Question

在Bash陷阱功能中使用内置调用者时，caller 0的结果给出错误的行号，始终给出1。例如：

#!/bin/bash
function foo {
    exit 1
}
function bar {
    foo
}
function err {
    (( i = 0 ))
    while caller $i; do
        (( ++i ))
    done
}
trap err EXIT
bar

给出以下输出：

1 foo ./test.sh
6 bar ./test.sh
15 main ./test.sh

虽然i > 0的输出是正确的，但在陷阱处理程序中使用caller 0时，似乎总是将1作为行号。有没有办法从陷阱处理程序中获取失败函数的实际行号？

Answer 1

这似乎是在3.2.57（1）之后的某个时间引入的错误 - 发布：

$ bash -version
GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin14)
Copyright (C) 2007 Free Software Foundation, Inc.

$ bash ./test.sh
3 foo ./test.sh
6 bar ./test.sh
15 main ./test.sh

$ /usr/local/bin/bash --version
GNU bash, version 4.3.30(1)-release (x86_64-apple-darwin14.1.0)
Copyright (C) 2013 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

$ /usr/local/bin/bash ./test.sh
1 foo ./test.sh
6 bar ./test.sh
15 main ./test.sh

bash项目似乎已经有bug report。

调用者0在陷阱处理程序中给出错误的行号

1 个答案: