bash 1：找不到命令
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1: command not found
提问by mcandre
I'm writing a divides-by-three function in Bash, and it won't let me set a variable to a number.
我正在 Bash 中编写一个除以三的函数，它不会让我将变量设置为数字。
fizzy.sh:
嘶嘶声：
#!/usr/bin/env sh
div3() {
return ` % 3 -eq 0`
}
d=div3 1
echo $d
Example:
例子：
$ ./fizzy.sh
./fizzy.sh: line 7: 1: command not found
回答by Keith Thompson
Bash functions normally "return" values by printing them to standard output, where the caller can capture them using
Bash 函数通常通过将值打印到标准输出来“返回”值，调用者可以使用
`func args ...`
or
或者
$(func args ...)
This makes functions work like external commands.
这使得函数像外部命令一样工作。
The returnstatement, on the other hand, sets the value of $?. Normally that's going to be set to 0 for success, 1 for failure, or some other value for a specified kind of failure. Think of it as a status code, not a general value. It's likely that this will only support values from 0 to 255.
return另一方面，该语句设置 的值$?。通常，它会被设置为 0 表示成功，1 表示失败，或者其他一些指定类型的失败值。将其视为状态代码，而不是一般值。这很可能只支持 0 到 255 之间的值。
Try this instead:
试试这个：
#!/bin/sh
div3() {
expr % 3 = 0 # prints "0" or "1"
}
d=$(div3 1)
echo $d
Note that I've also changed the shebang linefrom #!/usr/bin/env shto #!/bin/sh. The #!/usr/bin/envtrick is often used when invoking an interpreter (such as perl) that you want to locate via $PATH. But in this case, shwill alwaysbe in /bin/sh(the system would break in various ways if it weren't). The only reason to write #!/usr/bin/env shwould be if you wanted to use whatever shcommand happens to appear first in your $PATHrather than the standard one. Even in that case you're probably better of specifying the path to shdirectly.
请注意，我还将shebang 行从更改#!/usr/bin/env sh为#!/bin/sh。#!/usr/bin/env在调用perl要通过 定位的解释器（例如）时，经常使用该技巧$PATH。但在这种情况下，sh将始终存在/bin/sh（如果不是，系统会以各种方式中断）。编写的唯一原因#!/usr/bin/env sh是，如果您想使用sh出现在您的$PATH而不是标准命令中的任何命令。即使在这种情况下，您最好sh直接指定路径。
回答by thiton
The
这
d=div3 1
line is the culprit because you assign the string div3 to the env variable d and try to execute 1. To avoid this, use backticks to assign the result of the evaluation instead:
line 是罪魁祸首，因为您将字符串 div3 分配给 env 变量 d 并尝试执行 1。为避免这种情况，请使用反引号来分配评估结果：
d=`div3 1`
Then, another error occurs in your evaluation function. You need to run test on your arguments instead of trying to evaluate them as a command:
然后，您的评估函数中出现另一个错误。您需要对您的参数运行测试，而不是尝试将它们作为命令进行评估：
return `test $(( % 3)) -eq 0`
Still no output, but no errors either. What would your expected output be, actually?
仍然没有输出，但也没有错误。实际上，您的预期输出是什么？
回答by shellter
div3() {
return $(( ( % 3 ) == 0 ))
}
d=$(div3 1)
rc=$?
echo $d
echo $rc
(blank line)
0
Note, return just sets the value of $? for the function to return value, just print it.
注意， return 只是设置 $? 对于返回值的函数，只需打印它。
div3() {
printf $(( ( % 3 ) == 0 )) "\n"
}
