printf format specifiers

printf format specifiers

[Steve Dondley]

Hi,

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?

Thanks.

Re: printf format specifiers

[Der Herr Hofrat]

> Hi,
> 
> 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.

I guess you would see a difference if you would use ascii chars above 127 as 
the one would change sign the other would not - and hu would read a short
which is 2bytes instead of 4 but I don_t see why %hu should read one byte 
only.

hofrat

Re: printf format specifiers

[John Love-Jensen]

Hi Steve,

Let's assume that char is 8-bit, short is 16-bit, int is 32-bit, long is
64-bit, and long long is 128-bit.

When you put a char or short on the stack, they are expanded to int.

The int represents the platform's natural, native size... sometimes called a
"word".  Note:  "word" is platform dependent!

So if you had these:
char c = '1'; // 0x31, 8-bit, a quarter-word
short s = 0x1122; // 16-bit, a half-word
int i = 0x33445566; // 32-bit, a word
long d = 0x9988776655443322; // 64-bit, a double-word
long long q = 0xFFEEDDCCBBAA99887766554433221100; // 128-bit, a quad-word

If you do this:
printf("%d %d %d %ld %lld", c, s, i, d, q);

The stack will look like this:
02100010 : pointer to "%d %d %d %ld %lld"
00000031 : c, expanded to word-size
00001122 : s, expanded to word-size
33445566 : i
99887766 : d's MSW
55443322 : d's LSW
FFEEDDCC : q's MSW
BBAA9988 : q's 2SW
77665544 : q's 3SW
33221100 : q's LSW

The fictional platform presented is a big-endian platform.  Ala Motorola
680x0.

--Eljay

Re: printf format specifiers

[Segher Boessenkool]

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

Re: printf format specifiers

[Roberto Díaz]

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There was a list called tuxcprograming or something like that for C
programming doubts.. I am mention this because I see there are at times
some C programming doubts answered in this list which is supposed to be
a gcc list.. 

Anyway I dont care, it doesnt bother me this kind of messages.. but
maybe other do.


> 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
> 
> 
> 
-- 

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