(standard_in) 1: bash 脚本中的语法错误
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(standard_in) 1: syntax error in bash script
提问by Reluctant_Linux_User
I'm trying to generate some quasi random numbers to feed into a monte carlo simulation. I'm using bash. I seem to have hit a syntax error which I've narrowed down to being in this bit of code.
我正在尝试生成一些准随机数以输入蒙特卡罗模拟。我正在使用 bash。我似乎遇到了一个语法错误，我已将其范围缩小到这段代码中。
randno4=($RANDOM % 100001)
upper_limit4=$(echo "scale=10; 1*75.3689"|bc)
lower_limit4=$(echo "scale=10; 1*75.1689"|bc)
range4=$(echo "scale=10; $upper_limit4-$lower_limit4"|bc)
t_twall=`echo "scale=10; ${lower_limit4}+${range4}*${randno3}/100001" |bc`
echo "$t_twall"
Does anyone know why I the below output and not a value between 75.3689 and 75.1689 as that is what I would be expecting?
有谁知道为什么我在下面的输出而不是 75.3689 和 75.1689 之间的值，因为这是我所期望的？
(standard_in) 1: syntax error
回答by Gilles Quenot
The first line should looks like :
第一行应如下所示：
randno4=$((RANDOM % 100001))
(( ))is bash arithmetic, with the leading $, the value is substituted : $(( ))
(( ))是 bash 算术，以 开头$，值被替换：$(( ))
When you wrote
当你写
randno4=( )
you try to feed an ARRAY with a arithmetic expression with the wrong syntax.
您尝试使用语法错误的算术表达式为 ARRAY 提供数据。
See http://wiki.bash-hackers.org/syntax/arith_expr
见http://wiki.bash-hackers.org/syntax/arith_expr
And finally, like Etan Reisnersaid, You also use $randno3 in the t_twall assignment line which is undefined
最后，就像Etan Reisner所说的，您还在未定义的 t_twall 分配行中使用了 $randno3
