Usefulness of %hd in printf

[Old Wolf]

Is there any possible situation for printf where %hd causes a different
result to %d, and the corresponding argument was of type 'short int' ?

[user923005]

Old Wolf wrote:

Is there any possible situation for printf where %hd causes a different
result to %d, and the corresponding argument was of type 'short int' ?

Think about the scanf() family of functions.

Sure, the short will promote just like floats promote to double for the
same reason. But addresses do not promote. Hence, the need for a
width specifier for 'smaller than default promotion' types.

[Old Wolf]

user923005 wrote:

Old Wolf wrote:

Is there any possible situation for printf where %hd causes a different
result to %d, and the corresponding argument was of type 'short int' ?

Think about the scanf() family of functions.

I'm asking about the printf() family.

Are you saying that there is no difference in printf, but they added
support for %hd just to try and keep the printf family similar to the
scanf family?

[user923005]

Old Wolf wrote:

user923005 wrote:

Old Wolf wrote:

Is there any possible situation for printf where %hd causes a different
result to %d, and the corresponding argument was of type 'short int' ?

Think about the scanf() family of functions.

I'm asking about the printf() family.

Are you saying that there is no difference in printf, but they added
support for %hd just to try and keep the printf family similar to the
scanf family?

No. I am saying scanf() has to have it, while printf() could care
less.

We could just as easily complain about:

12.9: Someone told me it was wrong to use %lf with printf(). How can
printf() use %f for type double, if scanf() requires %lf?

A: It's true that printf's %f specifier works with both float and
double arguments. Due to the "default argument promotions"
(which apply in variable-length argument lists such as
printf's, whether or not prototypes are in scope), values of
type float are promoted to double, and printf() therefore sees
only doubles. (printf() does accept %Lf, for long double.)
See also questions 12.13 and 15.2.

References: K&R1 Sec. 7.3 pp. 145-47, Sec. 7.4 pp. 147-50; K&R2
Sec. 7.2 pp. 153-44, Sec. 7.4 pp. 157-59; ISO Sec. 7.9.6.1,
Sec. 7.9.6.2; H&S Sec. 15.8 pp. 357-64, Sec. 15.11 pp. 366-78;
CT&P Sec. A.1 pp. 121-33.

[user923005]

>From the C99 standard, here is the bit on default promotions for things
lacking prototypes and for variadic functions...

>From ©ISO/IEC ISO/IEC 9899:1999 (E), 6.5.2.2 Function calls:

"6 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. If the number
of arguments does not equal the number of parameters, the behavior is
undefined. If the function is defined with a type that includes a
prototype, and either the prototype ends with an ellipsis (, ...) or
the types of the arguments after promotion are not compatible with the
types of the parameters, the behavior is undefined. If the function is
defined with a type that does not include a prototype, and the types of
the arguments after promotion are not compatible with those of the
parameters after promotion, the behavior is undefined, except for the
following cases:
* one promoted type is a signed integer type, the other promoted type
is the corresponding unsigned integer type, and the value is
representable in both types;
* both types are pointers to qualified or unqualified versions of a
character type or void."

[Old Wolf]

user923005 wrote:

Old Wolf wrote:

user923005 wrote:

Old Wolf wrote:
Is there any possible situation for printf where %hd causes a different
result to %d, and the corresponding argument was of type 'short int' ?
>
Think about the scanf() family of functions.

I'm asking about the printf() family.

Are you saying that there is no difference in printf, but they added
support for %hd just to try and keep the printf family similar to the
scanf family?

No. I am saying scanf() has to have it, while printf() could care
less.

So you are saying it does make a difference to printf?
What difference?

Or do you mean "couldn't care less" ?

We could just as easily complain about:

12.9: Someone told me it was wrong to use %lf with printf(). How can
printf() use %f for type double, if scanf() requires %lf?

Huh? I'm not complaining. I'm asking about printf (NOT scanf). You
brought scanf into this.

A: It's true that printf's %f specifier works with both float and
double arguments. Due to the "default argument promotions"
(which apply in variable-length argument lists such as
printf's, whether or not prototypes are in scope), values of
type float are promoted to double, and printf() therefore sees
only doubles. (printf() does accept %Lf, for long double.)
See also questions 12.13 and 15.2.

This is irrelevant to my question. (If you don't know the answer then
just say so, or don't reply)

[P.J. Plauger]

"Old Wolf" <ol*****@inspir e.net.nzwrote in message
news:11******** **************@ d71g2000cwa.goo glegroups.com.. .

user923005 wrote:

>Old Wolf wrote:

user923005 wrote:
Old Wolf wrote:
Is there any possible situation for printf where %hd causes a
different
result to %d, and the corresponding argument was of type 'short
int' ?

