我在想编译汇编文件是否有一些等效的CPPPATH
?Scons;汇编器的构造变量包括目录;
CPPPATH=include_dirs
完全适用于.cpp
文件,但它似乎没有申请.S
文件。我得到像“没有这样的文件或目录”的错误信息
那么是否有某种'ASPATH'?
感谢
PS:至于解决方法目前我用ASFLAGS
与-I/my_path
,但我真的不喜欢它...
新增:
这种方法适用(ASFLAGS)
asm_dirs = '-I' + gen_build_dir + gen_file_dir + ' ' #relative path here
as_flags = asm_dirs + env['ASFLAGS']
t = env.Program('boot.elf', sources, CPPPATH=include_dirs, LIBPATH=lib_dirs, LIBS=libs, ASFLAGS=as_flags,
LINKFLAGS=link_flags)
这种方法没有(CPPPATH)
include_dirs += [ #add path to 'CPPPATH'
gen_build_dir + gen_file_dir + ' '
]
t = env.Program('boot.elf', sources, CPPPATH=include_dirs, LIBPATH=lib_dirs, LIBS=libs, LINKFLAGS=link_flags)
详细的脚本:
asm_cppflags = ('-S -Wno-invalid-offsetof ')
gen_build_dir = 'build/Italy4K6/rel/'
gen_file_dir = 'OS/UserAccess/'
gen_file_name = 'UserAccess_HandleSysCall_ConstGen'
src_full_name = gen_file_dir + gen_file_name + '.cpp'
obj_full_name = gen_build_dir + gen_file_dir + gen_file_name + '.o'
dst_full_name = gen_build_dir + gen_file_dir + gen_file_name + '.h'
a = env.Object(src_full_name, CPPFLAGS=asm_cppflags, CPPPATH=include_dirs)
gen = env.Command(target = "HEADER", source = "",
action = "cat " + obj_full_name + " | " + gen_script + " > " + dst_full_name)
Depends(gen, a)
asm_dirs = '-I' + gen_build_dir + gen_file_dir + ' '
as_flags = asm_dirs + env['ASFLAGS']
link_script = 'Targets/LinkerScripts/%s.ld' % product
link_flags = ('-nodefaultlibs -nostartfiles -Wl,-T -Wl,%s -Wl,-Map -Wl,LINK.MAP ' % link_script) + env['LINKFLAGS']
t = env.Program('boot.elf', sources, CPPPATH=include_dirs, LIBPATH=lib_dirs, LIBS=libs, ASFLAGS=as_flags,
LINKFLAGS=link_flags)
Depends(t, link_script)
您正在使用哪种汇编程序?哪个平台? – bdbaddog
汇编ARM,GNU GCC工具链,在mac os下。 – user3124812