Why is printf with a single argument (without conversion specifiers) deprecated? - c

In a book that I'm reading, it's written that printf with a single argument (without conversion specifiers) is deprecated. It recommends to substitute
printf("Hello World!");
with
puts("Hello World!");
or
printf("%s", "Hello World!");
Can someone tell me why printf("Hello World!"); is wrong? It is written in the book that it contains vulnerabilities. What are these vulnerabilities?

printf("Hello World!"); is IMHO not vulnerable but consider this:
const char *str;
...
printf(str);
If str happens to point to a string containing %s format specifiers, your program will exhibit undefined behaviour (mostly a crash), whereas puts(str) will just display the string as is.
Example:
printf("%s"); //undefined behaviour (mostly crash)
puts("%s"); // displays "%s\n"

printf("Hello world");
is fine and has no security vulnerability.
The problem lies with:
printf(p);
where p is a pointer to an input that is controlled by the user. It is prone to format strings attacks: user can insert conversion specifications to take control of the program, e.g., %x to dump memory or %n to overwrite memory.
Note that puts("Hello world") is not equivalent in behavior to printf("Hello world") but to printf("Hello world\n"). Compilers usually are smart enough to optimize the latter call to replace it with puts.

Further to the other answers, printf("Hello world! I am 50% happy today") is an easy bug to make, potentially causing all manner of nasty memory problems (it's UB!).
It's just simpler, easier and more robust to "require" programmers to be absolutely clear when they want a verbatim string and nothing else.
And that's what printf("%s", "Hello world! I am 50% happy today") gets you. It's entirely foolproof.
(Steve, of course printf("He has %d cherries\n", ncherries) is absolutely not the same thing; in this case, the programmer is not in "verbatim string" mindset; she is in "format string" mindset.)

I'll just add a bit of information regarding the vulnerability part here.
It's said to be vulnerable because of printf string format vulnerability. In your example, where the string is hardcoded, it's harmless (even if hardcoding strings like this is never fully recommended). But specifying the parameter's types is a good habit to take. Take this example:
If someone puts format string character in your printf instead of a regular string (say, if you want to print the program stdin), printf will take whatever he can on the stack.
It was (and still is) very used to exploit programs into exploring stacks to access hidden information or bypass authentication for example.
Example (C):
int main(int argc, char *argv[])
{
printf(argv[argc - 1]); // takes the first argument if it exists
}
if I put as input of this program "%08x %08x %08x %08x %08x\n"
printf ("%08x %08x %08x %08x %08x\n");
This instructs the printf-function to retrieve five parameters from the stack and display them as 8-digit padded hexadecimal numbers. So a possible output may look like:
40012980 080628c4 bffff7a4 00000005 08059c04
See this for a more complete explanation and other examples.

This is misguided advice. Yes, if you have a run-time string to print,
printf(str);
is quite dangerous, and you should always use
printf("%s", str);
instead, because in general you can never know whether str might contain a % sign. However, if you have a compile-time constant string, there's nothing whatsoever wrong with
printf("Hello, world!\n");
(Among other things, that is the most classic C program ever, literally from the C programming book of Genesis. So anyone deprecating that usage is being rather heretical, and I for one would be somewhat offended!)

Calling printf with literal format strings is safe and efficient, and there
exist tools to automatically warn you if your invocation of printf with user
provided format strings is unsafe.
The most severe attacks on printf take advantage of the %n format
specifier. In contrast to all other format specifiers, e.g. %d, %n actually
writes a value to a memory address provided in one of the format arguments.
This means that an attacker can overwrite memory and thus potentially take
control of your program. Wikipedia
provides more detail.
If you call printf with a literal format string, an attacker cannot sneak
a %n into your format string, and you are thus safe. In fact,
gcc will change your call to printf into a call to puts, so there litteraly
isn't any difference (test this by running gcc -O3 -S).
If you call printf with a user provided format string, an attacker can
potentially sneak a %n into your format string, and take control of your
program. Your compiler will usually warn you that his is unsafe, see
-Wformat-security. There are also more advanced tools that ensure that
an invocation of printf is safe even with user provided format strings, and
they might even check that you pass the right number and type of arguments to
printf. For example, for Java there is Google's Error Prone
and the Checker Framework.

