Sizeof and increment operators

[Anjali M]

#include <stdio.h>

int main()
{
int number = 10;
printf("Number is: %d\n", number);

printf("Sizeof of number is: %d\n", sizeof(number++ ));

printf("Number now is: %d\n", number);

return 0;
}
The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

Thanks,
Anjali.

[Richard Bos]

g_********@redi ffmail.com (Anjali M) wrote:

printf("Sizeof of number is: %d\n", sizeof(number++ )); The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

From the last public draft of the C99 Standard, 6.5.3.4#2:

# The sizeof operator... If the type of the operand is a variable length
# array tye, the operand is evaluated; otherwise, the operand is not
# evaluated and the result is an integer constant.

Richard

[smitha]

Hi,

The sizeof operator just returns the size of the type or expressions,

so i guess,it is not defined to allow airthmetic operations within it.

Bye,

Smitha

"Anjali M" <g_********@red iffmail.com> wrote in message
news:99******** *************** ***@posting.goo gle.com...

#include <stdio.h>

int main()
{
int number = 10;
printf("Number is: %d\n", number);

printf("Sizeof of number is: %d\n", sizeof(number++ ));

printf("Number now is: %d\n", number);

return 0;
}
The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

Thanks,
Anjali.

[Sumukh]

HI anjali,

Sizeof is an *operator* in C.
The operand expected in sizeof is either an identifier that is a
unary-expression, or a type-cast expression (that is, a type specifier
enclosed in parentheses). The unary-expression cannot represent a
bit-field object, an incomplete type, or a function designator.

A unary expression is an expression whose valuation is computed by
applying its (unary) operator to the valuation of its operand. There are
three kinds of unary operators:
(1) the logical complement unary operator "!",
(2) the plus unary operator "+", and
(3) the minus unary operator "-".

Hope this helps

sumukh
Anjali M wrote:

#include <stdio.h>

int main()
{
int number = 10;
printf("Number is: %d\n", number);

printf("Sizeof of number is: %d\n", sizeof(number++ ));

printf("Number now is: %d\n", number);

return 0;
}
The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

Thanks,
Anjali.

[Nitin Bhardwaj]

g_********@redi ffmail.com (Anjali M) wrote in message news:<99******* *************** ****@posting.go ogle.com>...

#include <stdio.h>

int main()
{
int number = 10;
printf("Number is: %d\n", number);

printf("Sizeof of number is: %d\n", sizeof(number++ ));

printf("Number now is: %d\n", number);

return 0;
}
The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

Thanks,
Anjali.

Hi Anjali,

The problem is not that of the ++ operator...but the 'sizeof'
operator.
The C Standard states that the operand of sizeof can be either a type
name or an expression.

e.g. sizeof (int) OR sizeof (a+b)

If it is an expression,then it will not be evaluated; See section
A.7.4.8 of K&R2 - [The C Programming Language 2/ed - Kernighan &
Ritchie ], Appendix A.

For the sake of convenience I'm reproducing the contents verbatim
here :

A.7.4.8 Sizeof Operator
The sizeof operator yields the number of bytes required to store an
object of the type of its operand. The operand is either an
expression,***_ which is not evaluated_**, or a parenthesized type
name. When sizeof is applied to a char, the result is 1; when applied
to an array, the result is the total number of bytes in the array.
When applied to a structure or union, the result is the number of
bytes in the object, including any padding required to make the object
tile an array: the size of an array of n elements is n times the size
of one element. The operator may not be applied to an operand of
function type, or of incomplete type, or to a bit-field. The result is
an unsigned integral constant; the particular type is
implementation-defined.

Hence your expresion number++ never gets evaluated.So the value of
number doesn't increase.

HTH
Nitin

[Nitin Bhardwaj]

"smitha" <sm*****@cisco. com> wrote in message news:<108909700 7.531247@sj-nntpcache-3>...

Hi,

The sizeof operator just returns the size of the type or expressions,

so i guess,it is not defined to allow airthmetic operations within it.
Wrong, the operand of sizeof operator is either type-names OR
expressions - which also includes arithmetic expressions.If it were
not defined for arithmetic expressions then the compiler must have
complained !!

You can even include a function call in the argument to sizeof !!!
Bye,

Smitha

"Anjali M" <g_********@red iffmail.com> wrote in message
news:99******** *************** ***@posting.goo gle.com...

#include <stdio.h>

int main()
{
int number = 10;
printf("Number is: %d\n", number);

printf("Sizeof of number is: %d\n", sizeof(number++ ));

printf("Number now is: %d\n", number);

return 0;
}
The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

Thanks,
Anjali.

[smitha]

yes... sizeof is an operator and not a function. In a function call each
argument will be evaluated but this is not the case in sizeof.

"Nitin Bhardwaj" <ni************ *@hotmail.com> wrote in message
news:17******** *************** ***@posting.goo gle.com...

"smitha" <sm*****@cisco. com> wrote in message

news:<108909700 7.531247@sj-nntpcache-3>...

Hi,

The sizeof operator just returns the size of the type or expressions,

so i guess,it is not defined to allow airthmetic operations within it.

Wrong, the operand of sizeof operator is either type-names OR
expressions - which also includes arithmetic expressions.If it were
not defined for arithmetic expressions then the compiler must have
complained !!

You can even include a function call in the argument to sizeof !!!

Bye,

Smitha

"Anjali M" <g_********@red iffmail.com> wrote in message
news:99******** *************** ***@posting.goo gle.com...

#include <stdio.h>

int main()
{
int number = 10;
printf("Number is: %d\n", number);

printf("Sizeof of number is: %d\n", sizeof(number++ ));

printf("Number now is: %d\n", number);

return 0;
}
The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

Thanks,
Anjali.

[Dan Pop]

In <1089097007.531 247@sj-nntpcache-3> "smitha" <sm*****@cisco. com> writes:

The sizeof operator just returns the size of the type or expressions,
Without evaluating them (except for VLAs in C99).
so i guess,it is not defined to allow airthmetic operations within it.

It is perfectly defined and allowed, they just won't be performed: the
compiler will merely figure out the type of the result.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Charles Richmond]

Anjali M wrote:

#include <stdio.h>

int main()
{
int number = 10;
printf("Number is: %d\n", number);

printf("Sizeof of number is: %d\n", sizeof(number++ ));

printf("Number now is: %d\n", number);

return 0;
}

The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

What's the matter??? Can't you *do* the job after you
suck it up from the U. S.???

--
+-------------------------------
| Charles and Francis Richmond
| richmond at plano dot net
| Re-Defeat Bush!!!
+-------------------------------

[Richard Bos]

Charles Richmond <ri******@plano .net> wrote:

What's the matter??? Can't you *do* the job after you
suck it up from the U. S.???

Got fired, didya?

Richard