问题描述
来自《 Bash初学者指南》表7-1 ，我们可以说，如果 FILE1 早于 FILE2 ，则 [FILE1 -ot FILE2] 将产生 True ，或者如果 FILE2 存在而 FILE1 不存在.如果缺少 FILE1 并且存在 FILE2 ，则在shell上执行此命令将返回 True .但是，当我在make脚本中使用它时，只要不存在 FILE1 且存在 FILE2 ，它就会产生 False .确切地说，我在make和shell脚本中使用了以下命令:
From Table 7-1 of Bash Guide for Beginners, we can say that [ FILE1 -ot FILE2 ] will yield True if FILE1 is older than FILE2, or if FILE2 exists and FILE1 does not. Executing this command on shell returns True if FILE1 is absent and FILE2 is present. However, when I made use of this in my make script, it yielded False whenever FILE1 was absent and FILE2 was present. To be precise, I made use of the following command in my make and shell scripts:
[FILE1 -ot FILE2]&&回声1 ||回声0
在shell上运行此命令时，只要期望它就会返回1.但是，仅当 FILE1 存在并且被发现早于 FILE2 时，通过make运行此命令的结果为1.在其他情况下，例如，每当未发现 FILE1 早于 FILE2 时，或者每当 FILE1 不存在且 FILE2 存在，上面的命令在我的make脚本中产生0.请注意，在我的make脚本中，我将上述命令的结果存储在全局变量中.稍后，该全局变量将在条件之一中用于条件.
Running this command on shell returned 1 whenever it was expected to. However, running this command via make resulted in 1 only when FILE1 existed and was found to be older than FILE2. In other scenarios, i.e. whenever FILE1 was not found to be older than FILE2 OR whenever FILE1 was absent and FILE2 existed, the above command yielded 0 in my make script. Note that in my make script, I'm storing the results of the above command in a global variable. This global variable is later used in one of the recipes by a conditional.
所以，我想知道这是make中的错误/功能还是我的最终可能出现的问题(由于某些拼写错误或其他可能的错误来源)?
So, I wish to know whether this is a bug/feature in make OR a possible issue from my end(due to some typo or other possible sources of error)?
注意:我正在Ubuntu 18.04.3上使用GNU Make 4.1
Note : I'm using GNU Make 4.1 on Ubuntu 18.04.3
推荐答案
尝试此实验(在脚本中):
Try this experiment (in a script):
在我测试的情况下:当file1不存在而当file2存在时:
In the case I tested: when file1 does not exist and when file2 does exist:
#!/bin/sh
[ file1 -ot file2 ] && echo "file1 older" || echo "file2 older"
使文件2变老
#!/bin/bash
[ file1 -ot file2 ] && echo "file1 older" || echo "file2 older"
使文件1变老
默认情况下，make将使用#！/bin/sh作为外壳，您可以通过以下方法对此进行更改: SHELL =/bin/bash 在您的makefile中
By default make will use #!/bin/sh as the shell, you can change this by: SHELL = /bin/bash in your makefile
我不确定为什么会有所不同-它只是sh和bash之间的不同实现方式...
I am not sure why this is different - its just a different implementation between sh and bash...
这篇关于执行[FILE1 -ot FILE2]在make和shell中产生不同的结果的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！