shell command to skip file in sequence

Question

I run a C++ code for several data files in sequence using:

for i in $(seq 0 100); do ./f.out c2.$(($i*40+495)).bin `c2.avg.$(($i*40+495)).txt; done`

Now if some input files are missing, say c2.575.bin is missing then the command is not executed for the rest of the files. How could I modify the shell command to skip the input files those are missing and move to the next input file?

Thanks.

Answer 1

in the loop, test if file exists before calling a program operating on that file:

for i in $(seq 0 100); do
  INPUT=c2.$(($i*40+495)).bin
  test -e $INPUT && ./f.out $INPUT c2.avg.$(($i*40+495)).txt
done

This way the ./f.out ... will be executed only for existing input files.

See man test for details.

BTW, the && notation is a shorthand for if. See help if or man sh.

Answer 2

You can use {0..100} instead of $(seq 0 100) for better readability. You can put the following code in a script and execute the script. e.g., runCode.bash

#!/bin/bash
for i in {0..100}
do
  # Assign a variable for the filenames
  ifn=c2.$(($i*40+495)).bin
  # -s option checks if the file exists and size greater than zero
  if [ -s "${ifn}" ]; then
     ./f.out "${ifn}" c2.avg.$(($i*40+495)).txt
  else
     echo "${ifn} No such file"
  fi
done

Change permission and execute the script.

chmod u+x runCode.bash`
./runCode.bash`