🚀 go-pugleaf

In article <vkohklaizj.fsf@atari.connectinc.com>,
Bryan Ischo <bji@connectinc.com> wrote:
>Just out of curiosity ... why can't the compiler be sure that 'flag' won't be
>modified by printf?

Because it could be modified by printf.

>After all, its type is declared as int, not int *, and
>the loop conditional is 'while(flag)' not 'while (*flag)'.

So?  &flag generates a pointer to it, through which it could be modified, and
any call to any function could possibly refer to it.  Trivally,
	printf("%n", &flag);
would be stupid but would modify flag.

>What I'm getting
>at is, the only way flag could be modified is if the programmer is assuming
>that the size of an int is the same as the size of an int * and is passing in
>an int in place of an int *, presumably one which has been casted from an
>int * somewhere earlier.

This is simply untrue.  The programmer could have put flag's address in a
global, and a function named printf might modify the contents.

At least, I think what you're saying is untrue; I can't honestly understand a
word of it.

>Shouldn't the compiler be smart enough to know that flag can't be modified
>even IF the printf is in the loop, given how smart it's been about everything
>else?

Some versions might be, but it's an unlikely assumption.

-s
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