A rather nasty aspect of printf is that even on platforms where the stray memory reads could only cause limited (and acceptable) harm, one of the formatting characters, %n, causes the next argument to be interpreted as a pointer to a writable integer, and causes the number of characters output thus far to be stored to the variable identified thereby. I've never used that feature myself, and sometimes I use lightweight printf-style methods which I've written to include only the features I actually use (and don't include that one or anything similar) but feeding standard printf functions strings received from untrustworthy sources may expose security vulnerabilities beyond the ability to read arbitrary storage.

Since no one has mentioned, I'd add a note regarding their performance.
Under normal circumstances, assuming no compiler optimisations are used (i.e. printf() actually calls printf() and not fputs()), I would expect printf() to perform less efficiently, especially for long strings. This is because printf() has to parse the string to check if there are any conversion specifiers.
To confirm this, I have run some tests. The testing is performed on Ubuntu 14.04, with gcc 4.8.4. My machine uses an Intel i5 cpu. The program being tested is as follows:
#include <stdio.h>
int main() {
int count = 10000000;
while(count--) {
// either
printf("qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM");
// or
fputs("qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM", stdout);
}
fflush(stdout);
return 0;
}
Both are compiled with gcc -Wall -O0. Time is measured using time ./a.out > /dev/null. The following is the result of a typical run (I've run them five times, all results are within 0.002 seconds).
For the printf() variant:
real 0m0.416s
user 0m0.384s
sys 0m0.033s
For the fputs() variant:
real 0m0.297s
user 0m0.265s
sys 0m0.032s
This effect is amplified if you have a very long string.
#include <stdio.h>
#define STR "qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM"
#define STR2 STR STR
#define STR4 STR2 STR2
#define STR8 STR4 STR4
#define STR16 STR8 STR8
#define STR32 STR16 STR16
#define STR64 STR32 STR32
#define STR128 STR64 STR64
#define STR256 STR128 STR128
#define STR512 STR256 STR256
#define STR1024 STR512 STR512
int main() {
int count = 10000000;
while(count--) {
// either
printf(STR1024);
// or
fputs(STR1024, stdout);
}
fflush(stdout);
return 0;
}
For the printf() variant (ran three times, real plus/minus 1.5s):
real 0m39.259s
user 0m34.445s
sys 0m4.839s
For the fputs() variant (ran three times, real plus/minus 0.2s):
real 0m12.726s
user 0m8.152s
sys 0m4.581s
Note: After inspecting the assembly generated by gcc, I realised that gcc optimises the fputs() call to an fwrite() call, even with -O0. (The printf() call remains unchanged.) I am not sure whether this will invalidate my test, as the compiler calculates the string length for fwrite() at compile-time.

For gcc it is possible to enable specific warnings for checking printf() and scanf().
The gcc documentation states:
-Wformat is included in -Wall. For more control over some aspects
of format checking, the options -Wformat-y2k,
-Wno-format-extra-args, -Wno-format-zero-length,
-Wformat-nonliteral, -Wformat-security, and -Wformat=2 are
available, but are not included in -Wall.
The -Wformat which is enabled within the -Wall option does not enable several special warnings that help to find these cases:
-Wformat-nonliteral will warn if you do not pass a string litteral as format specifier.
-Wformat-security will warn if you pass a string that might contain a dangerous construct. It's a subset of -Wformat-nonliteral.
I have to admit that enabling -Wformat-security revealed several bugs we had in our codebase (logging module, error handling module, xml output module, all had some functions that could do undefined things if they had been called with % characters in their parameter. For info, our codebase is now around 20 years old and even if we were aware of these kind of problems, we were extremely surprised when we enabled these warnings how many of these bugs were still in the codebase).

printf("Hello World\n")
automatically compiles to the equivalent
puts("Hello World")
you can check it with diassembling your executable:
push rbp
mov rbp,rsp
mov edi,str.Helloworld!
call dword imp.puts
mov eax,0x0
pop rbp
ret
using
char *variable;
...
printf(variable)
will lead to security issues, don't ever use printf that way!
so your book is actually correct, using printf with one variable is deprecated but you can still use printf("my string\n") because it will automatically become puts

