Question

If I try to print integers that are 6 or less digits long everything works perfect:

Example

for i in $(seq 123450 123451); do printf "%7.0f\n" $i; done

123450

123451

However, when integers are 7 or more digits long the last digit of the number is being output as ‘0’:

Example

for i in $(seq 1234560 1234561); do printf "%7.0f\n" $i; done

1234560

1234560

Question: what problem do I encounter and what should I do to be able to use “long” integers?

Answer 1

By default, seq prints its output in %g format:

$ seq 1234560 1234561
1.23456e+06
1.23456e+06

so use seq -f "%7.0f" instead:

$ seq -f "%7.0f" 1234560 1234561
1234560
1234561

or just seq -f "%.0f":

$ seq -f "%.0f" 1234560 1234561
1234560
1234561