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发表于 2011-11-18 09:32:41
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这压根和&&及||毫无关系,它们就是按照你想的那样短路求值。
你咋不看看let a++的返回码咧?- $ true
- $ echo $?
- 0
- $ a=0
- $ let a++
- $ echo $?
- 1
复制代码 为神马?因为a=0,所以a++求值为0,然后你看bash(1) let arg [arg ...]
Each arg is an arithmetic expression to be evaluated (see ARITHMETIC EVALUATION above). If the last arg evaluates to 0, let returns 1; 0
is returned otherwise. |
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