Should I use printf in my C++ code?

Question

So I generally use cout and cerr to write text to the console. However sometimes I find it easier to use the good old printf statement. I use it when I need to format the output.

So one example of where I would use this is:

// Lets assume that I'm printing coordinates... 
printf("(%d,%d)\n", x, y);

// To do the same thing as above using cout....
cout << "(" << x << "," << y << ")" << endl;

I know I can format output using cout but I already know how to use the printf. So is there any reason I shouldn't use the C printf statement?

Answer 1

I've read often on the net that printf is faster (but with no type safety) than the c++ standard stream.

So you should use the standard stream most of the time because the type safety will help make correct code. But if you're in a performance critical application (like a log system in an AAA game for example) you'll need to get most of the performance so using printf might help on this side.

Just don't forget that premature optimization if the root of all evils and that correct code is the priority over faster code.

Answer 2

No reason at all. I think it's just some strange ideology that drives people towards using only C++ libraries even though good old C libs are still valid. I'm a C++ guy and I use C functions a lot too. Never had any problems with them.

Answer 3

My students, who learn cin and cout first, then learn printf later, overwhelmingly prefer printf (or more usually fprintf). I myself have found the printf model sufficiently readable that I have ported it to other programming languages. So has Olivier Danvy, who has even made it type-safe.

Provided you have a compiler that is capable of type-checking calls to printf, I see no reason not to use fprintf and friends in C++.

Disclaimer: I am a terrible C++ programmer.

Answer 4

I use printf because I hate the ugly <<cout<< syntax.

Answer 5

I often "drop back" to using printf(), but more often snprintf() for easier formatted output. When programming in C++ I use this wrapper I wrote a while back, called like this (to use your example as above): cout << format("(%d,%d)\n", x, y);

Here's the header (stdiomm.h):

#pragma once

#include <cstdarg>
#include <string>

template <typename T>
std::basic_string<T> format(T const *format, ...);

template <typename T>
std::basic_string<T> vformat(T const *format, va_list args);

And the source (stdiomm.cpp):

#include "stdiomm.h"
#include <boost/scoped_array.hpp>
#include <cstdio>

template <>
std::wstring vformat(wchar_t const *format, va_list arguments)
{
#if defined(_WIN32)
    int required(_vscwprintf(format, arguments));
    assert(required >= 0);
    boost::scoped_array<wchar_t> buffer(new wchar_t[required + 1]);
    int written(vswprintf(buffer.get(), required + 1, format, arguments));
    assert(written == required);
    return std::wstring(buffer.get(), written);
#else
#   error "No implementation yet"
#endif
}

template <>
std::string vformat(char const *format, va_list arguments)
{
#if defined(_WIN32)
    int required(_vscprintf(format, arguments));
    assert(required >= 0);
    boost::scoped_array<char> buffer(new char[required + 1]);
    int written(vsnprintf(buffer.get(), required + 1, format, arguments));
    assert(written == required);
    return std::string(buffer.get(), written);
#else
    char *buffer;
    int printed = vasprintf(&buffer, format, arguments);
    assert(printed != -1);
    std::string retval(buffer, printed);
    free(buffer);
    return retval;      
#endif
}

template <typename T>
std::basic_string<T> format(T const *format, ...)
{
    va_list ap;
    va_start(ap, format);
    std::basic_string<T> retval(vformat(format, ap));
    va_end(ap);
    return retval;
}

template std::wstring format(wchar_t const *format, ...);
template std::string format(char const *format, ...);

Update

After reading some of the other answers, I might have to make a switch to boost::format() myself!

Answer 6

If you ever hope to i18n your program, stay away from iostreams. The problem is that it can be impossible to properly localize your strings if the sentence is composed of multiple fragments as is done with iostream.

Besides the issue of message fragments, you also have an issue of ordering. Consider a report that prints a student's name and their grade point average:

std::cout << name << " has a GPA of " << gpa << std::endl;

When you translate that to another language, the other language's grammar may need you to show the GPA before the name. AFAIK, iostreams has not way to reorder the interpolated values.

If you want the best of both worlds (type safety and being able to i18n), use Boost.Format.

Answer 7

Streams are the canonical way. Try making this code work with printf:

template <typename T>
void output(const T& pX)
{
    std::cout << pX << std::endl;
}

Good luck.

What I mean is, you can make operators to allow your types to be outputted to ostream's, and without hassle use it just like any other type. printf doesn't fit the the generality of C++, or more specifically templates.

There's more than usability. There's also consistency. In all my projects, I have cout (and cerr and clog) tee'd to also output to a file. If you use printf, you skip all of that. Additionally, consistency itself is a good thing; mixing cout and printf, while perfectly valid, is ugly.

If you have an object, and you want to make it output-able, the cleanest way to do this is overload operator<< for that class. How are you going to use printf then? You're going to end up with code jumbled with cout's and printf's.

If you really want formatting, use Boost.Format while maintaining the stream interface. Consistency and formatting.

Answer 8

Use boost::format. You get type safety, std::string support, printf like interface, ability to use cout, and lots of other good stuff. You won't go back.

Answer 9

Use printf. Do not use C++ streams. printf gives you much better control (such as float precision etc.). The code is also usually shorter and more readable.

Google C++ style guide agrees.

Do not use streams, except where required by a logging interface. Use printf-like routines instead.

There are various pros and cons to using streams, but in this case, as in many other cases, consistency trumps the debate. Do not use streams in your code.

Answer 10

On the whole I agree (hate the << syntax especially if you need complex formatting)

But I should point out the safety aspects.

printf("%x",2.0f)
printf("%x %x",2)
printf("%x",2,2)

Probably won't be noticed by the compiler but could crash your app.

Answer 11

Use whatever fits your needs and preferences. If you're comfortable with printf then by all means use it. If you're happier with iostreams stick to 'em. Mix and match as best fits your requirements. This is software, after all - there's better ways and worse ways, but seldom is there only ONE way.

Share and enjoy.

Answer 12

I do not like printf. Its lack of type-safety makes it dangerous to use, plus the need to remember format specifiers is a pain. The templated operators that smartly do the right thing are much better. So I always use the C++ streams in C++.

Granted, many people prefer printf, for other reasons, enumerated elsewhere.

Answer 13

I almost always use printf for temporary debugging statements. For more permanent code, I prefer the 'c' streams as they are The C++ Way. Although boost::format looks promising and might replace my stream usage (especially for complexly formatted output), probably nothing will replace printf for me for a long time.

Answer 14

streams are preferred in cpp as they adhere to the object oriented paradigm of cpp, beside being type safe.

printf , on the other hand is more of a functional approach.

only reason for not using printf in cpp code that i can think of is not being object oriented.

its more of a personal choice.

Answer 15

cout stinks.