Float is 0 after integer division

问题:

It might be a simple solution but I can not fix it.

I am dividing 2 integers :

finishedGameFinalScore = [score integerValue];
CGFloat interval = 2/finishedGameFinalScore;
NSLog(@"interval = %f",interval);

The log returns 0.000000

Is there a limit for decimal places? I need to preserve the decimal result.

Thanks Shani

回答1:

The reason your code doesn't work is that you're dividing an integer by another integer and then casting the result to a float.

So you have 2 (an integer) and some other number (also an integer). Then you divide 2 by this number - which is probably greater than 2. Let's say it's 3.

Integer sees 2/3 and he's like "0.66666667? Pshh, no one ever needs anything after the decimal point anyway". So he truncates it. You just have 0.

Then Integer gives the number to Mr. float and Mr float is super happy to get a number! He's all like "yay, a 0! I'm going to add ALL OF THE SIGNIFICANT DIGITS". And that's how you end up with 0.0000000.

So yeah, just cast to a float first. Or even a double!

回答2:

@Dustin said u will need to typecast your divider value to float as it goes in float it shows integer value

CASE 1: Typecast

NSString *score = @"3";
 int interval = [str intValue];
 CGFloat interval = (2/(float)interval);
 NSLog(@"interval = %.2f",interval);

CASE 2: No need for typecast

NSString *score = @"3";
 float interval = [str floatValue];
 CGFloat interval = (2/interval);
 NSLog(@"interval = %.2f",interval);

回答3:

Just add the f-hint to the number 2. in this case that will do the trick.

CGFloat interval = 2.0f/finishedGameFinalScore;

all the above/below answers are correct and fully explain why this work.

回答4:

just devide long value from 1.0 and assigned to float variable.

unsigned long l1 = 65536;
unsigned long l2 = 256;
float f = (l1/1.0)/(l2/1.0);

NSLog(@"%f",f);