Float to int type casting?

[Enekajmer]

Hi,

1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);
5 printf("%d\n", *(int *)&a);
6 return 0;
7 }

The output is:

0
1099694080

I could not understand why the first printf prints 0 and the other
,1099694080...

I think there is sth different from simple type casting.Is it an
undefined behaviour?
Please let me understand.

Regards.

[Artie Gold]

Enekajmer wrote:

Hi,

1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);
Here you're trying to print a float value with a conversion specifier
for int. The results are undefined.
5 printf("%d\n", *(int *)&a);
Here, you're reinterpreting the representation of the float. The results
are implementation defined.
6 return 0;
7 }

The output is:

0
1099694080

I could not understand why the first printf prints 0 and the other
,1099694080...

I think there is sth different from simple type casting.Is it an
undefined behaviour?

[the word is `something']

See above.

HTH,
--ag
--
Artie Gold -- Austin, Texas
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
"If you have nothing to hide, you're not trying!"

[Enekajmer]

Artie Gold wrote:

Enekajmer wrote:

Hi,

1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);

Here you're trying to print a float value with a conversion specifier
for int. The results are undefined.

5 printf("%d\n", *(int *)&a);

Here, you're reinterpreting the representation of the float. The results
are implementation defined.

6 return 0;
7 }

The output is:

0
1099694080

I could not understand why the first printf prints 0 and the other
,1099694080...

I think there is sth different from simple type casting.Is it an
undefined behaviour?

[the word is `something']

See above.

HTH,
--ag
--
Artie Gold -- Austin, Texas
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
"If you have nothing to hide, you're not trying!"

Thanks i learnt the types of behaviours but i want to learn causes in
more details.To give an example, there is no problem for the snippet
code below:
....
float a=17.5;
int b=a; (type casting from float to int)
printf("b-->%d",b); // as normally, this prints "b-->17"
....

according to this example i guessed the output of the code below as 17
too before i compiled. As contrast its output is 0...

printf("a-->%d",a) //a is in float type
So, what are the causes of this case?The IEEE-754?

Regards.

[Robin Haigh]

"Artie Gold" <ar*******@aust in.rr.com> wrote in message
news:44******** ****@individual .net...

Enekajmer wrote:

Hi,

1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);

Here you're trying to print a float value with a conversion specifier
for int. The results are undefined.

Actually, you're trying to print a double value, because the float is
promoted when it's not governed by a formal parameter in a prototype.

If it weren't for that, you'd probably get the same output from both
printfs.

--
RSH

[Vladimir S. Oka]

Enekajmer wrote:

Thanks i learnt the types of behaviours but i want to learn causes in
more details.To give an example, there is no problem for the snippet
code below:
...
float a=17.5;
int b=a; (type casting from float to int)
This is, as you rightly say, a cast. It /converts/ a float value in `a`
to it's closest representation as int. Hence, (int)17.5 == 17.
printf("b-->%d",b); // as normally, this prints "b-->17"
...

according to this example i guessed the output of the code below as 17
too before i compiled. As contrast its output is 0...

printf("a-->%d",a) //a is in float type
Artie led you off the track here by saying "conversion specifier". A s a
matter of fact "%d" is a "format specifier". It tells printf()
to /interpret/ whatever is stored in a corresponding parameter (in your
case `a`) as an int. This means that printf() blindly takes whatever is
the bit representation of `a` and interprets these bits as if they were
an `int`. Them actually being an `float` the result is undefined. In
your case, it happened to be 0, but the effect could have been much
worse. Beware the Undefined behaviour monster -- it may make demons
come out of your nose! ;-)
So, what are the causes of this case?The IEEE-754?

No, see paragraph above...

Cheers

Vladimir

--
To invent, you need a good imagination and a pile of junk.
-- Thomas Edison

[Jordan Abel]

On 2006-02-05, Enekajmer <en*******@gmai l.com> wrote:

Hi,

1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);
5 printf("%d\n", *(int *)&a);
6 return 0;
7 }

The output is:

0
1099694080

I could not understand why the first printf prints 0 and the other
,1099694080...

I think there is sth different from simple type casting.Is it an
undefined behaviour?
Yes. Both printf lines invoke undefined behavior.
Please let me understand.

hint: a is promoted to a double when passed to printf as a float.