Think about the scanf() family of functions.

I'm asking about the printf() family.

Are you saying that there is no difference in printf, but they added
support for %hd just to try and keep the printf family similar to the
scanf family?

No. I am saying scanf() has to have it, while printf() could care
less.

So you are saying it does make a difference to printf?
What difference?

Or do you mean "couldn't care less" ?

>We could just as easily complain about:

12.9: Someone told me it was wrong to use %lf with printf(). How can
printf() use %f for type double, if scanf() requires %lf?

Huh? I'm not complaining. I'm asking about printf (NOT scanf). You
brought scanf into this.

>A: It's true that printf's %f specifier works with both float and
double arguments. Due to the "default argument promotions"
(which apply in variable-length argument lists such as
printf's, whether or not prototypes are in scope), values of
type float are promoted to double, and printf() therefore sees
only doubles. (printf() does accept %Lf, for long double.)
See also questions 12.13 and 15.2.

This is irrelevant to my question. (If you don't know the answer then
just say so, or don't reply)

%hd prints out the int argument cast to short. So it scrapes off any bits
that can't be represented in a short.

P.J. Plauger
Dinkumware, Ltd.
http://www.dinkumware.com

[Yevgen Muntyan]

P.J. Plauger wrote:

"Old Wolf" <ol*****@inspir e.net.nzwrote in message
news:11******** **************@ d71g2000cwa.goo glegroups.com.. .

>user923005 wrote:

>>Old Wolf wrote:
user923005 wrote:
Old Wolf wrote:
>Is there any possible situation for printf where %hd causes a
>differen t
>result to %d, and the corresponding argument was of type 'short
>int' ?
Think about the scanf() family of functions.
I'm asking about the printf() family.

Are you saying that there is no difference in printf, but they added
support for %hd just to try and keep the printf family similar to the
scanf family?
No. I am saying scanf() has to have it, while printf() could care
less.

So you are saying it does make a difference to printf?
What difference?

Or do you mean "couldn't care less" ?

>>We could just as easily complain about:

12.9: Someone told me it was wrong to use %lf with printf(). How can
printf() use %f for type double, if scanf() requires %lf?

Huh? I'm not complaining. I'm asking about printf (NOT scanf). You
brought scanf into this.

>>A: It's true that printf's %f specifier works with both float and
double arguments. Due to the "default argument promotions"
(which apply in variable-length argument lists such as
printf's, whether or not prototypes are in scope), values of
type float are promoted to double, and printf() therefore sees
only doubles. (printf() does accept %Lf, for long double.)
See also questions 12.13 and 15.2.

This is irrelevant to my question. (If you don't know the answer then
just say so, or don't reply)

%hd prints out the int argument cast to short. So it scrapes off any bits
that can't be represented in a short.

So the following is a bug?

----------------------------
#include <stdio.h>

int main (void)
{
int a = 1<<30;
short b = a;
printf ("%d %hd %hd\n", a, a, b);
return 0;
}
----------------------------

It prints

1073741824 1073741824 0

i.e. it doesn't truncate 'a' (which of type int) when prints
it with %hd.

Thanks,
Yevgen

[Old Wolf]

Yevgen Muntyan wrote:

So the following is a bug?

----------------------------
#include <stdio.h>

int main (void)
{
int a = 1<<30;
short b = a;
printf ("%d %hd %hd\n", a, a, b);
return 0;
}
----------------------------

It prints

1073741824 1073741824 0

i.e. it doesn't truncate 'a' (which of type int) when prints
it with %hd.

You caused undefined behaviour by not passing an argument of
type 'short int' to %hd

[Old Wolf]

P.J. Plauger wrote:

Old Wolf wrote:
Is there any possible situation for printf where %hd causes a
different result to %d, and the corresponding argument was
of type 'short int' ?

%hd prints out the int argument cast to short. So it scrapes off any bits
that can't be represented in a short.

Well, when the short int is passed as argument, it is promoted to int
(according to the default argument promotions). So how can the int
have any bits that can't be represented in a short?

#include <stdio.h>
int main()
{
short x;
/* your code here */
printf("%hd\n%d \n", x, x);
return 0;
}

Is there any possible code you can insert where marked, that will
cause two different values to be printed (and the program to not
have undefined behaviour) ?