tryick use of sizeof again[Explaination Wanted]

[goacross]

char ch='a';

int v=sizeof ++ch;

cout<<ch<<endl;// output: 'a'

why not 'b'?
thanks

[Lew Pitcher]

On Sep 14, 11:16 am, goacr...@gmail. com wrote:

char ch='a';

int v=sizeof ++ch;

cout<<ch<<endl;// output: 'a'

why not 'b'

Who knows the vagarities of C++? Why not ask in comp.lang.c++ instead
of here? We don't deal with C++ here, only C

[jacob navia]

go******@gmail. com wrote:

char ch='a';

int v=sizeof ++ch;

cout<<ch<<endl;// output: 'a'

why not 'b'?
thanks

sizeof expressions are NOT evaluated. Only the type is important
to get the size of it.

[Charlie Gordon]

"Lew Pitcher" <lp******@teksa vvy.coma écrit dans le message de news:
11************* *********@w3g20 00...legro ups.com...

On Sep 14, 11:16 am, goacr...@gmail. com wrote:

>char ch='a';

int v=sizeof ++ch;

cout<<ch<<endl ;// output: 'a'

why not 'b'

Who knows the vagarities of C++? Why not ask in comp.lang.c++ instead
of here? We don't deal with C++ here, only C

Why be unnecessarily obtuse when the issue is not C++ specific.

sizeof expr computes the size in bytes of the object denoted by expr without
producing code for evaluating that expression. Therefore ch is not
incremented.

--
Chqrlie.

[Punkie]

jacob navia wrote:

go******@gmail. com wrote:

>char ch='a';

int v=sizeof ++ch;

cout<<ch<<endl ;// output: 'a'

why not 'b'?
thanks

sizeof expressions are NOT evaluated. Only the type is important
to get the size of it.

indeed. Consider sizeof to be

size_t sizeof(type)
and not
size_t sizeof(int)

[christian.bau]

On Sep 14, 6:19 pm, "Charlie Gordon" <n...@chqrlie.o rgwrote:

Why be unnecessarily obtuse when the issue is not C++ specific.

How do you know it is not C++ specific? For example, after

char c - '0';
c += sizeof ('0');

what is the value stored in c? There are subtle differences between
the sizeof operator in C and C++, and someone knowing absolutely
everything about C but nothing about C++ would most likely not know
the right answer to the problem I wrote.

[Keith Thompson]

go******@gmail. com writes:

char ch='a';

int v=sizeof ++ch;

cout<<ch<<endl;// output: 'a'

why not 'b'?

If you're going to post code here, please make sure it's C, not C++.
It's also helpful to post a (small) complete program rather than an
isolated snippet.

'cout', '<<', and 'endl' are specific to C++.

Your underlying question happens, in this particular case, to be
equally valid in both languages, but there are subtle differences
between C and C++ that even the C experts here may not be aware of.

Just this once, I'll show you a C version of your program.

#include <stdio.h>
int main(void)
{
char ch = 'a';
int v = sizeof ++ch;
printf("%c\n", ch);
return 0;
}

Next time, please either post C in the first place, or post to
comp.lang.c++. If you're having trouble knowing the difference, you
need to learn one language or the other (perhaps eventually both).
A newsgroup is not a good place to do that, but
alt.comp.lang.l earn.c-c++ might be more appropriate for your purposes
than comp.lang.c.

Several people have already given you the answer to your question --
which is unfortunate, since the question looks very much like
homework. An expression like 'sizeof ++ch' isn't something that you'd
write if your goal is to accomplish something. It was almost
certainly written by someone trying to make a specific point.

If you want us to do your homework for you, please post your
instructor's e-mail address so we can submit our solutions directly.
We're willing to provide hints if you've done some of the work
yourself and you're stuck on something, but we won't (usually) do it
for you.

--
Keith Thompson (The_Other_Keit h) 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."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[jacob navia]

christian.bau wrote:

On Sep 14, 6:19 pm, "Charlie Gordon" <n...@chqrlie.o rgwrote:

>Why be unnecessarily obtuse when the issue is not C++ specific.

How do you know it is not C++ specific? For example, after

char c - '0';
c += sizeof ('0');

what is the value stored in c? There are subtle differences between
the sizeof operator in C and C++, and someone knowing absolutely
everything about C but nothing about C++ would most likely not know
the right answer to the problem I wrote.

This is not the case for the problem at hand. Both C and C++
do not evaluate the expression of a sizeof.

[Martin Wells]

sizeof is a handy way of ensuring that an expression is legal without
actually evaluating the expression. For instance, you could have code
something like the following:

typedef ... SomeType;

void Func(SomeType const *p)
{
sizeof (p->val); /* Ensure that SomeType has a member called
val */
}

Martin

[Martin Wells]

There are subtle differences between

the sizeof operator in C and C++, and someone knowing absolutely
everything about C but nothing about C++ would most likely not know
the right answer to the problem I wrote.

There's no subtle differences between C and C++ in this respect. Or
even differences for that matter.

Martin