[Artie Gold]

Vladimir S. Oka wrote:

Enekajmer wrote:

Thanks i learnt the types of behaviours but i want to learn causes in
more details.To give an example, there is no problem for the snippet
code below:
...
float a=17.5;
int b=a; (type casting from float to int)

This is, as you rightly say, a cast. It /converts/ a float value in `a`
to it's closest representation as int. Hence, (int)17.5 == 17.

printf("b-->%d",b); // as normally, this prints "b-->17"
...

according to this example i guessed the output of the code below as 17
too before i compiled. As contrast its output is 0...

printf("a-->%d",a) //a is in float type

Artie led you off the track here by saying "conversion specifier". A s a
matter of fact "%d" is a "format specifier". It tells printf()

A slip of the fingers. Thanks for the correction.
--ag to /interpret/ whatever is stored in a corresponding parameter (in your
case `a`) as an int. This means that printf() blindly takes whatever is
the bit representation of `a` and interprets these bits as if they were
an `int`. Them actually being an `float` the result is undefined. In
your case, it happened to be 0, but the effect could have been much
worse. Beware the Undefined behaviour monster -- it may make demons
come out of your nose! ;-)

So, what are the causes of this case?The IEEE-754?

No, see paragraph above...

Cheers

Vladimir

--
Artie Gold -- Austin, Texas
http://goldsays.blogspot.com (new post 8/5)
http://www.cafepress.com/goldsays
"If you have nothing to hide, you're not trying!"

[CBFalconer]

Robin Haigh wrote:

"Artie Gold" <ar*******@aust in.rr.com> wrote in message

Enekajmer wrote:

1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);

Here you're trying to print a float value with a conversion specifier
for int. The results are undefined.

Actually, you're trying to print a double value, because the float is
promoted when it's not governed by a formal parameter in a prototype.

If it weren't for that, you'd probably get the same output from both
printfs.

It's even worse than that - he is using a variadic function without
a prototype, which makes everything undefined behaviour.

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson
More details at: <http://cfaj.freeshell. org/google/>
Also see <http://www.safalra.com/special/googlegroupsrep ly/>

[Flash Gordon]

Vladimir S. Oka wrote:

Enekajmer wrote:

Thanks i learnt the types of behaviours but i want to learn causes in
more details.To give an example, there is no problem for the snippet
code below:
...
float a=17.5;
int b=a; (type casting from float to int)

This is, as you rightly say, a cast. It /converts/ a float value in `a`
to it's closest representation as int. Hence, (int)17.5 == 17.

No, it's not a cast. It is, on the other hand, a conversion.

printf("b-->%d",b); // as normally, this prints "b-->17"
...

according to this example i guessed the output of the code below as 17
too before i compiled. As contrast its output is 0...

printf("a-->%d",a) //a is in float type

Artie led you off the track here by saying "conversion specifier". A s a
matter of fact "%d" is a "format specifier". It tells printf()
to /interpret/ whatever is stored in a corresponding parameter (in your
case `a`) as an int. This means that printf() blindly takes whatever is
the bit representation of `a` and interprets these bits as if they were
an `int`. Them actually being an `float` the result is undefined. In
your case, it happened to be 0, but the effect could have been much
worse. Beware the Undefined behaviour monster -- it may make demons
come out of your nose! ;-)

Actually, it does not take the bit pattern and reinterpret it. It
assumes that the parameter is an int and tries to read that and this is
undefined behaviour (i.e. anything can happen) because the standard says
it is.

There are calling conventions where even varargs parameters are passed
in appropriate registers, so in this case it could read whatever the
first integer register happens to contain which could have absolutely
nothing to do with the contents of a. The specific implementation I'm
thinking of (a real one) can actually cause even more fun. If the
parameter is a double it gets passed in a floating point register, but
also converted to an int and passed in an integer register, so you would
actually get the result the OP expected. However, when you pass an int
that is *only* passed in an integer register, so if you try to print an
int using the format specifier for a double, it will try to print
whatever happened to be in the floating point register.

The implementation happens to be the latest version of Visual Studio
when given appropriate options.

The moral of this is *never* try to say what C implementations do when
you invoke undefined behaviour, for it's ways are stranger than your
imagination can conceive.

It is reasonable to wonder why something is undefined, and sometimes
useful to know how undefined behaviour can manifest, but only if you
accept that literally *anything* can cause a different behaviour,
including it deciding to show a picture of your boss in the nude.

So, what are the causes of this case?The IEEE-754?

No, see paragraph above...

Indeed.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.

[Jack Klein]

On Sun, 05 Feb 2006 11:16:49 -0600, Artie Gold
<ar*******@aust in.rr.com> wrote in comp.lang.c:

Enekajmer wrote:

Hi,

1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);

Here you're trying to print a float value with a conversion specifier
for int. The results are undefined.

5 printf("%d\n", *(int *)&a);

Here, you're reinterpreting the representation of the float. The results
are implementation defined.

[snip]

No, undefined. There are only two defined ways to look at the bits of
an object of type float, those are using an lvalue of type float, or
using an array of unsigned characters the size of a float.

--
Jack Klein
Home: http://JK-Technology.Com
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