Argc and Argv in Bash
23,335
Ok, try like this
define a variable ARGC=$#
and you if statement will look like
if [ $ARGC -ne $MAX_ARGS ]; then
Legend:
-ne = not equal
-gt = greater than
-eq = equal to
Author by
Admin
Updated on July 09, 2022Comments
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Admin almost 2 years
I written the following script which gets name of a file and then assemble and link the file. But it doesn't work. What is the problem with it?
EXPECTED_ARGS=2 if [ $# -ne $EXPECTED_ARGS ] then echo "[+] Assembling with Nasm." nasm -f elf32 $1 -o $1.o echo "[+] Linking ..." ld $1.o -o $1 echo "[+] Done!" else printf "\nInvalid number of arguments, please check the inputs and try again\n" fi;
When I run it without passing any args, it doesn't shows following error:
printf "\nInvalid number of arguments, please check the inputs and try again\n"