Makefile - 使用 ifeq 的运行时决策

Question

我正在尝试解决一个特定问题，其中变量在一个配方中分配，然后在其他配方中进行解释，所有这些都是在运行时进行的。据我了解， ifeq 条件在解析过程中进行评估，这对我不起作用，因为其中一些总是错误的。有没有办法实现我正在尝试做的事情（预期输出如下）？如果需要，我会提供更多信息。

我在 Linux Mint 17.1 上使用 make 版本 3.81。

这是我目前所拥有的：

fourth =

all: check valueOfFourth definitionOfFourth
.PHONY: all

check:
    @echo "TEST"$(cnt)
ifeq ($(first),$(second))
    @echo "1. First condition"
    $(eval fourth = "first")
else ifeq ($(first),$(third))
    @echo "1. Second condition"
    $(eval fourth = "second")
else
    @echo "1. Conditions weren't met"
endif

valueOfFourth:
ifeq ($(fourth),"first")
    @echo "2. First"
else ifeq ($(fourth),"second")
    @echo "2. Second"
else
    @echo "2."
endif

definitionOfFourth:
ifeq ($(fourth),)
    @echo "3. Variable is not defined"
else
    @echo "3. Variable is defined"
endif

它是这样调用的：

make cnt="1" first="x" second="x" third="y" && printf "\n" && \
make cnt="2" first="x" second="y" third="x" && printf "\n" && \
make cnt="3" first="x" second="y" third="z"

预期输出：

TEST1
1. First condition
2. First
3. Variable is defined

TEST2
1. Second condition
2. Second
3. Variable is defined

TEST3
1. Conditions weren't met
2. 
3. Variable is not defined

实际输出：

TEST1
1. First condition
2. 
3. Variable is not defined

TEST2
1. Second condition
2. 
3. Variable is not defined

TEST3
1. Conditions weren't met
2. 
3. Variable is not defined

很明显，只有“检查”目标做了它应该做的事情，其他两个根本不起作用。

Answer 1

我仍然不完全清楚这些目标应该如何交互（它们之间的任何交互通常都是一个坏主意，因为并行执行意味着如果没有通过它们之间的先决条件显式排序，则无法保证执行顺序）。但是假设非并行 make 并且每个目标都应该输出其中一条输出线，我相信这可以满足您的需求。

.PHONY: all
all: check valueOfFourth definitionOfFourth

ifeq ($(first),$(second))
    fourth = First
    condmsg = $(fourth) condition
else ifeq ($(first),$(third))
    fourth = Second
    condmsg = $(fourth) condition
else
    condmsg = Conditions weren'\''t met
endif

check:
    @echo 'TEST$(cnt)'
    @echo '1. $(condmsg)'

valueOfFourth:
    @echo '2. $(fourth)'

definitionOfFourth:
ifeq ($(fourth),)
    @echo "3. Variable is not defined"
else
    @echo "3. Variable is defined"
endif