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std::string formatting like sprintf

I have to format std::string with sprintf and send it into file stream. How can I do this?

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Max Frai Avatar asked Feb 26 '10 14:02

Max Frai


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1 Answers

Modern C++ makes this super simple.

C++20

C++20 introduces std::format, which allows you to do exactly that. It uses replacement fields similar to those in python:

#include <iostream> #include <format>   int main() {     std::cout << std::format("Hello {}!\n", "world"); } 

Code from cppreference.com, CC BY-SA and GFDL

Check out the compiler support page to see if it's available in your standard library implementation. As of 2021-11-28, partial support is available in Visual Studio 2019 16.10, which was released on 2021-05-25 and Clang 14, which is tracked here. In all other cases, you can resort to the C++11 solution below, or use the {fmt} library, which has the same semantics as std::format.


C++11

With C++11s std::snprintf, this already became a pretty easy and safe task.

#include <memory> #include <string> #include <stdexcept>  template<typename ... Args> std::string string_format( const std::string& format, Args ... args ) {     int size_s = std::snprintf( nullptr, 0, format.c_str(), args ... ) + 1; // Extra space for '\0'     if( size_s <= 0 ){ throw std::runtime_error( "Error during formatting." ); }     auto size = static_cast<size_t>( size_s );     std::unique_ptr<char[]> buf( new char[ size ] );     std::snprintf( buf.get(), size, format.c_str(), args ... );     return std::string( buf.get(), buf.get() + size - 1 ); // We don't want the '\0' inside } 

The code snippet above is licensed under CC0 1.0.

Line by line explanation:

Aim: Write to a char* by using std::snprintf and then convert that to a std::string.

First, we determine the desired length of the char array using a special condition in snprintf. From cppreference.com:

Return value

[...] If the resulting string gets truncated due to buf_size limit, function returns the total number of characters (not including the terminating null-byte) which would have been written, if the limit was not imposed.

This means that the desired size is the number of characters plus one, so that the null-terminator will sit after all other characters and that it can be cut off by the string constructor again. This issue was explained by @alexk7 in the comments.

int size_s = std::snprintf( nullptr, 0, format.c_str(), args ... ) + 1; 

snprintf will return a negative number if an error occurred, so we then check whether the formatting worked as desired. Not doing this could lead to silent errors or the allocation of a huge buffer, as pointed out by @ead in the comments.

if( size_s <= 0 ){ throw std::runtime_error( "Error during formatting." ); } 

Because we know that size_s can't be negative, we use a static cast to convert it from a signed int to an unsigned size_t. This way, even the most pedantic compiler won't complain about the conversions that would otherwise happen on the next lines.

size_t size = static_cast<size_t>( size_s ); 

Next, we allocate a new character array and assign it to a std::unique_ptr. This is generally advised, as you won't have to manually delete it again.

Note that this is not a safe way to allocate a unique_ptr with user-defined types as you can not deallocate the memory if the constructor throws an exception!

std::unique_ptr<char[]> buf( new char[ size ] ); 

In C++14, you could instead use make_unique, which is safe for user-defined types.

auto buf = std::make_unique<char[]>( size ); 

After that, we can of course just use snprintf for its intended use and write the formatted string to the char[].

std::snprintf( buf.get(), size, format.c_str(), args ... ); 

Finally, we create and return a new std::string from that, making sure to omit the null-terminator at the end.

return std::string( buf.get(), buf.get() + size - 1 ); 

You can see an example in action here.


If you also want to use std::string in the argument list, take a look at this gist.


Additional information for Visual Studio users:

As explained in this answer, Microsoft renamed std::snprintf to _snprintf (yes, without std::). MS further set it as deprecated and advises to use _snprintf_s instead, however _snprintf_s won't accept the buffer to be zero or smaller than the formatted output and will not calculate the outputs length if that occurs. So in order to get rid of the deprecation warnings during compilation, you can insert the following line at the top of the file which contains the use of _snprintf:

#pragma warning(disable : 4996) 

Final thoughts

A lot of answers to this question were written before the time of C++11 and use fixed buffer lengths or vargs. Unless you're stuck with old versions of C++, I wouldn't recommend using those solutions. Ideally, go the C++20 way.

Because the C++11 solution in this answer uses templates, it can generate quite a bit of code if it is used a lot. However, unless you're developing for an environment with very limited space for binaries, this won't be a problem and is still a vast improvement over the other solutions in both clarity and security.

If space efficiency is super important, these two solution with vargs and vsnprintf can be useful. DO NOT USE any solutions with fixed buffer lengths, that is just asking for trouble.

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iFreilicht Avatar answered Oct 10 '22 21:10

iFreilicht