Pass condition into a function - bash

Question

I´m trying to pass to a function a condition to be applied in a while loop, and if it´s true increase a variable of that condition.
If that´s possible on bash?. Here an example code
a(){
condition=$1
echo "condition:$condition"
while [[ $condition ]] # --> run the condition
do
num1=num1+1
echo "increase num1:$num1" # num1 is a var of the condition
done
}
b(){
num1=1
num2=3
`a $num1 < $num2`
}
b

Answer 1

You can also use as below;
#! /bin/bash
a(){
condition="(( \$num $2 $3 ))"
num=$1
while eval $condition
do
(( num++ ))
echo "increase num1:$num"
done
}
b(){
num1=1
num2=5
a $num1 '<' $num2
}
b

Answer 2

Instead of eval, I would just pass the name of a function which a can call.
a () {
while "$1"; do
...
done
}
b(){
num1=1
num2=3
c () (( num1 < num2 ))
a c
}
b
(Keep in mind that functions are always global; c is not local to the body of b.)

Answer 3

You could do something like below :
a(){
num1="$1";
num2="$3";
if [ "$2" = "<" ]
then
op="-lt"
elif [ "$2" = ">" ]
then
op="-gt"
else
op="-eq"
fi
condition="$num1 $op $num2"
while eval "[ $condition ]" # --> evaluate the condition
do
((num1++)) #num1=num1+1 is in the original is wrong.
echo "increased num1:$num1"
condition="$num1 $op $num2" #rebuild the condition
done
}
b(){
num1=1
num2=3
a "$num1" '<' "$num2" #quote the params, else '<' stands for redirection
}
b

Answer 4

You can try with:
#!/bin/bash
a(){
condition="expr $1 > /dev/null"
while eval $condition; do
echo "num1=$num1 num2=$num2"
let num1++
done
}
b(){
num1=1
num2=3
a '$num1 \< $num2'
}
b
remember the command expr 1 < 3 produces an error:
$ expr 1 < 3
bash: 3: No such file or directory
this is the reason to write \< instead of <