PREPROCESSOR: Multiple expansions.

[nobrow]

(This is a contrived example)

With the following preprocessor lines:

#define SIZEOF_mytype 4

#define hton4b htonl

#define OUT(TYPE, VAR) \
OUT1(SIZEOF_ ## TYPE, VAR)

#define OUT1(SIZE, VAR) \
hton ## SIZE ## b(VAR)

I type:

OUT(mytype, t)

and want to get:

htonl(t)

but actually get:

htonSIZEOF_mytypeb(t)

I tried various intermediate macros to catenate SIZEOF_ and TYPE to try
get it to convert to 4 but the closest I have gotten is:

hton 4 b(t)

Whats the trick?

Thanks.

[Ancient_Hacker]

well, you're pushing the non-barfing limits of macro usage, but, ok,

does it work any better if you take out the spaces around the ##'s?

[nobrow]

Ancient_Hacker wrote:

well, you're pushing the non-barfing limits of macro usage, but, ok,

does it work any better if you take out the spaces around the ##'s?

I did as it happens (spaces surrounding glue have no effect as far as I
can see).

Calm prevailed and I went through it slowly and carefully, resulting in
the following which behaves as expected.

#define SIZEOF_mytype 4

#define hton4b htonl

#define CAT(A, B) A##B

#define OUT2(ARG, VAR) \
CAT(hton, ARG)

#define OUT1(SIZE, VAR) \
OUT2(CAT(SIZE,b(VAR)), VAR)

#define OUT(TYPE, VAR) \
OUT1(SIZEOF_##TYPE, VAR)

(Incidentally, the result is the same when using "A ## B" and "SIZEOF_
## TYPE").

Thanks for the reply anyway.

[Ancient_Hacker]

This works:
#define SIZEOF_mytype 4

#define hton4b htonl

#define OUT(TYPE, VAR) OUT1(SIZEOF_##TYPE, VAR)

#define OUT1(SIZE,VAR) expand(hton,SIZE,VAR)

#define expand(x,y,z) x##y##b(z)
OUT(mytype, t)

[nobrow]

Ancient_Hacker wrote:

This works:
#define SIZEOF_mytype 4

#define hton4b htonl

#define OUT(TYPE, VAR) OUT1(SIZEOF_##TYPE, VAR)

#define OUT1(SIZE,VAR) expand(hton,SIZE,VAR)

#define expand(x,y,z) x##y##b(z)
OUT(mytype, t)

Thanks.