Makefile 依赖中的变量展开问题 - V2EX
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liyongjun0803
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Makefile 依赖中的变量展开问题

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  •   liyongjun0803 2020-12-16 17:16:39 +08:00 1229 次点击
    这是一个创建于 1762 天前的主题,其中的信息可能已经有所发展或是发生改变。

    Makefile:

    TARGET = test1 TARGET += test3 all: $(TARGET) @echo $^ @echo $(TARGET) test1: test3: TARGET += test2 

    执行结果:

    $ make test1 test3 test1 test3 test2 

    Makefile 中,等号赋值的变量,会在整个 Makefile 文件展开后,再决定变量的值,如 @echo $(TARGET) 打印 test1 test3 test2 。
    但是,不解的是,为什么依赖中的 TARGET 仅仅为 test1 test3 ?

    2 条回复    2020-12-17 09:26:24 +08:00
    SJ2050cn
        1
    SJ2050cn  
       2020-12-16 18:04:51 +08:00   1
    make 会先从上向下执行所有没缩进的语句,注意,是从头上下,在执行依赖项的时候,后面的赋值语句还没执行,接着,会执行缩进中的内容,所以你的 @echo $(TARGET)会执行全部赋值完后的结果。我们可以做一个小实验,
    ```
    Ranep = test1
    TARGET = $(Ranep) test3


    all: $(TARGET)
    @echo $^
    @echo $(TARGET)
    test1:
    test2:
    test3:

    Ranep = test2
    ```
    你可以看看输出结果你就知道了。总之,就是 make 会先执行没缩进的,再执行缩进的。
    liyongjun0803
        2
    liyongjun0803  
    OP
       2020-12-17 09:26:24 +08:00
    @SJ2050cn 学到了,感谢!
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