Having troubles formatting scanf while keeping my code understandable - c

Question

I need a function to read a file name, with a max length of MAX_FILE_NAME_SIZE, which is a symbolic constant, I did this the following way:
char * readFileName()
{
char format[6];
char * fileName = malloc(MAX_FILE_NAME_SIZE * sizeof(fileName[0]));
if(fileName== NULL)
return NULL;
sprintf(format, "%%%ds", MAX_FILE_NAME_SIZE-1);
scanf(format, fileName);
fileName= realloc(fileName, strlen(fileName)*sizeof(fileName[0]));
return fileName;
}
I'd really like to get read of the sprintf part (and also the format vector), what's the cleanest and most efficient way to do this?

Answer 1

Solution
You can make a little Preprocessor hack:
#define MAX_BUFFER 30
#define FORMAT(s) "%" #s "s"
#define FMT(s) FORMAT(s)
int main(void)
{
char buffer[MAX_BUFFER + 1];
scanf(FMT(MAX_BUFFER), buffer);
printf("string: %s\n", buffer);
printf("length: %d\n", strlen(buffer));
return 0;
}
The FORMAT and FMT macros are necessary for the preprocessor to translate them correctly. If you call FORMAT directly with FORMAT(MAX_BUFFER), it will translate into "%" "MAX_BUFFER" "s" which is no good.
You can verify that using gcc -E scanf.c. However, if you call it through another macro, which will effectively resolve the macro names for you and translate to "%" "30" "s", which is a fine format string for scanf.
Edit
As correctly pointed out by #Jonathan Leffler in the comments, you can't do any math on that macro, so you need to declare buffer with plus 1 character for the NULL terminating byte, since the macro expands to %30s, which will read 30 characters plus the null byte.
So the correct buffer declaration should be char buffer[MAX_BUFFER + 1];.
Requested Explanation
As asked in the comments, the one macro version won't work because the preprocessor operator # turns an argument into a string (stringification, see bellow). So, when you call it with FORMAT(MAX_BUFFER), it just stringifies MAX_BUFFER instead of macro-expanding it, giving you the result: "%" "MAX_BUFFER" "s".
Section 3.4 Stringification of the C Preprocessor Manual says this:
Sometimes you may want to convert a macro argument into a string constant. Parameters are not replaced inside string constants, but you can use the ‘#’ preprocessing operator instead. When a macro parameter is used with a leading ‘#’, the preprocessor replaces it with the literal text of the actual argument, converted to a string constant. Unlike normal parameter replacement, the argument is not macro-expanded first. This is called stringification.
This is the output of the gcc -E scanf.c command on a file with the one macro version (the last part of it):
int main(void)
{
char buffer[30 + 1];
scanf("%" "MAX_BUFFER" "s", buffer);
printf("string: %s\n", buffer);
printf("length: %d\n", strlen(buffer));
return 0;
}
As expected. Now, for the two levels, I couldn't explain better than the documentation itself, and in the last part of it there's an actual example of this specific case (two macros):
If you want to stringify the result of expansion of a macro argument, you have to use two levels of macros.
#define xstr(s) str(s)
#define str(s) #s
#define foo 4
str (foo)
==> "foo"
xstr (foo)
==> xstr (4)
==> str (4)
==> "4"
s is stringified when it is used in str, so it is not macro-expanded first. But s is an ordinary argument to xstr, so it is completely macro-expanded before xstr itself is expanded (see Argument Prescan). Therefore, by the time str gets to its argument, it has already been macro-expanded.
Resource
The C Preprocessor