Beside the other well-explained answers with any side-concerns covered, I would like to give a precise and concise answer to the provided question.
Why is printf with a single argument (without conversion specifiers) deprecated?
A printf function call with a single argument in general is not deprecated and has also no vulnerabilities when used properly as you always shall code.
C Users amongst the whole world, from status beginner to status expert use printf that way to give a simple text phrase as output to the console.
Furthermore, Someone have to distinguish whether this one and only argument is a string literal or a pointer to a string, which is valid but commonly not used. For the latter, of course, there can occur inconvenient outputs or any kind of Undefined Behavior, when the pointer is not set properly to point to a valid string but these things can also occur if the format specifiers are not matching the respective arguments by giving multiple arguments.
Of course, It is also not right and proper that the string, provided as one and only argument, has any format or conversion specifiers, since there is no conversion going to be happen.
That said, giving a simple string literal like "Hello World!" as only argument without any format specifiers inside that string like you provided it in the question:
printf("Hello World!");
is not deprecated or "bad practice" at all nor has any vulnerabilities.
In fact, many C programmers begin and began to learn and use C or even programming languages in general with that HelloWorld-program and this printf statement as first ones of its kind.
They wouldn´t be that if they were deprecated.
In a book that I'm reading, it's written that printf with a single argument (without conversion specifiers) is deprecated.
Well, then I would take the focus on the book or the author itself. If an author is really doing such, in my opinion, incorrect assertions and even teaching that without explicitly explaining why he/she is doing so (if those assertions are really literally equivalent provided in that book), I would consider it a bad book. A good book, as opposed to that, shall explain why to avoid certain kind of programming methods or functions.
According to what I said above, using printf with only one argument (a string literal) and without any format specifiers is not in any case deprecated or considered as "bad practice".
You should ask the author, what he meant with that or even better, mind him to clarify or correct the relative section for the next edition or imprints in general.

Related

In C, why is %s working without giving it a value?

According to my knowledge and some threads like this, if you want to print strings in C you have to do something like this:
printf("%s some text", value);
And the value will be displayed instead of %s.
I wrote this code:
char password[] = "default";
printf("Enter name: \n");
scanf("%s", password);
printf("%s is your password", password); // All good - the print is as expected
But I noticed that I can do the exact same thing without the value part and it will still work:
printf("%s is your password");
So my question is why does the %s placeholder get a value without me giving it one, and how does it know what value to give it?
This is undefined behavior, anything can happen included something that looks like correct. But it is incorrect.
Your compiler can probably tell you the problem if you use correct options.
Standard says (emphasized is mine):
7.21.6.1 The fprintf function
The fprintf function writes output to the stream pointed to by stream,
under control of the string pointed to by format that specifies how
subsequent arguments are converted for output. If there are
insufficient arguments for the format, the behavior is undefined. If
the format is exhausted while arguments remain, the excess arguments
are evaluated (as always) but are otherwise ignored. The fprintf
function returns when the end of the format string is encountered.
The printf() function uses a C language feature that lets you pass a variable number of arguments to a function. (Technically called 'variadic functions' - https://en.cppreference.com/w/c/variadic - I'll just say 'varargs' for short.)
When a function is called in C, the arguments to the function are pushed onto the stack(*) - but the design of the varargs feature provides no way for the called function to know how many parameters were passed in.
When the printf() function executes, it scans the format string, and the %s tells it to look for a string in the next position in the variable argument list. Since there are no more arguments in the list, the code 'walks off the end of the array' and grabs the next thing it sees in memory. I suspect what's happening is that the next location in memory still has the address of password from your prior call to scanf, and since that address points to a string, and you told printf to print a string, you got lucky, and it worked.
Try putting another function call (for example: printf("%s %s %s\n","X","Y","Z") in between your call to scanf("%s", password); and printf("%s is your password"); and you will almost certainly see different behavior.
Free Advice: C has a lot of sharp corners and undefined bits, but a good compiler (and static analysis or 'lint' tool) can warn you about a lot of common errors. If you are going to work in C, learn how to crank your compiler warnings to the max, learn what all the errors and warnings mean (as they happen, not all at once!) and force yourself to write C code that compiles without any warnings. It will save you a lot of unnecessary hassle.
(*) generalizing here for simplicity - sometimes arguments can be passed in registers, sometimes things are inlined, blah blah blah.
So, there are a lot of posts telling that you shouldn't do printf("%s is your password");, and that you were just lucky. I guess from your question that you somewhat knew that. But few are telling you the probable reason for why you were lucky.
To understand what probably happened, we have to understand how function parameters are passed. The caller of a function must put the parameters on an agreed upon place for the function to find the parameters. So for parameters 1...N we call these places r1 ... rN. (This kind of agreement is part of something we call a "Function Calling Convention")
That means that this code:
scanf("%s", password);
printf("%s is your password",password);
may be turned into this pseudo-code by the compiler
r1="%s";
r2=password;
call scanf;
r1="%s is your password";
r2=password;
call printf;
If you now remove the second parameter from the printf call, your pseudo-code will look like this:
r1="%s";
r2=password;
call scanf;
r1="%s is your password";
call printf;
Be aware that after call scanf;, r2 might be unmodified and still be set to password, therefore call printf; "works"
You might think that you have discovered a new way to optimize code, by eliminating one of the r2=password; assignments. This might be true for old "dumb" compilers, but not for modern ones.
Modern compilers will already do this when it is safe. And it is not always safe. Reasons for why it isn't safe might be thatscanf and printf have different calling conventions, r2 might have been modified behind your back, etc..
To better get a feeling of what the compiler is doing, I recommend to look at the assembler output from your compiler, at different optimization levels.
And please, always compile with -Wall. The compiler is often good at telling you when you are doing dumb stuff.

