[SSI-users] compilation error
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From: Barry, C. <cb...@in...> - 2004-03-23 03:11:32
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All, I'm attempting to build a new SSI kernel on a newly installed Debian = system. I'm following kvaneeshs' instructions, and have searched this = list for solutions, but it appears that my particular issue has not come = up. Here is the error from the compile: make[2]: Entering directory `/usr/src/linux-2.4.20-30.9/ipc' gcc -D__KERNEL__ -I/usr/src/linux-2.4.20-30.9/include -Wall = -Wstrict-prototypes -Wno-trigraphs -O2 -fno-strict-aliasing -fno-common = -pipe -mpreferred-stack-boundary=3D2 -march=3Di586 -nostdinc = -iwithprefix include -DKBUILD_BASENAME=3Dutil -c -o util.o util.c gcc -D__KERNEL__ -I/usr/src/linux-2.4.20-30.9/include -Wall = -Wstrict-prototypes -Wno-trigraphs -O2 -fno-strict-aliasing -fno-common = -pipe -mpreferred-stack-boundary=3D2 -march=3Di586 -nostdinc = -iwithprefix include -DKBUILD_BASENAME=3Dmsg -c -o msg.o msg.c msg.c: In function `sysvipc_msg_read_proc': msg.c:1423: parse error before `int' msg.c:1434: `size' undeclared (first use in this function) msg.c:1434: (Each undeclared identifier is reported only once msg.c:1434: for each function it appears in.) msg.c:1435: `node_id_pairs' undeclared (first use in this function) msg.c:1441: `tmp_pairs' undeclared (first use in this function) msg.c:1443: `id_count' undeclared (first use in this function) msg.c:1455: `node_num' undeclared (first use in this function) msg.c:1457: `ipc_id' undeclared (first use in this function) msg.c:1464: `tmsq' undeclared (first use in this function) make[2]: *** [msg.o] Error 1 make[2]: Leaving directory `/usr/src/linux-2.4.20-30.9/ipc' make[1]: *** [first_rule] Error 2 make[1]: Leaving directory `/usr/src/linux-2.4.20-30.9/ipc' make: *** [_dir_ipc] Error 2 When I look in msg.c, the offending code appears to be PRINT_HEADER, = which points to a sprintf statement. Has this code recently changed? And = if so, does anyone have a clue as to what could be wrong with it? I = *thought* sprintf took 3 args - the offending code appears to only = provide two. Thanks, and I can't wait to get this going! -- Chris |