printf("....%d",sizeof((int)(double)(char) i))

[aarklon@gmail.com]

Hi all,

why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
size of int ???

is it because sizeof doesn't evaluate its operand....???

[Joachim Schmitz]

Re: printf(".. ..%d",size of((int)(double )(char) i))

aarklon@gmail.c om wrote:

Hi all,
>
why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
size of int ???
>
is it because sizeof doesn't evaluate its operand....???

It is because your code is just the longer version of
printf("....%d" ,sizeof(int))

Bye, Jojo

[Richard Heathfield]

Re: printf(".. ..%d",size of((int)(double )(char) i))

aarklon@gmail.c om said:

Hi all,
>
why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
size of int ???

What did you want it to give the size of?

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Ivar]

Re: printf(&quot;.. ..%d&quot;,size of((int)(double )(char) i))

On Apr 15, 1:47 pm, "Joachim Schmitz" <nospam.j...@sc hmitz-digital.de>
wrote:

aark...@gmail.c om wrote:

Hi all,

>

why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
size of int ???

>

is it because sizeof doesn't evaluate its operand....???

>
It is because your code is just the longer version of
printf("....%d" ,sizeof(int))
>
Bye, Jojo

By doing printf("....%d" ,sizeof((int)(d ouble)(char) i)), you're
finally casting i to an int, so it would print the sizeof(int) value.
Correct me if I'm wrong

[Peter Nilsson]

Re: printf(&quot;.. ..%d&quot;,size of((int)(double )(char) i))

aark...@gmail.c om wrote:

why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives
the size of int ???
>
is it because sizeof doesn't evaluate its operand....???

With the exception of variable length arrays in C99, yes.

The object i doesn't even have to be initialised. Even the
following is valid...

sizeof( (int) (int (*)(void)) -1 )

....even though there is no well defined conversion involved
at any stage.

--
Peter

[Joachim Schmitz]

Re: printf(&quot;.. ..%d&quot;,size of((int)(double )(char) i))

Ivar wrote:

On Apr 15, 1:47 pm, "Joachim Schmitz" <nospam.j...@sc hmitz-digital.de>
wrote:

>aark...@gmail. com wrote:

>>Hi all,

>>

>>why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
>>size of int ???

>>

>>is it because sizeof doesn't evaluate its operand....???

>>
>It is because your code is just the longer version of
>printf("....%d ",sizeof(in t))
>>
>Bye, Jojo

>
By doing printf("....%d" ,sizeof((int)(d ouble)(char) i)), you're
finally casting i to an int, so it would print the sizeof(int) value.

Exactly, and that's what I said...

Correct me if I'm wrong

No need 8-)

Bye, Jojo