问题
In a Bash script I am using a simple for loop, that looks like:
for i in $(seq 1 1 500); do
echo $i
done
This for loop works fine. However, when I would like to use a sequence of larger numbers (e.g. 10^8 to 10^12), the loop won't seem to start.
for i in $(seq 100000000 1 1000000000000); do
echo $i
done
I cannot imagine, that these numbers are too large to handle. So my question: am I mistaken? Or might there be another problem?
回答1:
The problem is that $(seq ...) is expanded into a list of words before the loop is executed. So your initial command is something like:
for i in 100000000 100000001 100000002 # all the way up to 1000000000000!
The result is much too long, which is what causes the error.
One possible solution would be to use a different style of loop:
for (( i = 100000000; i <= 1000000000000; i++ )) do
echo "$i"
done
This "C-style" construct uses a termination condition, rather than iterating over a literal list of words.
Portable style, for POSIX shells:
i=100000000
while [ $i -le 1000000000000 ]; do
echo "$i"
i=$(( i + 1 ))
done
来源：https://stackoverflow.com/questions/52609966/for-loop-over-sequence-of-large-numbers-in-bash