printf argument

[john]

Hello,

Supposed sizeof(short) == 2 and sizeof(int) == 4 and argument passing in
a stack.

When doing:

int i1; int i2;
printf("%d %d", i1, i2); // OK. 2 arg of size 4

int i; short s;
printf("%d %d", i, s); // 1 arg of size 2 and one of 4
This works but why ?

How printf can correctly unwind the arguments from the stack ? That is,
how can it knows it has to get 4 bytes for the first arg and 2 for the
second.

I thought it was based on the format but both are '%d' in this case ?

thanks,
john

[Keith Thompson]

john <jo*******@nospam.teleme.com> writes:

Supposed sizeof(short) == 2 and sizeof(int) == 4 and argument passing
in a stack.

When doing:

int i1; int i2;
printf("%d %d", i1, i2); // OK. 2 arg of size 4

int i; short s;
printf("%d %d", i, s); // 1 arg of size 2 and one of 4
This works but why ?

How printf can correctly unwind the arguments from the stack ? That
is, how can it knows it has to get 4 bytes for the first arg and 2 for
the second.

I thought it was based on the format but both are '%d' in this case ?

Both arguments are passed as int (which is why printf doesn't have a
format for short). Arguments to printf (after the format string)
undergo the "default argument promotions"; types shorter than int are
promoted to int or unsigned int, and float is promoted to double.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Richard Bos]

john <jo*******@nospam.teleme.com> wrote:

Supposed sizeof(short) == 2 and sizeof(int) == 4 and argument passing in
a stack.

When doing:

int i1; int i2;
printf("%d %d", i1, i2); // OK. 2 arg of size 4
Yes (assuming your suppositions), but...
int i; short s;
printf("%d %d", i, s); // 1 arg of size 2 and one of 4
....no.
This works but why ?

How printf can correctly unwind the arguments from the stack ? That is,
how can it knows it has to get 4 bytes for the first arg and 2 for the
second.

It doesn't. For variable argument lists, any of the variable arguments
which is narrower than an int, for integer types, or a double, for FP
types, gets widened to int (unsigned where appropriate) resp. double
before being passed to the function. printf() gets an int.
This rule also applies to functions without a prototype in place (i.e.,
with an old-style declaration, or with no declaration), but in genereal
you shouldn't write any functions without a prototype, these days.

Richard