How to correctly match the format string in the corresponding argument
The C standard library function
printf() takes a format string and subsequent arguments of various types. On an architecture where arguments are passed on the stack, the format string tells the
printf() function what types it should interpret the blob of arguments with. On an architecture where
printf()‘s arguments are passed by (possibly specialized) registers, the format string tells
printf() in what registers it should look for them.
For this reason, the arguments of
printf() after the first one should match the indications of the format string. The C11 standard leaves no room for deviation:
184.108.40.206:9 […] If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.
A common C pattern used for printing bytes is the following:
unsigned char u = …;
%x is well-known for expecting an
unsigned int. The problem in the above code snippet is that
printf() is not passed an
unsigned int. “Default arguments promotions” are applied to
printf()‘s arguments after the format string. On most architectures, an
unsigned char is promoted to
int can represent all the values contained in
This mistake is quite harmless, because the types
unsigned int are guaranteed to have the same representation, and it seems unlikely that an ABI would pass
unsigned int in different registers. However, if you like writing for the C standard instead of for the couple of platforms that happen to be popular at the moment, you had better use the line below instead:
printf("%02x", (unsigned int) u);
This time, the type of the second argument clearly matches the format string in the first argument.