difference in expected argument type and provided argument in printf

Question

I wanted to ask whether the following statement where printf() has been provided with a type int where it expected a type char will invoke UNDEFINED BEHAVIOUR. If not, what exactly is the step will be taken to make it compatible with expected type. Will a be shortened to char type??

   int a = 65;   
   printf("%c", a);

Answer 1

Variable numbers of arguments

SUMMARY: In order to access the arguments within the called function, the functions declared in the header file must be included. This introduces a new type, called a va_list, and three functions that operate on objects of this type, called va_start, va_arg, and va_end.

#include <stdarg.h>
type va_arg(va_list ap, type);

Each call to this macro will extract the next argument from the argument list as a value of the specified type. The va_list argument must be the one initialized by va_start. If the next argument is not of the specified type, the behaviour is undefined.

The behaviour is also undefined if va_arg is called when there were no further arguments.

Answer 2

Variable arguments are subject to the default argument promotion, eg char will be promoted to int, float to double (that's the reason why you only have a single format specifier to print both single and double precision floating point values).

So passing an int instead of a char is perfectly valid and even desirable, as character literals are of type int anyway. According to the C99 spec, section 7.19.6.1 §8, on seeing the conversion specifier %c, printf() expects an argument of type int and will then go on and cast this value to unsigned char.

This means the following is guaranteed to output a, as conversion of signed to unsigned types is well-defined:

int a = 'a' + UCHAR_MAX + 1;
printf("%c", a);