Size of extern int x;

[skishorev]

If i declared extern int x; in linux environment. what is the size?

 

[Chris McDonald]

If i declared extern int x; in linux environment. what is the size?

Why do you suspect that it'll be anything other than sizeof(int) ?

 

[skishorev]

Hi,

I am asking that whether it will create memory at the time of
decalaration of extern int x;

 

[Vladimir Oka]

I am asking that whether it will create memory at the time of
decalaration of extern int x;

Quote context. Read <http://cfaj.freeshell.org/google/>.

You did not make yourself clear the first time.

For the `extern` declaration, no memory need be allocated, as it just
tells the compiler not to worry about it, as `x` is guaranteed to exits
(as a variable of type `int`) somewhere else. Exactly where this
`somewhere else` is physically, will be discovered at link phase (if
not, linking will fail).

 

[Richard Heathfield]

(e-mail address removed) said:

If i declared extern int x; in linux environment. what is the size?

No storage is reserved, so there isn't a size.

 

[jaysome]

(e-mail address removed) said:


No storage is reserved, so there isn't a size.

One may interpret the OP's question as: "What is sizeof x?" In which
case, the following code should answer that question.

#include <limits.h>
#include <stdio.h>
extern int x;
int main(void)
{
const int INT_BITS = (int)(sizeof(int) * CHAR_BIT);
printf
(
"sizeof x = %d, same as sizeof(int) = %d\n",
(int)sizeof x,
(int)sizeof(int)
);
printf
(
"type int is %d bit%s\n",
INT_BITS,
INT_BITS == 1 ? "" : "s"
);
return 0;
}
int x = 0;

If you can spot how a standard-compliant compiler will never achieve
100% decision coverage in the above code, then you win a prize.

 

[Barry Schwarz]

snip code

If you can spot how a standard-compliant compiler will never achieve
100% decision coverage in the above code, then you win a prize.

What does the phrase "decision coverage" mean?


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