Hello,
Here is the program
#include stdio
int main(void)
{
const int num = 100;
int *ip;
ip = (int *)#
*ip = 200;
printf("value of num is %d(%d) \n", num, *ip);
}
Output:
value of num is 100(200)
The output says that *ip is changed and 'num' is unchanged. How is this
possible when both of them point to the same memory location? My wild guess
says that this trick is handled at the compiler level. Am I correct?
Even when the memory location is accessable and the contents changed the
'const integer' is unaffected.
Thanks
Nov 14 '05
83  3070 
On Thu, 08 Jan 2004 14:57:21 +0530, Vijay Kumar R Zanvar wrote: On my system the output is
value of num is 200(200)
on my system is 200(200) too, using gcc.
I think that being an undefined behavior according to the standard,
as Martin pointed out, the developers of the compiler used by user
have made up a mess. gcc handles it logically, as I have explained
(hope to be correct) in my earlier followup.
Now, user, which compiler have you used? I'm curious about it.
Bye
Daniele "tinybyte" Milan
user wrote: Hello,
Please don't top-post. It's rude.
I am sorry. After reading the post myself, I felt that the post is
incomplete in itself.
Well when I tried this exercise I expected the code to fail in the
compilation phase with an error ( not a warning). It did not, and the output
made me think how is this possible to have 2 different value when I am
pointing to the same memory location? I exptected the output to 2 100s or 2
200s. Is 'num' variable pointing some where ????
My question is how is the output different? Is the program performing an
undefined behavior?
Yes. This should be fairly obvious.
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
tinybyte wrote: On Thu, 08 Jan 2004 15:02:12 +0530, user wrote:
Well when I tried this exercise I expected the code to fail in the
compilation phase with an error ( not a warning). It did not, and the output
made me think how is this possible to have 2 different value when I am
pointing to the same memory location? I exptected the output to 2 100s or 2
200s. Is 'num' variable pointing some where ????
I'll try to explain this way:
#include <stdio.h>
int
main (void)
{
const int num = 100;
int *ip;
ip = (int *) #
*ip = 200;
printf ("value of num is %d(%d) \n", num, *ip);
printf ("address of ip is 0x%p\n", ip);
printf ("address of num is 0x%p\n", &num);
These last two statements invoke undefined behavior. %p is for void
pointers only. It's also a little silly to add your own 0x prefix. On
some implementations you'll get output like
0x0x034aff30
}
this will print out the addresses the pointers are pointing to.
The output is as expected: the address is the same.
Now, I'm not sure of what i'm about to say, but well, correct me if
I am wrong :)
There's no good reason to try to explain undefined behavior. In this
case, the compiler probably eliminated a fetch from memory by using the
value 100 directly, since it "knew" that's what the stored value would be.
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
>When /can/ the address of a named object change?
sorry I wanted to say the adress that the pointer points to If you assign anything to int *ip it will be assigned to an adress which
the computer will find in order to store it there.
I'm not sure about this but I think that when you write int *p=20; the
compiler will handle the memmory allocation of the int which has '20' value.
--
---------------------------
Time to get it done!
Remove (d*elete*) to reply
On Thu, 08 Jan 2004 10:39:30 +0000, Kevin Goodsell wrote: There's no good reason to try to explain undefined behavior. In this
case, the compiler probably eliminated a fetch from memory by using the
value 100 directly, since it "knew" that's what the stored value would be.
There's always a good reason to try to know how things really works.
Now we had discovered that that compiler is broken. gcc handles the
thing correctly, as it should be. Just saying "according to the
standard" isn't enough. According to the standard that piece of code
is broken, but it actually executes, and sometimes properly.
You're not compiling programs with the standard, but with compilers.
Daniele
On Thu, 08 Jan 2004 10:39:30 +0000, Kevin Goodsell wrote: These last two statements invoke undefined behavior. %p is for void
pointers only. It's also a little silly to add your own 0x prefix. On
I admit it's silly :)
Undefined behavior? then cast that pointers to void and you're done.
...
printf ("address of ip is %p\n", (void *)ip);
printf ("address of num is %p\n", (void *)&num);
...
Daniele
Frane Roje wrote: I'm not sure about this but I think that when you write int *p=20; the
compiler will handle the memmory allocation of the int which has '20' value.
No, it will not. Unless You make p point somewhere decent first the
above is wrong.
--
Thomas
tinybyte <ne******@box.i t> wrote: On Thu, 08 Jan 2004 14:57:21 +0530, Vijay Kumar R Zanvar wrote:
On my system the output is
value of num is 200(200)
on my system is 200(200) too, using gcc.
I think that being an undefined behavior according to the standard,
as Martin pointed out, the developers of the compiler used by user
have made up a mess.
No, the compiler used by the OP is quite as logical as gcc, and possibly
better at optimising - it seems to have diked out all references to the
const int and replaced them with the constant value, which is both a
sensible optimisation, and potentially quite a valuable one.
gcc handles it logically,
s/logically/lazily/, IYAM.
Richard
tinybyte <ne******@box.i t> wrote: On Thu, 08 Jan 2004 10:39:30 +0000, Kevin Goodsell wrote: There's no good reason to try to explain undefined behavior. In this
case, the compiler probably eliminated a fetch from memory by using the
value 100 directly, since it "knew" that's what the stored value would be.
There's always a good reason to try to know how things really works.
Now we had discovered that that compiler is broken. gcc handles the
thing correctly, as it should be.
Wrong, and wrong. The OP's compiler is quite correct, and gcc handles
this code as it _may_, not as it _should_ be handled. The code invokes
undefined behaviour - changing the value of a const object - and _any_
behaviour after that is correct.
Just saying "according to the
standard" isn't enough. According to the standard that piece of code
is broken, but it actually executes, and sometimes properly.
For _some_ values of properly. _My_ value of properly is the reverse -
I'd rather have a compiler that compiles correct code to a fast
executable than one that compiles incorrect code to appear to be
correct. Hence, I'd prefer the OP's compiler to gcc.
You're not compiling programs with the standard, but with compilers.
A meaningless statement, really.
Richard
tinybyte wrote: On Thu, 08 Jan 2004 10:39:30 +0000, Kevin Goodsell wrote:
There's no good reason to try to explain undefined behavior. In this
case, the compiler probably eliminated a fetch from memory by using the
value 100 directly, since it "knew" that's what the stored value would be.
There's always a good reason to try to know how things really works.
Now we had discovered that that compiler is broken.
No we have not.
gcc handles the
thing correctly, as it should be.
No it does not. There is no right way and there is no wrong way.
Just saying "according to the
standard" isn't enough.
Yes it is.
According to the standard that piece of code
is broken, but it actually executes, and sometimes properly.
There is no properly.
You're not compiling programs with the standard, but with compilers.
Which conform to a standard. When the standard says that something is
undefined you cannot rely on the behaviour of any compiler when using
that something. The result will be different between different compilers
hence rendering your code unportable. The result will be different
between different versions of the same compiler. The result will even be
different for the same compiler invoked with different flags.
Try compiling with the optimising flags for gcc and be prepared to learn
a valuable lesson.
--
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