Why does one need to use two kinds of sizeof operator:
* sizeof unary-expression,
* sizeof (type-name)
?
Their behavior seem not to be different (see an example below).
------ C++ code ------
#include <iostream>
using namespace std;
int main()
{
int x = 100;
cout << sizeof ++x << endl; // expression sizeof
cout << x << endl;
x = 100;
cout << endl;
cout << sizeof (++x) << endl; // type sizeof
cout << x << endl;
return 0;
}
----------------------
------ Run ------
4
100
4
100
-----------------
Alex Vinokur
email: alex DOT vinokur AT gmail DOT com http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn 15  2247 
Alex Vinokur wrote: Why does one need to use two kinds of sizeof operator:
* sizeof unary-expression,
* sizeof (type-name)
?
Because you might want to use it on a value to find out its size, or you
want to use it on a type to find out its size.
Their behavior seem not to be different (see an example below).
------ C++ code ------
#include <iostream>
using namespace std;
int main()
{
int x = 100;
cout << sizeof ++x << endl; // expression sizeof
cout << x << endl;
x = 100;
cout << endl;
cout << sizeof (++x) << endl; // type sizeof
(++x) is not a type. int would be a type.
cout << x << endl;
return 0;
}
----------------------
------ Run ------
4
100
4
100
"Rolf Magnus" <ra******@t-online.de> wrote in message news:e6******** *****@news.t-online.com... Alex Vinokur wrote:
[snip]
int main()
{
int x = 100;
cout << sizeof ++x << endl; // expression sizeof
cout << x << endl;
x = 100;
cout << endl;
cout << sizeof (++x) << endl; // type sizeof
(++x) is not a type. int would be a type.
[snip]
I think, ++x is parsed as typename in sizeof (++x).
--
Alex Vinokur
email: alex DOT vinokur AT gmail DOT com http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn
Alex Vinokur wrote: "Rolf Magnus" <ra******@t-online.de> wrote in message news:e6******** *****@news.t-online.com... Alex Vinokur wrote:
[snip]
int main()
{
int x = 100;
cout << sizeof ++x << endl; // expression sizeof
cout << x << endl;
x = 100;
cout << endl;
cout << sizeof (++x) << endl; // type sizeof
(++x) is not a type. int would be a type.
[snip]
I think, ++x is parsed as typename in sizeof (++x).
No it is a unary-expression because a primary-expression is also a
unary-expression and ( expression ) is a primary-expression and clearly
++x is an expression.
Markus Schoder wrote: Alex Vinokur wrote: "Rolf Magnus" <ra******@t-online.de> wrote in message news:e6******** *****@news.t-online.com... Alex Vinokur wrote:
[snip] >
> int main()
> {
> int x = 100;
>
> cout << sizeof ++x << endl; // expression sizeof
> cout << x << endl;
>
> x = 100;
> cout << endl;
> cout << sizeof (++x) << endl; // type sizeof
(++x) is not a type. int would be a type.
[snip]
I think, ++x is parsed as typename in sizeof (++x).
No it is a unary-expression because a primary-expression is also a
unary-expression and ( expression ) is a primary-expression and clearly
++x is an expression.
What is the difference between
sizeof ++x
and
sizeof (++x)
?
Alex Vinokur
email: alex DOT vinokur AT gmail DOT com http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn
Alex Vinokur wrote: What is the difference between
sizeof ++x
and
sizeof (++x)
?
Nothing. sizeof takes an expression, and parens in expressions may be
optional.
sizeof(int) is the size of the typecast to int. (int)0.
--
Phlip http://c2.com/cgi/wiki?ZeekLand <-- NOT a blog!!!
Phlip wrote: Alex Vinokur wrote:
What is the difference between
sizeof ++x
and
sizeof (++x)
?
Nothing. sizeof takes an expression, and parens in expressions may be
optional.
So, why do we need two kinds of sizeof operator?
sizeof(int) is the size of the typecast to int. (int)0.
[snip]
Alex Vinokur
email: alex DOT vinokur AT gmail DOT com http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn
Alex Vinokur wrote: Phlip wrote: Alex Vinokur wrote:
What is the difference between
sizeof ++x
and
sizeof (++x)
?
Nothing. sizeof takes an expression, and parens in expressions may be
optional.
So, why do we need two kinds of sizeof operator?
One takes an expression and returns the size of the expression's type
and one takes a type directly (which must be put in parentheses). It is
mostly a matter of convenience.
Alex Vinokur schrieb: Phlip wrote: Alex Vinokur wrote:
What is the difference between
sizeof ++x
and
sizeof (++x)
?
Nothing. sizeof takes an expression, and parens in expressions may be
optional.
So, why do we need two kinds of sizeof operator?
int x;
1) sizeof(int)
2) sizeof(x) (or sizeof x)
Which one do you think is redundant?
Thomas This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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According to my analysis, operator sizeof, (type) and * have the same
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Roy
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