sizeof operator

[JS]

I read in K&R page 204 that sizeof use on a char returns 1. But when I write
the following I get 4!

int main(void){
printf("%d\n",s izeof('g'));

}

Was it not supposed to return 1?

[Ben Pfaff]

"JS" <[e-mail] m> writes:

I read in K&R page 204 that sizeof use on a char returns 1. But when I write
the following I get 4!

printf("%d\n",s izeof('g'));

In C, character constants have type `int', so sizeof 'g' is equal
to sizeof(int).
--
Ben Pfaff
email: bl*@cs.stanford .edu
web: http://benpfaff.org

[JS]

"Ben Pfaff" <bl*@cs.stanfor d.edu> skrev i en meddelelse
news:87******** ****@benpfaff.o rg...

"JS" <[e-mail] m> writes:

I read in K&R page 204 that sizeof use on a char returns 1. But when I write the following I get 4!

printf("%d\n",s izeof('g'));

In C, character constants have type `int', so sizeof 'g' is equal
to sizeof(int).

Why do they write char if they mean int?

[Alex Fraser]

"JS" <[e-mail] m> wrote in message news:d1******** **@news.net.uni-c.dk...

I read in K&R page 204 that sizeof use on a char returns 1. But when I
write the following I get 4!

int main(void){
printf("%d\n",s izeof('g'));

}

Was it not supposed to return 1?

No, a character in single quotes, as in 'g', is a constant of type signed
int in C (I think this is different in C++).

Try:

int main(void) {
char c;
printf("%d\n", (int)sizeof c);
return 0;
}

Alex

[Joona I Palaste]

JS <[e-mail] m> scribbled the following:

"Ben Pfaff" <bl*@cs.stanfor d.edu> skrev i en meddelelse
news:87******** ****@benpfaff.o rg...

"JS" <[e-mail] m> writes:

> I read in K&R page 204 that sizeof use on a char returns 1. But when I write > the following I get 4!
>
> printf("%d\n",s izeof('g'));
In C, character constants have type `int', so sizeof 'g' is equal
to sizeof(int).

Why do they write char if they mean int?

They don't. sizeof(char) is 1, but sizeof('g') is sizeof(int), which is
4 on your platform. 'g' is not a char. It's an int. Like Ben Pfaff
said, character constants are ints, not chars.

--
/-- Joona Palaste (pa*****@cc.hel sinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"The day Microsoft makes something that doesn't suck is probably the day they
start making vacuum cleaners."
- Ernst Jan Plugge

[Martin Ambuhl]

JS wrote:

I read in K&R page 204 that sizeof use on a char returns 1. But when I write
the following I get 4!

int main(void){
printf("%d\n",s izeof('g'));

}

Was it not supposed to return 1?

You have stumbled upon a difference between C and C++.
In C, literal 'g' is an int, while in C++ literal 'g' is a char.
Here are the C and C++ forms of the code, with the output of each.

#include <stdio.h>
#define COMPILER "<your compiler here> (C)"

int main(void)
{
printf(COMPILER " %d\n", (int) sizeof 'g');
return 0;
}

#include <cstdio>
using namespace std;
#define COMPILER "<your compiler here> (C++)"

int main(void)
{
printf(COMPILER " %d\n", (int) sizeof 'g');
return 0;
}

[outputs]
gcc (C) 4
bcc (C) 4
gcc (C++) 1
bcc (C++) 1

[Ben Pfaff]

"JS" <[e-mail] m> writes:

"Ben Pfaff" <bl*@cs.stanfor d.edu> skrev i en meddelelse
news:87******** ****@benpfaff.o rg...

"JS" <[e-mail] m> writes:

> I read in K&R page 204 that sizeof use on a char returns 1. But when I write > the following I get 4!
>
> printf("%d\n",s izeof('g'));

In C, character constants have type `int', so sizeof 'g' is equal
to sizeof(int).

Why do they write char if they mean int?

When do K&R do that?
--
Ben Pfaff
email: bl*@cs.stanford .edu
web: http://benpfaff.org

[Tim Rentsch]

"JS" <[e-mail] m> writes:

I read in K&R page 204 that sizeof use on a char returns 1. But when I write
the following I get 4!

int main(void){
printf("%d\n",s izeof('g'));

}

Was it not supposed to return 1?

Contrary to what you might expect, a character constant such as 'g'
is an int value. So sizeof('g') == sizeof(int).

If you do sizeof( (char)'g' ) you should get 1.

[Mark Odell]

Alex Fraser wrote:

No, a character in single quotes, as in 'g', is a constant of type signed
int in C (I think this is different in C++).

Try:

int main(void) {
char c;
printf("%d\n", (int)sizeof c);
Don't you mean:
printf("%u\n", sizeof c);
return 0;
}

Alex

[Christopher Benson-Manica]

Ben Pfaff <bl*@cs.stanfor d.edu> spoke thus:

In C, character constants have type `int', so sizeof 'g' is equal
to sizeof(int).

(semi OT) Why did the authors of the standard make this arguably
non-intuitive stipulation?

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.