What is the underlying difference between printf(s) and printf("%s", s)?

The question is plain and simple, s is a string, I suddenly got the idea to try to use printf(s) to see if it would work and I got a warning in one case and none in the other.
char* s = "abcdefghij\n";
printf(s);
// Warning raised with gcc -std=c11:
// format not a string literal and no format arguments [-Wformat-security]
// On the other hand, if I use
char* s = "abc %d efg\n";
printf(s, 99);
// I get no warning whatsoever, why is that?
// Update, I've tested this:
char* s = "random %d string\n";
printf(s, 99, 50);
// Results: no warning, output "random 99 string".
So what's the underlying difference between printf(s) and printf("%s", s) and why do I get a warning in just one case?
In the first case, the non-literal format string could perhaps come from user code or user-supplied (run-time) data, in which case it might contain %s or other conversion specifications, for which you've not passed the data. This can lead to all sorts of reading problems (and writing problems if the string includes %n — see printf() or your C library's manual pages).
In the second case, the format string controls the output and it doesn't matter whether any string to be printed contains conversion specifications or not (though the code shown prints an integer, not a string). The compiler (GCC or Clang is used in the question) assumes that because there are arguments after the (non-literal) format string, the programmer knows what they're up to.
The first is a 'format string' vulnerability. You can search for more information on the topic.
GCC knows that most times the single argument printf() with a non-literal format string is an invitation to trouble. You could use puts() or fputs() instead. It is sufficiently dangerous that GCC generates the warnings with the minimum of provocation.
The more general problem of a non-literal format string can also be problematic if you are not careful — but extremely useful assuming you are careful. You have to work harder to get GCC to complain: it requires both -Wformat and -Wformat-nonliteral to get the complaint.
From the comments:
So ignoring the warning, as if I really know what I am doing and there will be no errors, is one or another more efficient to use or are they the same? Considering both space and time.
Of your three printf() statements, given the tight context that the variable s is as assigned immediately above the call, there is no actual problem. But you could use puts(s) if you omitted the newline from the string or fputs(s, stdout) as it is and get the same result, without the overhead of printf() parsing the entire string to find out that it is all simple characters to be printed.
The second printf() statement is also safe as written; the format string matches the data passed. There is no significant difference between that and simply passing the format string as a literal — except that the compiler can do more checking if the format string is a literal. The run-time result is the same.
The third printf() passes more data arguments than the format string needs, but that is benign. It isn't ideal, though. Again, the compiler can check better if the format string is a literal, but the run-time effect is practically the same.
From the printf() specification linked to at the top:
Each of these functions converts, formats, and prints its arguments under control of the format. The format is a character string, beginning and ending in its initial shift state, if any. The format is composed of zero or more directives: ordinary characters, which are simply copied to the output stream, and conversion specifications, each of which shall result in the fetching of zero or more arguments. The results are undefined if there are insufficient arguments for the format. If the format is exhausted while arguments remain, the excess arguments shall be evaluated but are otherwise ignored.
In all these cases, there is no strong indication of why the format string is not a literal. However, one reason for wanting a non-literal format string might be that sometimes you print the floating point numbers in %f notation and sometimes in %e notation, and you need to choose which at run-time. (If it is simply based on value, %g might be appropriate, but there are times when you want the explicit control — always %e or always %f.)
The warning says it all.
First, to discuss about the issue, as per the signature, the first parameter to printf() is a format string which can contain format specifiers (conversion specifier). In case, a string contains a format specifier and the corresponding argument is not supplied, it invokes undefined behavior.
So, a cleaner (or safer) approach (of printing a string which needs no format specification) would be puts(s); over printf(s); (the former does not process s for any conversion specifiers, removing the reason for the possible UB in the later case). You can choose fputs(), if you're worried about the ending newline that automatically gets added in puts().
