years later: C++20 std::format

Finally, this will be the superior choice once you can use it:

#include <format> #include <string> int main() { std::cout << std::format("{:4.3}\n{:4.3}\n", 3.1, 3.1415); }

Expected output:

 3.1 3.14

This will therefore completely overcome the madness of modifying std::cout state.

The existing fmt library implements it for before it gets official support: https://github.com/fmtlib/fmt Install on Ubuntu 22.04:

sudo apt install libfmt-dev

Modify source to replace:

<format> with <fmt/core.h>
std::format to fmt::format

main.cpp

#include <iostream> #include <fmt/core.h> int main() { std::cout << fmt::format("{:4.3}\n{:4.3}\n", 3.1, 3.1415); }

and compile and run with:

g++ -std=c++11 -o main.out main.cpp -lfmt ./main.out

Output:

 3.1 3.14

Old answer

All answers given are good since there seems to be no really standard way of doing what I want, I'll vote you all up, and accept my own sum up course of action:

if the string is short enough to estimate its size by hand, do it, multiply your estimative by 4, and allocate it statically.
if you can get away with stringstream + setprecision, do it since it is standard
if not, and you are willing to write and include a short helper function based on snprintf/check overflow/dynamic allocation, do it and put it into your project "utils" file
finnally consider which dependency is less restrictive for your project (maybe you are already using one of them):
- if boost is less restrictive, use Boost.Format
- if GNU/BSD extensions are less restrictive, use asprintf

Kerrek SB · Accepted Answer · 2012-08-20 14:49:50Z

If you literally want to use your syntax and forgo type safety, I'd write a little wrapper class that wraps snprintf. I'd make it start out with a local, automatic buffer of some small size (say 2048?) and call snprintf once. If it succeeds, return a std::string created from that buffer. If you overran, allocate a std::string of the correct size and repeat the snprintf.

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