Re: printf format specifiers

[Segher Boessenkool <segher at koffie dot nl>]

Hi Steve,

> I have the following line in one of my programs:
> 
> printf("%d\n", *string);
> 
> *string is a pointer to a string.  The above line prints out the ASCII
> decimal equivalent of the character that the pointer is pointing to.  This
> is what I was looking to accomplish.
> 
> My question is why?  Why wouldn't I need to use the %hu (unsigned short
> integer) format specifier?  When I do use the %hu, I get precisely the same
> results.  This despite the fact that %d reads an entire word and %hu reads a
> single byte.
> 
> Is this some compiler magic going on here?

printf() is declared something like

	int printf(const char *format, ...);

The ellipsis (dots) here means the function takes "variable arguments".
All arguments passed "through the ellipsis" are subject to the default integer
conversions:

	- if all values representable by the type of the argument are representable
	  by the type "int", the argument is converted to "int";
	- otherwise, if all values representable by the type of the argument are
	  representable by the type "unsigned int", the argument is converted to
	  "unsigned int";
	- otherwise, the argument is passed as-is.

In your case, "string" is a pointer to char, so *string is a char, which matches
the first case above, so it is passed as int, so the format specifier "%d" works
just fine.  If you use "%hu", printf() still reads an int, but converts it to
an unsigned short.  Now either you didn't test this with negative char values,
or your char happens to be an unsigned type; otherwise, "%hu" wouldn't have
given the same output as "%d".

Btw, if you wanted to print it as a char value, instead of a short value, you
should have used "%hhd" or "%hhu".

Hope this clarifies things for you,


Segher