How to use bc in a bash while loop condition statement? - bash

Question

I'm writing what should be a simple bash script to calculate the minimum value p needs to be in the formula C=1-(p^n). C and p are input by the user and are floating point numbers.
For example if C=0.36 and p=0.8, then my program should return 2 for the value of n, since 1-(0.8^2)=0.36
My code is below. Please note: I have put |bc in many different places in my while loop condition statement and each time I get an error. The last place I tried was, as you can see, between the two closing brackets.
I receive the following error:
operand expected (error token is ".36")
#!/bash/bin
echo "Enter I/O fraction time:"
read ioINPUT
echo "Enter expect CPU utilization:"
read uINPUT
n=1
function cpuUTIL {
x='scale=3; 1-$ioINPUT**$n'|bc
}
while [[ cpuUTIL -lt $uINPUT ]|bc]; do
n+=1
x='scale=3; 1-$ioINPUT**$n'|bc
done
echo $c
How do I properly use bc in a bash script while loop condition statement?

Answer 1

A few issues have been pointed out in comments. Here is a fixed up version of what you were trying:
#!/bin/bash
cpuUTIL () {
bc <<< "scale = 3; 1 - $ioINPUT ^ $n"
}
read -p "Enter I/O fraction time: " ioINPUT
read -p "Enter expect CPU utilization: " uINPUT
n=1
while (( $(bc <<< "$(cpuUTIL) < $uINPUT") )); do
(( ++n ))
x=$(bc <<< "scale = 3; 1 - $ioINPUT ^ $n")
done
echo $x
Comments:
#!/bin/bash, not #!/bash/bin
echo and read can be shortened to read -p
function cpuUTIL is less portable than cpuUTIL ()
Single quotes prevent parameter expansion; you have to use double quotes for your bc commands
I've used here-strings (<<<) instead of pipes to avoid creating a subshell
"Power of" in bc is ^ and not **
The while condition is complex: is uses nested command substitution within (( )), which translates the output of the bc statement (1 if the comparison holds, 0 if it doesn't) to true/false for Bash
n+=1 appends the string 1 to the contents of variable n (unless it was declared as an integer with declare -i, but that's rather exotic), so you have to use something like an arithmetic context, (( ++n ))
The final statement should probably be echo $x
But, as demonstrated by karakfa's answer, you can do it directly. In bc, it would look like this:
$ p=0.8
$ C=0.36
$ bc -l <<< "scale = 3; l(1-$C) / l($p)"
2.000
bc -l defines the math library, which contains (among a few others) the natural logarithm l(x).

Answer 2

you can iterate and find a value, but why? here is a better method
awk -v p=0.8 -v C=0.36 'BEGIN{print log(1-C)/log(p)}'
also the value will be, in general floating point, you can get the floor by wrapping the log ratio with int()

Answer 3

Using GNU bc:
#!/usr/bin/bc -ql
scale=3
print "Enter I/O fraction time: "
p=read()
print "Enter expected CPU utilization: "
c=read()
l(1-c)/l(p)
quit
Save the above to foo.bc, run chmod +x foo.bc, then execute with ./foo.bc.
Or without prompts, from bash:
bc -l <<< "p=0.8;c=0.36;scale=3;l(1-c)/l(p)"