Problem:
Ask user how many number he wants to add and then add them.
Sample Input:
Taken as <STDIN>
Sample Output:
Sum is xyz
Solution:
print "How many numbers you want to add\n";
my $len = <STDIN>;
my $tmp = 1;
my $sum = 0;
while($tmp <= $len)
{
print "Enter number $tmp\n";
my $var = <STDIN> ;
$sum = $sum+$var;
++$tmp;
}
print "Sum is $sum";
Tips:
Ask user how many number he wants to add and then add them.
Sample Input:
Taken as <STDIN>
Sample Output:
Sum is xyz
Solution:
print "How many numbers you want to add\n";
my $len = <STDIN>;
my $tmp = 1;
my $sum = 0;
while($tmp <= $len)
{
print "Enter number $tmp\n";
my $var = <STDIN> ;
$sum = $sum+$var;
++$tmp;
}
print "Sum is $sum";
Tips:
- <STDIN> is used to take user input and save it in a variable.
Two suggested changes: First, rather than make your user count the number of numbers in their list, just read values until you get a blank line. Second, add a \n to your result print stmt, so that the prompt doesn't print right after your output.
ReplyDeleteMy version:
#! /usr/bin/perl
use strict;
#print "How many numbers you want to add\n";
#my $len = ;
my $tmp = 1;
my $sum = 0;
print "Enter number $tmp\n";
my $var = ;
chomp $var;
while(length($var) > 0)
{
$sum = $sum+$var;
++$tmp;
print "Enter number $tmp\n";
$var = ;
chomp $var;
}
print "Sum is $sum\n";
@robert: problem statement required count of numbers, so i coded this way
DeleteAlso, what's wrong with
ReplyDelete$sum += $var;
@Dave: nothing wrong, i just want to make it simple.
Delete