bash set -e 导致for循环随机退出

Question

当我使用 set -e 选项时，下面的简化循环随机退出。如果我删除 set -e 选项，它总是会完成。如果可能的话，我想使用 set -e 选项，但到目前为止，我对它退出的原因以及每次我运行它时随机循环迭代发生的原因感到茫然（试试吧！）。如您所见，仅有的命令是 let 和 echo。为什么 let 或 echo 命令 return 会随机出现一个非零代码，或者是其他原因？

#!/bin/bash

#   Do Release configuration builds so we can set the build parameters

set -e
CFG=Release

for CASE in {0..511}
do

#   CASE [0...511] iterate
#   MMMM [2...255] random test cases
#   NNNN [1..MMMM) random test cases
#   RRRR [0...255] random test cases
#   XXXX [0...255] random test cases
#   DSXX [1...128] random test cases
#   OASM [1...255] random test cases
#   OLSM [1...255] random test cases

    let "MMMM = $RANDOM % 254 + 2"
    let "NNNN = $RANDOM % ($MMMM - 1) + 1"
    let "RRRR = $RANDOM % 256"
    let "XXXX = $RANDOM % 256"
    let "DSXX = $RANDOM % 128 + 1"
    let "OASM = $RANDOM % 255 + 1"
    let "OLSM = $RANDOM % 255 + 1"

    echo CFG = $CFG, CASE = $CASE, MMMM = $MMMM, NNNN = $NNNN, RRRR = $RRRR, XXXX = $XXXX, DSXX = $DSXX, OASM = $OASM, and OLSM = $OLSM

#   Some other stuff (build and test), that is not causing the problem, goes here

done

#   Some other stuff, that is not causing the problem, goes here

exit 0

Answer 1

将 || true 附加到您的 let 命令或使用 $((...)) 进行计算。

来自help let：

Exit Status: If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise.

bash set -e 导致for循环随机退出

bash set -e causing random exit from for loop

bash

random

set

echo

let