I've check to see if the program is recieving number and names and it is. I'm wondering why doesn't it print "fred" when input number is 1?
number = ARGF.read.chomp
names = %w{fred betty barney}
if number == 1
puts names[0]
elsif number == 2
puts names[1]
elsif number == 3
puts name[2]
end
number is likely a string here, but you are comparing it with an integer.
1 == '1' # false
try
number = ARGF.read.chomp.to_i # note the to_i here
names = %w{fred betty barney}
if number == 1
puts names[0]
elsif number == 2
puts names[1]
elsif number == 3
puts names[2]
end
Also, you can use a case/when statement when you want to take a different path based on multiple values of a single variable. This is usually the cleaner way to handle this type of flow.
number = ARGF.read.chomp.to_i
names = %w{fred betty barney}
case number
when 1
puts names[0]
when 2
puts names[1]
when 3
puts names[2]
end
Or in this case, the even more simple:
number = ARGF.read.chomp.to_i
names = %w{fred betty barney}
puts names[number-1]
should work.
I know you're learning if...else, but keep in mind you can also do this:
number = ARGF.read.chomp.to_i - 1
names = %w{fred betty barney}
puts names[number]
number is a string, not an integer. You can either convert to an integer:
number = ARGF.read.chomp.to_i
or you can test against strings instead:
if number == "1"
...
...
...
The number is a string. You can check that that by printing its class like this:
p number.class
You need to convert number to an integer like this:
number = ARGF.read.chomp.to_i
Keep in mind though that to_i would return 0 for invalid string. Do it only when you are sure about the incoming data.
Try this on Codepad.
Related
This code is supposed to update the total within the loop, but when I put the total when its broken it doesn't update.
total = 0
while true do
puts "Give me a number"
input = gets.chomp
if input == Integer
total += input
elsif input == "stop"
puts total
break
end
end
input = gets.chomp will result String class. So your logic on if input == Integer it will never be reached. you need to convert it to integer using to_i and input == Integer i never used that kind of syntax to check the classes, i rather use input.is_a?(String). but if you convert to integer first it will never check stop string condition. so maybe
total = 0
while true do
puts "Give me a number"
input = gets.chomp
if input == "stop"
puts total
break
end
total += input.to_i
end
As mentioned in the above comment by mu is too short and dedypuji's answer you have a couple of issue. Here is another variation that I think will work and I think is a little more ruby idiomatic.
total = 0
loop do
print "Give me a number: "
input = gets
break if /stop|^$/ =~ input
total += input.to_i
end
puts total
I have this loop:
puts "Welcome to the Loop Practice Problems"
puts " Write a number between 1 and 10, but not 5 or else...."
ans = gets.chomp!
if ans < 1
puts "Tf bruh bruh"
elsif ans > 10
puts "Now you just playin"
elsif x == 5
print "You wildin B"
else
puts "Fosho that's all I require"
end
It doesn't run properly, and I'm trying to understand why. If you can help me with this, I'd appreciate it.
If you know a good site for practice problems, I'd love to try it. I checked out Coderbyte and Code Kata, but the way they're set up doesn't look right, and they don't have questions to solve for fundamentals.
The issue here is that you're not converting ans to a number, but you're comparing it to one. ans is going to be a string.
In Ruby, when you compare a number to a string, Ruby says that the two aren't equal:
"1" == 1
=> false
You can reproduce the problem with this code:
puts "Welcome to the Loop Practice Problems"
puts " Write a number between 1 and 10, but not 5 or else...."
ans=gets.chomp!
p ans
The p method will output an "inspected" version of that object, (it's the same as doing puts ans.inspect). This will show it wrapped in quotes, which indicates that it's a string.
You probably want to do this:
ans = gets.chomp!.to_i
The to_i method here will convert the number to an integer, and then your comparisons will work correctly.
