Same pointer parameter const and non-const

Is this program guaranteed to print 10, or can the compiler
get 'confused' by the const and assume the variable
still has its initial value?

#include <stdio.h>

int func(int *A, int const *B)
{
*A = 10;
return *B;
}

int main()
{
int X = 5;
printf("%d\n", func(&X, &X));
}

Old Wolf

On 2009-11-03, Old Wolf <> wrote:
> Is this program guaranteed to print 10, or can the compiler
> get 'confused' by the const and assume the variable
> still has its initial value?

I would say it has to print 10. There's no restrict anywhere in sight,
there's questionable aliasing, so I think the compiler's obliged to
assume that modifications of other objects with compatible types could
potentially modify the thing pointed to by B.

-s
--
Copyright 2009, all wrongs reversed. Peter Seebach / usenet-
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
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Seebs

"Old Wolf" <> wrote in message
news:b5d2ab96-b143-4bb8-8da4-...
> Is this program guaranteed to print 10, or can the compiler
> get 'confused' by the const and assume the variable
> still has its initial value?
>
> #include <stdio.h>
>
> int func(int *A, int const *B)
> {
> *A = 10;
> return *B;
> }
>
> int main()
> {
> int X = 5;
> printf("%d\n", func(&X, &X));
> }

The const will only mean that func won't set B, it won't assume anything
else. The compiler won't get confused either, it will optimise away that
**** completely.

DanDanDan

"DanDanDan" <> writes:

> "Old Wolf" <> wrote in message
> news:b5d2ab96-b143-4bb8-8da4-...
>> Is this program guaranteed to print 10, or can the compiler
>> get 'confused' by the const and assume the variable
>> still has its initial value?
>>
>> #include <stdio.h>
>>
>> int func(int *A, int const *B)
>> {
>> *A = 10;
>> return *B;
>> }
>>
>> int main()
>> {
>> int X = 5;
>> printf("%d\n", func(&X, &X));
>> }
>
> The const will only mean that func won't set B, it won't assume anything
> else.

This wording is a little confusing. B is not const so the function is
permitted to set B (it doesn't, but it might). In fact, there isn't a
single const object anywhere in the program. The const is simply a
promise that func won't use B to change the int that B points to.

<snip>
--
Ben.

Ben Bacarisse