Re: calculator script

The problem is ,in the following .

$nbArguments = $#ARGV + 1; 
print "number of arguments: $nbArgumentsn"; 
exit(1) unless $nbArguments == 5;
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Here you are checking the number of arguments is five.
So it will fail if you are giving input in the following format.
1 + 2 3
So that time you need to check the number of arguments is equal to four.
If you want to give input like,
1 + 2 + 3
then that time you need to check the number of arguments is equal to five.And you need to change the following part of code also.
your code,

$a = $ARGV[0]; 
$operation = $ARGV[1]; 
$b = $ARGV[2]; 
$c = $ARGV[3]
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Change into,

$val1 = $ARGV[0]; 
$operation1 = $ARGV[1]; 
$val2 = $ARGV[2]; 
$operation2 = $ARGV[3]
$val3=$ARGV[4]
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These changes are needed only in the case of if you are giving input in the format 1 + 2 + 3 .

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