Why does this compile with no warnings, what, if anything, does it mean,
and why does the program print out 1?
#include <stdio.h>
int main(void)
{
printf("%d\n", sizeof(int()));
return 0;
}
I know I probably "meant" sizeof (int), not sizeof (int()). I'm just
curious.
Thanks in advance for any explanations. 11 2175
Ben C wrote: Why does this compile with no warnings, what, if anything, does it mean, and why does the program print out 1?
#include <stdio.h>
int main(void) { printf("%d\n", sizeof(int())); return 0; }
I know I probably "meant" sizeof (int), not sizeof (int()). I'm just curious.
Good one. ;-)
'int ()' is actually a function type. (A function taking no argument
and returning an int.)
sizeof on a function type is undefined behavior. Whether it returns
1 or 10000, doesn't matter. It makes no sense.
With a appropriate compiler options, you should get a warning:
for instance, with GCC:
gcc -Wall -pedantic
Test1.c: In function `main':
Test1.c:5: warning: invalid application of `sizeof' to a function type
Ben C wrote: Why does this compile with no warnings, what, if anything, does it mean, and why does the program print out 1?
#include <stdio.h>
int main(void) { printf("%d\n", sizeof(int())); return 0; }
I know I probably "meant" sizeof (int), not sizeof (int()). I'm just curious.
Maybe your warning level is not high enough? For me gcc with -Wall
-ansi and -pedantic options reports:
2.c: In function `main':
2.c:5: warning: invalid application of `sizeof' to a function type
But it compiles and prints 1 as output when run.
Ben C wrote: Why does this compile with no warnings,
Because you have not got your diagnositics turned on.
gcc, even with the source language specified as gnu99,
warns
In function 'main':
5: warning: invalid application of 'sizeof' to a function type
5: warning: format '%d' expects type 'int', but argument 2 has type
'long unsigned int'
what, if anything, does it mean,
nothing, except as a non-portable extension provided by your implementation
and why does the program print out 1?
because that's what you implementation decided to interpret this
non-portable construct as meaning. #include <stdio.h>
int main(void) { printf("%d\n", sizeof(int())); return 0; }
I know I probably "meant" sizeof (int), not sizeof (int()). I'm just curious.
Thanks in advance for any explanations.
Bill collectors write "thanks in advance." It is language used to
create dominance in the expectation of compliance. Polite human beings
do not use it.
On 2006-03-25, Guillaume <> wrote: Ben C wrote: Why does this compile with no warnings, what, if anything, does it mean, and why does the program print out 1?
#include <stdio.h>
int main(void) { printf("%d\n", sizeof(int())); return 0; } [...] 'int ()' is actually a function type. (A function taking no argument and returning an int.)
sizeof on a function type is undefined behavior. Whether it returns 1 or 10000, doesn't matter. It makes no sense.
Of course, a function type! I was using -Wall but not -pedantic. Thanks.
The original example (which is on comp.programmin g) was this:
typedef void (*voidf)();
... sizeof (voidf()) ...
Well, not sizeof (voidf()) actually, but va_arg(ap, voidf()).
Here voidf() is also a function type-- a function that returns a pointer
to a function that returns void and takes no arguments that takes no
arguments. Surely not what was intended.
Ben C wrote: Of course, a function type! I was using -Wall but not -pedantic. Thanks.
Yet another one of these "GCCisms". I don't think this should be allowed
at all. A warning doesn't seem enough to me here: and you only get one
if you go "-pedantic" with GCC. Weird.
Some other C compilers report this construct as an error, and I think
they should as well.
Martin Ambuhl wrote: Ben C wrote: Why does this compile with no warnings,
Because you have not got your diagnositics turned on. gcc, even with the source language specified as gnu99, warns
In function 'main': 5: warning: invalid application of 'sizeof' to a function type 5: warning: format '%d' expects type 'int', but argument 2 has type 'long unsigned int'
what, if anything, does it mean,
nothing, except as a non-portable extension provided by your implementation and why does the program print out 1?
because that's what you implementation decided to interpret this non-portable construct as meaning.
#include <stdio.h>
int main(void) { printf("%d\n", sizeof(int())); return 0; }
I know I probably "meant" sizeof (int), not sizeof (int()). I'm just curious.
Thanks in advance for any explanations.
Bill collectors write "thanks in advance." It is language used to create dominance in the expectation of compliance. Polite human beings do not use it.
Who's 'Bill Collectors' - any relation to 'Bill Posters?
--
==============
*Not a pedant*
==============
"Martin Ambuhl" <ma*****@earthl ink.net> wrote in message
news:le******** ***********@new sread1.news.atl .earthlink.net. .. Thanks in advance for any explanations.
Bill collectors write "thanks in advance." It is language used to create dominance in the expectation of compliance. Polite human beings do not use it.
Wow. And, Plauger told me, "My, you do have a twisty way of thinking, don't
you?"
I may not be the correct person from your perspective to defend BenC, but, I
haven't seen a harsh word yet. I'd say it's a polite statement and nothing
more.
Martin Ambuhl wrote:
[snip] Bill collectors write "thanks in advance." It is language used to create dominance in the expectation of compliance. Polite human beings do not use it.
Really?
I guess that people like myself and probably the OP who have never
talked to or heard from a bill collector should be expected to know
better. Where did we go wrong?
--
jay
Martin Ambuhl wrote: Bill collectors write "thanks in advance." It is language used to create dominance in the expectation of compliance. Polite human beings do not use it.
I'm going to have to call BS that. I don't believe it's in any way
impolite.
Brian
--
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won't shut up.
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