bash, bc modulo 不适用于 -l 标志
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bash, bc modulo does not work with -l flag
提问by Saad Attieh
So I am trying to use bcto calculate some logarithms but I also need to use it to calculate the modulus for something. Whilst making my script, I launched bcto test it.
所以我试图用它bc来计算一些对数，但我也需要用它来计算某些东西的模数。在制作我的脚本时，我开始bc测试它。
Without any flags, bc <<< "3%5"of course returns 3.
没有任何标志，bc <<< "3%5"当然返回3.
But with bc -l(loads math library so I can compute logarithms) any calculation of a%breturns 0where aand bcan be any number but 0.
但是使用bc -l（加载数学库以便我可以计算对数）任何a%b返回的计算，0其中a和b可以是任何数字，但0.
What's happening?
发生了什么？
回答by gniourf_gniourf
That's because, from the manual:
那是因为，从手册中：
expr % expr
The result of the expression is the "remainder" and it is com‐
puted in the following way. To compute a%b, first a/b is com‐
puted to scale digits. That result is used to compute a-(a/b)*b
to the scale of the maximum of scale+scale(b) and scale(a). If
scale is set to zero and both expressions are integers this
expression is the integer remainder function.
When you run bcwith the -lflag, scaleis set to 20. To fix this:
当您bc使用-l标志运行时，scale设置为20。要解决此问题：
bc -l <<< "oldscale=scale; scale=0; 3%5; scale=oldscale; l(2)"
We first save scalein variable oldscale, then set scaleto 0to perform some arithmetic operations, and to compute a lnwe set scaleback to its old value. This will output:
我们首先保存scale在变量中oldscale，然后设置scale为0执行一些算术运算，并计算ln我们设置scale回它的旧值。这将输出：
3
.69314718055994530941
as wanted.
随心所欲。
回答by afenster
According to the bcmanual,
根据bc手册，
expr % expr
The result of the expression is the "remainder" and it is computed
in the following way. To compute a%b, first a/b is computed to
scale digits. That result is used to compute a-(a/b)*b to the
scale of the maximum of scale+scale(b) and scale(a). If scale is
set to zero and both expressions are integers this expression is
the integer remainder function.
So what happens is that it tries to evaluate a-(a/b)*busing the current scalesettings. The default scaleis 0 so you get the remainder. When you run bc -lyou get scale=20and the expression a-(a/b)*bevaluates to zero when using 20 fractional digits.
因此，它会尝试a-(a/b)*b使用当前scale设置进行评估。默认scale值为 0，因此您可以获得余数。当您运行时，当您使用 20 个小数位时，bc -l您会得到scale=20并且表达式的a-(a/b)*b计算结果为零。
To see how it works, try some other fractions:
要查看它是如何工作的，请尝试其他分数：
$ bc -l
1%3
.00000000000000000001
To make a long story short, just compare three outputs:
长话短说，只需比较三个输出：
Default scalewith -lenabled (20):
默认scale以-l启用（20）：
scale
20
3%5
0
1%4
0
Let's set scaleto 1:
让我们设置scale为 1：
scale=1
3%5
0
1%4
.2
Or to zero (default without -l):
或为零（默认没有-l）：
scale=0
3%5
3
1%4
1
回答by Thor
You could define a function that works in math mode by temporarily setting scaleto zero.
您可以通过临时设置scale为零来定义一个在数学模式下工作的函数。
I have bcaliased like this:
我有这样的bc别名：
alias bc='bc -l ~/.bcrc'
Thus ~/.bcrcis evaluated before any other expressions, so you can define functions in ~/.bcrc. For example a modulus function:
因此~/.bcrc在任何其他表达式之前计算，因此您可以在~/.bcrc. 例如模数函数：
define mod(x,y) {
tmp = scale
scale = 0
ret = x%y
scale = tmp
return ret
}
Now you can do modulo like this:
现在你可以像这样做模：
echo 'mod(5,2)' | bc
Output:
输出：
1
回答by mpiffault
man bc :
男子公元前：
If bc is invoked with the -l option, a math library is preloaded and the default scale is set to 20.
如果使用 -l 选项调用 bc，则会预加载数学库并将默认比例设置为 20。
So maybe you should set the scale to 0 :
所以也许你应该将比例设置为 0 ：
#bc
scale=0
10%3
1
回答by Unbeliever
For what it's worth, when I use bc -l, I have the following functions defined:
对于它的价值，当我使用时bc -l，我定义了以下函数：
define trunc(x) {auto s; s=scale; scale=0; x=x/1; scale=s; return x}
define mod(x,y) {return x-(y*trunc(x/y))}
That should give you a proper MODfunction, while keeping your scale intact. Of course, it won't help if you NEED to use the %operator for some reason.
这应该给你一个适当的MOD功能，同时保持你的秤完好无损。当然，如果您%出于某种原因需要使用运算符，那也无济于事。
(That TRUNCfunction is quite handy too, forming the basis for many other useful functions that are outside the scope of this answer.)
（该TRUNC功能也非常方便，构成了本答案范围之外的许多其他有用功能的基础。）
