Question

Is the following code correct?

char mychar = 200;
printf("%x", mychar);

According to http://www.cplusplus.com/reference/clibrary/cstdio/printf/ %x expects an integer (4 bytes with my compiler) and I only pass 1 byte here. Since printf makes use of varargs, my fear is that this only works because of byte alignment on the stack (i.e. a char always uses 4 bytes when pushed on the stack).

I think it would be better to write:

char mychar = 200;
printf("%x", static_cast<int>(mychar));

Do you think the first code is safe anyway? If not, do you think I could get a different output if I switch to a bigendian architecture?

Answer 1

In your example, the argument is actually of type int. mychar is promoted to int due to the default argument promotions.

(C99, 6.5.2.2p6) “If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument promotions.“

and (emphasis mine):

(C99, 6.5.2.2p7) “If the expression that denotes the called function has a type that does include a prototype, the arguments are implicitly converted, as if by assignment, to the types of the corresponding parameters, taking the type of each parameter to be the unqualified version of its declared type. The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument
promotions are performed on trailing arguments.”

Note that technically the x conversion specifier requires an unsigned int argument but int and unsigned int types are guaranteed to have the same representation.