Why can't gcc find the random() interface when -std=c99 is set?

Question

I do "#include <stdlib.h>" at the top of the source.

Example compilation:

/usr/bin/colorgcc -std=c99 -fgnu89-inline  -g -Wall -I/usr/include -I./ -I../ -I../../ -I../../../ -I../../../../    -O3 -o f8  f8.c
In file included from f8.c:7:
ctype-cmp.c: In function ‘randomized’:
ctype-cmp.c:48: warning: implicit declaration of function ‘random’
ctype-cmp.c: In function ‘main’:
ctype-cmp.c:153: warning: implicit declaration of function ‘srandom’
ais@xcalibur:t$ 

When I turn off -std=c99, the function isfinite() can not be found. So I do want to use -std=c99 for this and other reasons. Is there some trick I'm missing?

Answer 1

man srandom says that the function is not part of C99 but part of POSIX.

Activate _BSD_SOURCE or _XOPEN_SOURCE >= 500 or any other suitable feature test macro that declares the srandom/random function (see man feature_test_macros and man srandom).

This one has good chances, but you need to figure out the macros that are defined/not defined implicitly thereby too by reading the manpages above.

/usr/bin/colorgcc -std=c99 -D_XOPEN_SOURCE=600 -fgnu89-inline -g -Wall 
    -I/usr/include -I./ -I../ -I../../ -I../../../ -I../../../../ -O3 -o f8  f8.c