Question

MAC_BOOST_PATH = -L/opt/local/lib
LINUX_BOOST_PATH = -L/usr/lib/
DEFAULT_PATH = -L/usr/local/lib
BOOST_PATH = $(DEFAULT_PATH)

ifeq ($(UNAME), Darwin)
BOOST_PATH = MAC_BOOST_PATH
@echo Compiling for Mac OS X
@echo 
endif
ifeq ($(UNAME), Linux)
BOOST_PATH = LINUX_BOOST_PATH
@echo Compiling for Linux
@echo 
endif

echo不是打印，BOOST_PATH没有改变，我不认为......所以......我不确定我在这里做错了什么... = \

Answer 1

您没有在任何地方定义UNAME变量。你可能想要这样的东西：

UNAME = $(shell uname)

Answer 2

您不能将命令放在Makefile中，与目标无关。您需要引入一个显示操作系统的目标。而且，你缺少'$（）'。使用例如

UNAME=$(shell uname)
MAC_BOOST_PATH = -L/opt/local/lib
LINUX_BOOST_PATH = -L/usr/lib/
DEFAULT_PATH = -L/usr/local/lib
BOOST_PATH = $(DEFAULT_PATH)

ifeq ($(UNAME),Darwin)
BOOST_PATH=$(MAC_BOOST_PATH)
endif
ifeq ($(UNAME),Linux)
BOOST_PATH=$(LINUX_BOOST_PATH)
endif

all: showos

showos: 
  @echo compiling for $(UNAME)

Answer 3

我认为您需要移除@符号才能显示回声

Answer 4

正如Erik所指出的，Make不会执行独立于目标的命令。但它可以评估没有目标的函数：

UNAME = whatever

ifeq ($(UNAME), Darwin)
BOOST_PATH = MAC_BOOST_PATH
$(info Compiling for Mac OS X)
$(info )  # note the space
endif
ifeq ($(UNAME), Linux)
BOOST_PATH = LINUX_BOOST_PATH
$(info Compiling for Linux)
$(info )
endif

Makefile：if语句工作的问题=（

4 个答案: