What is the difference between printf() and puts() in C?

Question

I know you can print with printf() and puts(). I can also see that printf() allows you to interpolate variables and do formatting.

Is puts() merely a primitive version of printf(). Should it be used for every possible printf() without string interpolation?

Solution 1

puts is simpler than printf but be aware that the former automatically appends a newline. If that's not what you want, you can fputs your string to stdout or use printf.

Solution 2

(This is pointed out in a comment by Zan Lynx, but I think it deserves an answer - given that the accepted answer doesn't mention it).

The essential difference between puts(mystr); and printf(mystr); is that in the latter the argument is interpreted as a formatting string. The result will be often the same (except for the added newline) if the string doesn't contain any control characters (%) but if you cannot rely on that (if mystr is a variable instead of a literal), you should not use it.

So, it's generally dangerous - and conceptually wrong - to pass a dynamic string as single argument of printf:

char * myMessage;
// ... myMessage gets assigned at runtime, unpredictable content
printf(myMessage);  // <--- WRONG! (what if myMessage contains a '%' char?)
puts(myMessage);    // ok 
printf("%s\n",myMessage); // ok, equivalent to the previous, perhaps less efficient

The same applies to fputs vs fprintf (but fputs doesn't add the newline).

Solution 3

Besides formatting, puts returns a nonnegative integer if successful or EOF if unsuccessful; while printf returns the number of characters printed (not including the trailing null).

Solution 4

In simple cases, the compiler converts calls to printf() to calls to puts().

For example, the following code will be compiled to the assembly code I show next.

#include <stdio.h>
main() {
    printf("Hello world!");
    return 0;
}

push rbp
mov rbp,rsp
mov edi,str.Helloworld!
call dword imp.puts
mov eax,0x0
pop rbp
ret

In this example, I used GCC version 4.7.2 and compiled the source with gcc -o hello hello.c.

Solution 5

In my experience, printf() hauls in more code than puts() regardless of the format string.

If I don't need the formatting, I don't use printf. However, fwrite to stdout works a lot faster than puts.

static const char my_text[] = "Using fwrite.\n";
fwrite(my_text, 1, sizeof(my_text) - sizeof('\0'), stdout);

Note: per comments, '\0' is an integer constant. The correct expression should be sizeof(char) as indicated by the comments.

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Comments

[Anthony Forloney]

puts() vs printf() - C/C++ Answers

[Zan Lynx]

Just a note on using printf instead of puts: never, ever do a printf(variable) to print a string. Use puts(variable) or printf("%s', variable). There's a security risk in using a variable format string: if the variable can be written by an attacker they can attack the program by using format strings.

[Antony Hatchkins]

"fwrite to stdout works a lot faster than puts." - What could possibly be the reason?

[zubergu]

And what about new line that puts places in stdout?

[Wiz]

@AntonyHatchkins It's typically not "a lot" faster. puts(), however, does have to perform an strlen() call every time on your string whereas if the size is known with fwrite() it can be avoided. That's pretty much the only real contributer to a performance difference.

[franklin]

In what way would using printf() be less efficient? At run time? At compile time?

[leonbloy]

@franklin at runtime, because printf needs to parse the format string. However, this should normally be irrelevant. Further, a clever compiler could optimize this, and replace the printf with call to puts

[Rafael Almeida]

It should have been printf("Hello world!\n"); gcc indeed translates that to puts. Since it's an old message, I will edit it myself.

[Koray Tugay]

How did you read the assembly code after compiling the C code?

[Bradley Garagan]

This answer is incorrect. '\0' has type int, so on most systems this will print Using fwrit. If you want to print 1 less byte, just use 1. sizeof (char), which is likely what you intended here, is guaranteed to be 1.

[Erutan409]

I think it's also important to mention the additional arguments printf take for adding additional variables into the outputted string.

[schaiba]

@KorayTugay : the -save-temps option for gcc does that

[Ivan Kaloyanov]

You could also use a tool like gdb to disassemble a binary.

[Nico Haase]

Please add some explanation to your answer such that others can learn from it - do you have reliable sources for that claim? Or some reasons to explain this difference?

[Lou]

When I've tested this, puts returns the number of characters in the string including the terminating character, whereas printf returns the number of characters printed without the terminating character. Which is technically consistent with your definition, as the former is a non-negative integer, but I'm not sure whether that's what you meant to say.

[diverger]

@lou, are you sure the extra including come from the null terminator but not the '\n'?

[Tom Kuschel]

I do not believe that puts() takes more time than printf(). Note that puts automatically appends a newline.

[Alex]

Also $ objdump -d printf.exe -M intel

[Alex]

By the way in gcc 7.5.0 assembly looks like = call printf@PLT