That said, regarding the warning option, -Wformat-security from the online gcc manual
At present, this warns about calls to printf and scanf functions where the format string is not a string literal and there are no format arguments, as in printf (foo);. This may be a security hole if the format string came from untrusted input and contains %n.
In your first case, there's only one argument supplied to printf(), which is not a string literal, rather a variable, which can be very well generated/ populated at run time, and if that contains unexpected format specifiers, it may invoke UB. Compiler has no way to check for the presence of any format specifier in that. That is the security problem there.
In the second case, the accompanying argument is supplied, the format specifier is not the only argument passed to printf(), so the first argument need not to be verified. Hence the warning is not there.
Update:
Regarding the third one, with excess argument that required by the supplied format string
printf(s, 99, 50);
quoting from C11, chapter §7.21.6.1
[...] If the format is exhausted while arguments remain, the excess arguments are
evaluated (as always) but are otherwise ignored. [...]
So, passing excess argument is not a problem (from the compiler perspective) at all and it is well defined. NO scope for any warning there.
There are two things in play in your question.
The first is covered succinctly by Jonathan Leffler - the warning you're getting is because the string isn't literal and doesn't have any format specifiers in it.
The other is the mystery of why the compiler doesn't issue a warning that your number of arguments doesn't match the number of specifiers. The short answer is "because it doesn't," but more specifically, printf is a variadic function. It takes any number of arguments after the initial format specification - from 0 on up. The compiler can't check to see if you gave the right amount; that's up to the printf function itself, and leads to the undefined behavior that Joachim mentioned in comments.
EDIT:
I'm going to give further answer to your question, as a means of getting on a small soapbox.
What's the difference between printf(s) and printf("%s", s)? Simple - in the latter, you're using printf as it's declared. "%s" is a const char *, and it will subsequently not generate the warning message.
In your comments to other answers, you mentioned "Ignoring the warning...". Don't do this. Warnings exist for a reason, and should be resolved (otherwise they're just noise, and you'll miss warnings that actually matter among the cruft of all the ones that don't.)
Your issue can be resolved in several ways.
const char* s = "abcdefghij\n";
printf(s);
will resolve the warning, because you're now using a const pointer, and there are none of the dangers that Jonathan mentioned. (You could also declare it as const char* const s, but don't have to. The first const is important, because it then matches the declaration of printf, and because const char* s means that characters pointed to by s can't change, i.e. the string is a literal.)
Or, even simpler, just do:
printf("abcdefghij\n");
This is implicitly a const pointer, and also not a problem.
So what's the underlying difference between printf(s) and printf("%s", s)
"printf(s)" will treat s as a format string. If s contains format specifiers then printf will interpret them and go looking for varargs. Since no varargs actually exist this will likely trigger undefined behaviour.
If an attacker controls "s" then this is likely to be a security hole.
printf("%s",s) will just print what is in the string.
and why do I get a warning in just one case?
Warnings are a balance between catching dangerous stupidity and not creating too much noise.
C programmers are in the habbit of using printf and various printf like functions* as generic print functions even when they don't actually need formatting. In this environment it's easy for someone to make the mistake of writing printf(s) without thinking about where s came from. Since formatting is pretty useless without any data to format printf(s) has little legitimate use.
printf(s,format,arguments) on the other hand indicates that the programmer deliberately intended formatting to take place.
Afaict this warning is not turned on by default in upstream gcc, but some distros are turning it on as part of their efforts to reduce security holes.
* Both standard C functions like sprintf and fprintf and functions in third party libraries.
The underlying reason: printf is declared like:
int printf(const char *fmt, ...) __attribute__ ((format(printf, 1, 2)));
This tells gcc that printf is a function with a printf-style interface where the format string comes first. IMHO it must be literal; I don't think there's a way to tell the good compiler that s is actually a pointer to a literal string it had seen before.
Read more about __attribute__ here.

Is there a function to count number of parameters for a printf format string?

