pre-processor string replacement

[sam_cit]

Hi Everyone,

I have the following question regarding string replacement
pre-processor statement,

#define status "23"
#define status "44"

int main()
{
printf("%s\n", status);
}

this works fine and prints 44

where as

#define status "23"
#define status1 status

int main()
{
printf("%s\n", status);
}

prints 23, string replacement occurs on the pre-processor statement
itself, which is not in case 1, why is it so?
Can anyone help me on this?

[sam_cit]

>

#define status "23"
#define status1 status

int main()
{
printf("%s\n", status);
}

Sorry i mean

int main()
{
printf("%s\n", status1);
}

[Ian Collins]

sa*****@yahoo.c o.in wrote:

Hi Everyone,

I have the following question regarding string replacement
pre-processor statement,

#define status "23"
#define status "44"

int main()
{
printf("%s\n", status);
}

this works fine and prints 44

where as

#define status "23"
#define status1 status

int main()
{
printf("%s\n", status);
}

prints 23, string replacement occurs on the pre-processor statement
itself, which is not in case 1, why is it so?
Can anyone help me on this?

What did you expect to see?

--
Ian Collins.

[sam_cit]

#define status "23"
#define status "44"

What did you expect to see?

--
Ian Collins.

i expect first statement to replace status of second statement with
23, so it would be

#define "23" "43"...

which isn't the case, so i though string replacement never happens in
pre-processor statement, however the second example i pointed proves
that understanding is not correct.

[Ian Collins]

sa*****@yahoo.c o.in wrote:

>

>>#define status "23"
#define status "44"

What did you expect to see?

--
Ian Collins.

i expect first statement to replace status of second statement with
23, so it would be

#define "23" "43"...

which isn't the case, so i though string replacement never happens in
pre-processor statement, however the second example i pointed proves
that understanding is not correct.

I see. In the first example you are redefining status (your compiler
did warn you, didn't it?).

--
Ian Collins.

[sam_cit]

I see. In the first example you are redefining status (your compiler
did warn you, didn't it?).

--
Ian Collins.

Yes it did, but my question is had the pre-processor replaced status
with "23" compiler wouldn't have known... why did the pre-processor
choose not to replace status of the second #define with "23"

#define status "23"
#define status "44"

[Richard Heathfield]

sa*****@yahoo.c o.in said:

Hi Everyone,

I have the following question regarding string replacement
pre-processor statement,

#define status "23"
#define status "44"

int main()
{
printf("%s\n", status);
}

this works fine and prints 44

This time, maybe. Next time it might not. Your program exhibits undefined
behaviour. Get the easy stuff right before worrying about the hard stuff.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[Ian Collins]

sa*****@yahoo.c o.in wrote:

>>I see. In the first example you are redefining status (your compiler
did warn you, didn't it?).

Please trim signatures.

>
Yes it did, but my question is had the pre-processor replaced status
with "23" compiler wouldn't have known... why did the pre-processor
choose not to replace status of the second #define with "23"

Because you are redefining the token, not using it. As Richard pointed
out, this is undefined.

Get your code to compile without warnings.

--
Ian Collins.

[Richard Heathfield]

Ian Collins said:

sa*****@yahoo.c o.in wrote:

>>>I see. In the first example you are redefining status (your compiler
did warn you, didn't it?).

Please trim signatures.

>>
Yes it did, but my question is had the pre-processor replaced status
with "23" compiler wouldn't have known... why did the pre-processor
choose not to replace status of the second #define with "23"

Because you are redefining the token, not using it. As Richard pointed
out, this is undefined.

Actually, what I was referring to was the call to a variadic function
without a valid prototype in scope.

Get your code to compile without warnings.

....after cranking up the warning level to a reasonable height.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[Mark McIntyre]

On 14 Jan 2007 00:10:18 -0800, in comp.lang.c , sa*****@yahoo.c o.in
wrote:

>

>
#define status "23"
#define status "44"

What did you expect to see?

--
Ian Collins.

i expect first statement to replace status of second statement with
23, so it would be

#define "23" "43"...

No, the preprocessor doesn't work like that. All you've done is
redefine status.
--
Mark McIntyre

"Debugging is twice as hard as writing the code in the first place.
Therefore, if you write the code as cleverly as possible, you are,
by definition, not smart enough to debug it."
--Brian Kernighan