You have to convert input string type object into integer type
ans = gets.chomp!.to_i #input string convert into integer.
if ans < 1
puts "Tf bruh bruh"
elsif ans > 10
puts "Now you just playin"
elsif x == 5
print "You wildin B"
else
puts "Fosho that's all I require"
end
def Summer
#summer = true
puts "Your fruit are ripe for the picking."
if #tree_age == 1..5 && #tree_age > 0
#oranges = 5
elsif #tree_age == 6..15
#oranges = 20
else
#oranges = 50
end
end
I'm trying to ensure a tree between a certain age range gives x oranges, however I'm stuck with the following error referring to my elsif statement:
Orange_tree.rb:14: warning: integer literal in conditional range
I have also tried using an if greater than && less than conditional statement, can somebody please explain what this error means, and how to reach my solution.
You have a few problems:
You'll want to put your ranges in parenthesis when other operators or methods are nearby. Your current error comes from Ruby parsing elsif #tree_age == 6..15 differently than you expect - it's treating it as (1 == 6)..15, and false..15 obviously doesn't make any sense.
To test a number is within a range, use (1..5) === num, not num == (1..5). Range#=== is defined to test that the Range includes the right hand side, while Fixnum#== and Fixnum#=== both just test that the right hand side is numerically equivalent.
You don't need to test #tree_age > 0. You're already testing that it's in 1..5.
You could also consider a case statement for this, which can be a bit easier to read. case does its comparisons using ===.
#oranges = case #tree_age
when 1..5 then 5
when 6..15 then 20
else 50
end
You should use include? instead of == to determine if the given number is within the range:
def Summer
#summer = true
puts "Your fruit are ripe for the picking."
if (1..5).include?(#tree_age) && #tree_age > 0
#oranges = 5
elsif (6..15).include? #tree_age
#oranges = 20
else
#oranges = 50
end
end
==:
Returns true only if obj is a Range, has equivalent begin and end
items (by comparing them with ==), and has the same exclude_end?
setting as the range.
Which is obviously not the case.
The problem is with the lines that say == with a range.
if ( 10 == 1..11) # throws integer literal in conditional range warning
puts "true"
end
If you did this instead
if ( 10.between?(1, 11))
puts "true"
end
I am writing a code that is supposed to read from STDIN exactly n number of times.So lets say 3 times.What is the best way to do that ?
I tried this
counter = 0
while sentence = gets.chomp && counter < 3 do
...
counter += 1
end
but for some strange reason, sentence variable inside loop is Boolean ?
You can do as below:
n.times { sentence = gets.chomp }
or
n.times do
sentence = gets.chomp
# your code here
end
Operator precedence. The line:
while sentence = gets.chomp && counter < 3 do
Is being interpretted as
while sentence = ( gets.chomp && counter < 3 ) do
So, you could do this:
while ( sentence = gets.chomp ) && counter < 3 do
That explains why you got true or false values into sentence, and the third option should fix this, so your code is very close to working. However, it is probably more usual in Ruby to see solutions like Babai's
I'm messing around in Ruby some more. I have a file containing a class with two methods and the following code:
if __FILE__ == $0
seq = NumericSequence.new
puts "\n1. Fibonacci Sequence"
puts "\n2. Pascal\'s Triangle"
puts "\nEnter your selection: "
choice = gets
puts "\nExcellent choice."
choice = case
when 1
puts "\n\nHow many fibonacci numbers would you like? "
limit = gets.to_i
seq.fibo(limit) { |x| puts "Fibonacci number: #{x}\n" }
when 2
puts "\n\nHow many rows of Pascal's Triangle would you like?"
n = gets.to_i
(0..n).each {|num| seq.pascal_triangle_row(num) \
{|row| puts "#{row} "}; puts "\n"}
end
end
How come if I run the code and supply option 2, it still runs the first case?
Your case syntax is wrong. Should be like this:
case choice
when '1'
some code
when '2'
some other code
end
Take a look here.
You also need to compare your variable against strings, as gets reads and returns user input as a string.
Your bug is this: choice = case should be case choice.
You're providing a case statement with no "default" object, so the first clause, when 1, always returns true.
Effectively, you've written: choice = if 1 then ... elsif 2 then ... end
And, as Mladen mentioned, compare strings to strings or convert to int: choice = gets.to_i