Say, if I do the following:
const char *strFmt = "Value=%d, str=%s, v=%I64d. Is 100%% true";
printf(strFmt, 1, "name", -1);
Is there a function that simply returns how many parameters are needed for strFmt without filling them in?
The usual reason for wanting the number of required parameters is to check one's code. C runtime libraries do this as a side-effect of the printf function. Source for those (as well as for similar printf-like functions such as StrAllocVsprintf in Lynx) can be found easily enough.
Because printf is a varargs function, there is no predefined prototype which tells the "right" number of parameters.
If there are too few parameters, bad things can happen ("undefined behavior"), while extra ones may be ignored. Most modern compilers provide some help, in the form of compile-time warnings.
There is no standard function to count the number of required parameters. The question has been asked before, e.g.,
Linux script that counts number of printf parameters for each occurrence
Passing too many arguments to printf
For whatever historical reason, the function also has (apparently) not been provided by any compiler suite. In most cases, this is because printf formats are usually checked by the compiler (or static analysis tools such as lint). Not all compilers do these checks. A few can be told using that some functions aside from printf and its standard relatives are "printf-like". Visual Studio lacks that feature, but can do some printf-checking (not a lot according to Is there any way to make visual C++ (9.0) generate warnings about printf format strings not matching type of printf's args?).
A string constant is not going to be associated by the compiler with the printf function. You could make it a #define and reuse the value in an actual printf function to get whatever checking the compiler can do.
For instance
#define strFmt "Value=%d, str=%s, v=%I64d. Is 100%% true"
printf(strFmt, 1, "name", -1);
The problem itself is portable: if your program also is portable, you can check it with other tools than the Visual Studio compiler.

Printf with no arguments explanation

I understand that if printf is given no arguments it outputs an unexpected value.
Example:
#include <stdio.h>
int main() {
int test = 4 * 4
printf("The answer is: %d\n");
return 0;
}
This returns a random number. After playing around with different formats such as %p, %x etc, it doesn't print 16(because I didn't add the variable to the argument section) What i'd like to know is, where are these values being taken from? Is it the top of the stack? It's not a new value every time I compile, which is weird, it's like a fixed value.
printf("The answer is: %d\n");
invokes undefined behavior. C requires a conversion specifier to have an associated argument. While it is undefined behavior and anything can happen, on most systems you end up dumping the stack. It's the kind of trick used in format string attacks.
It is called undefined behavior and it is scary (see this answer).
If you want an explanation, you need to dive into implementation specific details. So study the generated source code (e.g. compile with gcc -Wall -Wextra -fverbose-asm + your optimization flags, then look into the generated .s assembly file) and the ABI of your system.
The printf function will go looking for the argument on the stack, even if you don't supply one. Anything that's on there will be used, if it can't find an integer argument. Most times, you will get nonsensical data. The data chosen varies depending on the settings of your compiler. On some compilers, you may even get 16 as a result.
For example:
int printf(char*, int d){...}
This would be how printf works(not really, just an example). It doesn't return an error if d is null or empty, it just looks on the stack for the argument that's supposed to be there to display.
Printf is a variable argument function. Most compilers push arguments onto the stack and then call the function, but, depending on machine, operating system, calling convention, number of arguments, etc, there are also other values pushed onto the stack, which might be constant in your function.
Printf reads this area of memory and returns it.

C char array is not a string pattern?

I am having a compilation error on the following code:
printf((char *) buffer);
and the error message that I am getting is:
cc1: format not a string literal and no format arguments...
I suspect there are some libraries that I forgot to install, as I was able to compile and run the code without an error on the other machine...
PS: The question rises with the fact that I was able to run the same code on some other machine... I suspect a difference in gcc version might cause a problem like this?
Newer GCC versions try to parse the format string passed to printf and similar functions and determine if the argument list correctly matches the format string. It can't do this because you've passed it a buffer for the first argument, which would normally be a format string.
Your code is not incorrect C, it's just a poor usage of C. As others mentioned you should use "%s" as a format string to print a single string. This protects you from a class of errors that involve percentage signs in your string, if you don't control the input. It's a best practice to never pass anything but a string literal as the first argument to the printf or sprintf family of functions.
try
printf("%s", (char*) buffer);
;-)
This warning is generated by gcc if
-Wformat-nonliteral
is set. It's not part of -Wall or -Wextra (at least for version 4.4.0), so just drop it if you want the code to compile warning-free.
This is a warning for your safety, not an error. This new compiler is apparently being more strict about it. I don't think it's actually illegal in C, so the compiler should have an option to disable treating this as an error.
However, you pretty much never want to pass anything other than a string literal as the first argument to printf. The reason that doing so is such a horrible idea that the compiler has a special built-in check to warn you about it is this: Suppose the non-literal string that you pass as the first argument to printf happens to contain printf formatting characters. printf is then going to try to access second, third, fourth, etc, arguments that you didn't actually pass in, and may well crash your program by trying to do so. If the non-literal first argument is actually user-supplied, then the problem is even worse since a malicious user could crash your program at will.

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