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Allow to force the number of parallel make job
This is particularly useful when using distcc
tags/v0.2.16.rc1
Jerome Robert
9 years ago
parent
b9b52c295d
commit
ba024fcfc0
3 changed files
with
11 additions
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1 deletions
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+4
-0
Makefile.rule
+4
-0
Makefile.system
+3
-1
getarch.c
+ 4
- 0
Makefile.rule
View File
@@ -108,6 +108,10 @@ NO_AFFINITY = 1
# Don't use parallel make.
# NO_PARALLEL_MAKE = 1
# Force number of make jobs. The default is the number of logical CPU of the host.
# This is particularly useful when using distcc
# MAKE_NB_JOBS = 2
# If you would like to know minute performance report of GotoBLAS.
# FUNCTION_PROFILE = 1
+ 4
- 0
Makefile.system
View File
@@ -139,6 +139,10 @@ NO_PARALLEL_MAKE=0
endif
GETARCH_FLAGS += -DNO_PARALLEL_MAKE=$(NO_PARALLEL_MAKE)
ifdef MAKE_NB_JOBS
GETARCH_FLAGS += -DMAKE_NB_JOBS=$(MAKE_NB_JOBS)
endif
ifeq ($(HOSTCC), loongcc)
GETARCH_FLAGS += -static
endif
+ 3
- 1
getarch.c
View File
@@ -998,7 +998,9 @@ int main(int argc, char *argv[]){
#endif
#endif
#if NO_PARALLEL_MAKE==1
#ifdef MAKE_NB_JOBS
printf("MAKE += -j %d\n", MAKE_NB_JOBS);
#elif NO_PARALLEL_MAKE==1
printf("MAKE += -j 1\n");
#else
#ifndef OS_WINDOWS
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