bash脚本上的奇怪错误

Question

警告：我最近才刚开始学习bash，并尝试做一个可计算项的递归函数...所以...

x0 = 0 x1 = 1 xm = 3 * xm-1-2 * xm-2

我到目前为止编写的函数是：

#!/bin/bash
calculate()
{
 if [ $1 -eq 0 ]
 then
 echo "0"
 fi
 if [ $1 -eq 1 ]
 then
 echo "1"
 fi
 if [ $1 -ge 1 ]
 then
 let var1 = `calculate [ $1-1 ]`;
 let var2 = `calculate [ $1-2 ]`;
 let var3 = 3*var1-2*var2;
 echo var3
fi
}
calculate 3

但是我遇到了一些奇怪的错误……不确定我是否正确执行了……谁能告诉我是什么原因导致这些问题并更正了我的代码，从而可以正常工作？ 非常感谢。

错误：

TP1p1.sh: line 4: [: [: integer expression expected
TP1p1.sh: line 8: [: [: integer expression expected
TP1p1.sh: line 12: [: [: integer expression expected
TP1p1.sh: line 14: let: =: syntax error: operand expected (error token is "=")
TP1p1.sh: line 4: [: [: integer expression expected
TP1p1.sh: line 8: [: [: integer expression expected
TP1p1.sh: line 12: [: [: integer expression expected
TP1p1.sh: line 15: let: =: syntax error: operand expected (error token is "=")
TP1p1.sh: line 16: let: =: syntax error: operand expected (error token is "=")

Answer 1

不太确定您的计算结果，但是您在语法上清理的基本脚本就是这个：

#!/bin/bash
calculate() {
 if [ $1 -eq 0 ]; then
    echo -n "0"
 elif [ $1 -eq 1 ]; then
    echo -n "1"
 elif [ $1 -ge 1 ]; then
    var1=$( calculate $(($1-1)) )
    var2=$( calculate $(($1-2)) )
    var3=$((3*(var1-2)*var2))
    echo $var3
fi
}
calculate 5