How can I force Ruby numbers to behave like integers, not fixnums?

I have:

steven$ irb
ruby-1.9.2-p180 :001 > foo = "256MB"
 => "256MB" 
ruby-1.9.2-p180 :002 > interim_result = foo.slice(/\d+/).to_i
 => 256 
ruby-1.9.2-p180 :003 > interim_result.class
 => Fixnum 
ruby-1.9.2-p180 :004 > result = interim_result/1028
 => 0 

I want result to be 0.25. How can I make this happen?

Is it necessary/possible to force interim_result.class to be integer?

Please note the following doesn't give the desired result of 0.25:

ruby-1.9.2-p180 :002 > interim_res开发者_如何学运维ult = foo.slice(/\d+/).to_f
 => 256.0 
ruby-1.9.2-p180 :003 > result = interim_result/1028
 => 0.2490272373540856 
ruby-1.9.2-p180 :004 > result.round_to(2)
NoMethodError: undefined method `round_to' for 0.2490272373540856:Float

Thanks.

---

Yes, the easiest way would be to call to_f instead of to_i when assigning a value to interim_result.

---

I'm not quit sure what you mean behave like an integer. IF that was completely true you would have nothing after a decimal; however, I believe you can use round() to achieve what you want.

 ruby-1.9.2-p180-mri :001 > a = 0.2490272373540856 
 => 0.2490272373540856 
 ruby-1.9.2-p180-mri :002 > a.round(2)
 => 0.25 
 ruby-1.9.2-p180-mri :003 > 

---

Just a litte explanation what's going wrong in your example.

interim_result is correct 256. and the this happens:

256 / 1024 #-> 0

/ with two fixnums (integer) is kind of modula division. To show it: you get the first value of:

256.divmod(1024)#-> [0, 256]
256.divmod(1024).first #-> 0

Some solutions (I changed your entry values. your 256 / 1028 is not 0.25, but 24.9..... See Codeglots answer):

256 / 1024.0 #-> 0.25

Perhaps Rational is another nice solution:

256 / Rational(1024,1) #-> (1/4)
256 * Rational(1,1024) #-> (1/4)