命令替换不适用于 makefile 中的 echo [英] command substitution doesn't work with echo in makefile

问题描述

这是一个在我的 bash shell 中运行良好的简单命令:

Here's a simple command that works fine in my bash shell:

echo "Created at: $(date)" >> README.md

它将 Created at: Wed Jan 24 10:04:48 STD 2018 附加到 README.md.

It appends Created at: Wed Jan 24 10:04:48 STD 2018 to README.md.

然而，ii 我在我的 makefile 中包含相同的命令，行为是不同的.

However, ii I include the same command in my makefile, the behavior is different.

制作文件:

README.md:
    echo "Created at: $(date)" >> README.md

运行 make README.md 会将命令替换视为一个空字符串，如下所示:

Running make README.md will treat the command substitution as an empty string like this:

echo "Created at: " >> README.md

附加到 README.md 的是 Created at:.

What's appended to README.md is Created at:.

如何使用 makefile 中的 echo 进行命令替换以正确输出?

How do I get command substitution to output properly with echo in a makefile?

推荐答案

如果你想让 Make 调用的 shell 收到以下内容:

If you want the shell that Make invokes to receive the following:

echo "Created at: $(date)" >> README.md

然后，您需要在规则中使用另一个 $ 对 $ 进行转义:

Then, you need to escape the $ with another $ inside the rule:

README.md:
    echo "Created at: $$(date)" >> README.md

否则，Make 的变量 date 被扩展，这将是 echo 作为参数得到的，因为 $(date) 在 makefile 中展开变量 date.

Otherwise, the Make's variable date is expanded and that will be what echo gets as argument, since $(date) in a makefile expands the variable date.

请注意，如果 Make 的变量 date 定义如下，它将按预期工作，而无需在规则中引用 $:

Note that, if the Make's variable date is defined as below, it will however work as expected without quoting the $ in the rule:

date = $$(date)

README.md:
    echo "Created at: $(date)" >> README.md

原因是变量 date(在规则的配方中使用)将被 Make 扩展为 $(date) 并将传递给 shell.

The reason is that the variable date (used in the rule's recipe) will be expanded by Make to $(date) and that will be passed to the shell.

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