Precedence of sizeof

[Frederick Gotham]

sizeof has the same precedence as a cast, and they both bind from right to
left. The following won't compile for me with gcc:

int main(void)
{
sizeof(double)5 ;

return 0;
}

Has it got something to do with "globbing"?

--

Frederick Gotham

[Cong Wang]

On Oct 6, 9:39 pm, Frederick Gotham <fgotha...@SPAM .comwrote:

sizeof has the same precedence as a cast, and they both bind from right to
left. The following won't compile for me with gcc:

int main(void)
{
sizeof(double)5 ;

return 0;
}

Has it got something to do with "globbing"?

--

Frederick Gotham

"sizeof ((double)5)" works well.

[yuhaikuo]

Cong Wang wrote:

On Oct 6, 9:39 pm, Frederick Gotham <fgotha...@SPAM .comwrote:

sizeof has the same precedence as a cast, and they both bind from right to
left. The following won't compile for me with gcc:

int main(void)
{
sizeof(double)5 ;

return 0;
}

Has it got something to do with "globbing"?

--

Frederick Gotham

"sizeof ((double)5)" works well.

[Frederick Gotham]

Cong Wang posted:

"sizeof ((double)5)" works well.

Yes, I realise that. My question pertains to operator precedence and
associativity -- specifically, why:

sizeof(double)5

is interpretted by gcc as:

(sizeof(double) )5

rather than:

sizeof((double) 5)
--

Frederick Gotham

[Antti-Juhani Kaijanaho]

Frederick Gotham <fg*******@SPAM .comkirjoitti 06.10.2006:

sizeof has the same precedence as a cast

No it doesn't.

Has it got something to do with "globbing"?

Unlikely.

[Richard Tobin]

In article <M0************ *******@news.in digo.ie>,
Frederick Gotham <fg*******@SPAM .comwrote:

>sizeof has the same precedence as a cast

The standard does not describe C's syntax in terms of operator
precedence; that's just a handy way of summarizing some common cases.

The relevant part of the grammar is:

unary-expression:
[...]
unary-operator cast expression
sizeof unary-expression
sizeof ( type-name )

cast-expression:
unary-expression
( type-name ) cast expression

So in:

sizeof(double)5 ;

"(double)5" is not a unary expression, so only "sizeof ( type-name )"
is possible, and that of course doesn't work because of the "5"
afterwards.

-- Richard

[lovecreatesbea...]

On Oct 6, 9:39 pm, Frederick Gotham <fgotha...@SPAM .comwrote:

sizeof has the same precedence as a cast, and they both bind from right to
left. The following won't compile for me with gcc:

int main(void)
{
sizeof(double)5 ;

return 0;
}

The code: sizeof(double)5 ; in the example mean: sizeof(double) [X] 5;,
in the example the sequence [X] is none. It is trying to connect two
numeric expressions without operators. This is not correct. You may
mean some similar code like the following.

int main(void){
/*sizeof(double) 5;*/
sizeof(double) * 5;
sizeof((double) 5);

return 0;
}

[Robert Gamble]

On Oct 6, 9:39 am, Frederick Gotham <fgotha...@SPAM .comwrote:

sizeof has the same precedence as a cast

No, the sizeof operator has a higher precedence than the cast operator.

and they both bind from right to
left. The following won't compile for me with gcc:

int main(void)
{
sizeof(double)5 ;

return 0;
}

As it shouldn't because it doesn't make any sense.

Has it got something to do with "globbing"?

I have no idea what you mean by "globbing" in this context, your
expression is evaluated as:
(sizeof(double) ) 5;
which is a syntax error and is why it doesn't compile.

Robert Gamble

[Frederick Gotham]

Richard Tobin posted:

unary-expression:
[...]
unary-operator cast expression
sizeof unary-expression
sizeof ( type-name )

cast-expression:
unary-expression
( type-name ) cast expression

So in:

> sizeof(double)5 ;

"(double)5" is not a unary expression, so only "sizeof ( type-name )"
is possible, and that of course doesn't work because of the "5"
afterwards.

So is a unary expression something simple like a literal or the name of an
object? Does that mean that the following should be bogus:

int main(void)
{
sizeof(5+4);
return 0;
}

It compiles just fine with gcc.

--

Frederick Gotham

[Frederick Gotham]

lovecreatesbea. [e-mail] posted:

The code: sizeof(double)5 ; in the example mean: sizeof(double) [X] 5;,
in the example the sequence [X] is none. It is trying to connect two
numeric expressions without operators. This is not correct. You may
mean some similar code like the following.

int main(void){
/*sizeof(double) 5;*/
sizeof(double) * 5;
sizeof((double) 5);

return 0;
}

No. As I've already mentioned, this entire thread is to do with operator
precedence and associativity.

--

Frederick Gotham