Bash / SH，Same命令输出不同？

$ cat a.sh
#!/bin/bash

echo -n "apple" | shasum -a 256

$ sh -x a.sh
+ echo -n apple
+ shasum -a 256
d9d20ed0e313ce50526de6185500439af174bf56be623f1c5fe74fbb73b60972  -
$ bash -x a.sh
+ echo -n apple
+ shasum -a 256
3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b  -

And the last one is correct. Why is that? and how to solve it?

最后一个是正确的。这是为什么？以及如何解决？

1 个解决方案

#1

Per POSIX, echo supports no options.

根据POSIX，echo不支持任何选项。

Therefore, when echo -n is run with sh, it outputs literal -n instead of interpreting -n as the no-trailing-newline option:

因此，当echo -n与sh一起运行时，它输出literal -n而不是解释-n作为no-trailing-newline选项：

$ sh -c 'echo -n "apple"'
-n apple                  # !! Note the -n at the beginning.

^{Note: Not all sh implementations behave this way; some, such as on Ubuntu (where dash acts as sh), do support the -n option, but the point is that you cannot rely on that, if your code must run on multiple platforms.}

注意：并非所有sh实现都以这种方式运行;一些，比如Ubuntu（其中dash用作sh），确实支持-n选项，但重点是如果你的代码必须在多个平台上运行，你就不能依赖它。

The portable POSIX-compliant way to print to stdout is to use the printf utility:

符合POSIX标准的便携式打印到stdout的方法是使用printf实用程序：

printf %s "apple" | shasum -a 256

#1