向变量加1不能按预期工作（Bash算法）

如果我将以下内容写入bash终端：

A="0012"
B=$((A+1))
echo $B

我得到11，而不是我期望的13！

我已经四处搜寻，无法解释，也无法弄清楚如何增加数字。（我实际上想以B =“ 0013”结尾，并且每次将其用作备份前缀时都加1）

这是因为以开头的数字0被视为八进制bash，因此它正在做八进制（基数8）加法。要获得此结构的十进制加法，您需要显式定义Base或不一起使用00。

对于十进制，底数为10，表示为10#：

$ A="10#0012"
$ echo $((A+1))
13

您可以尝试使用以下命令获得答案：

A="0012"
echo $A + 1 | bc

有关bc命令的更多信息，请参见此处。

bc 手册页：

NAME
       bc - An arbitrary precision calculator language

SYNTAX
       bc [ -hlwsqv ] [long-options] [  file ... ]

DESCRIPTION
       bc is a language that supports arbitrary precision numbers with interactive execution of statements.  There are some similarities
       in the syntax to the C programming language.  A standard math library is available by command line  option.   If  requested,  the
       math  library is defined before processing any files.  bc starts by processing code from all the files listed on the command line
       in the order listed.  After all files have been processed, bc reads from the standard input.  All code is executed as it is read.
       (If a file contains a command to halt the processor, bc will never read from the standard input.)

       This  version of bc contains several extensions beyond traditional bc implementations and the POSIX draft standard.  Command line
       options can cause these extensions to print a warning or to be rejected.  This document describes the language accepted  by  this
       processor.  Extensions will be identified as such.

另一种方法是将变量保留为整数，并在最后将它们转换为字符串：

A=12
B=$((A+1))
echo $B
13
C=$( printf '%04d' $B )
echo $C
0013

在我习惯于BASIC编程时，这种在数学中使用整数并将其转换为答案的字符串的方式对我来说更加直观。我很欣赏Bash没有像C和BASIC这样的变量类型，但是假装它确实让我高兴。