Set Limit For Value, iPhone

Question

I have the following code:

float valueCalculated = (val1 / val2) * 100;

What I want to be able to do is to cap the maximum value of valueCalculated to 100.

I believe I could do this using some sort of if statement, but this would mean many more lines of code. Edit// This is not the case, see the answers below.

Thanks,

Stu

Answer 1

您是说这样的“多行代码”？

if (valueCalculated > 100) { valueCalculated = 100; }

Answer 2

float valueCalculated = MIN((val1 / val2) * 100, 100);

Answer 3

float valueCalculated = (float temp = (val1 / val2) * 100) > 100 ? 100 : temp;

but I guess a function would be better, you can do with function or a simple macro:

#define MAX(m, toset) toset = (toset > m ? m : toset)

MAX(100, valueCalculated);

I guess I understood better your question:

valueCalculated = MAX(100, (val1 / val2) * 100);

where

#define MAX(max, val) (((val) > (max)) ? (max) : (val))

hope it helps,
Joe

Answer 4

If I understand the question then...

float valueCalculated = fminf((val1 / val2) * 100, 100);

...should do what your asking.

However, are you worried about negative values, ie could val1 or val2 be negative and do you want to limit possible values for valueCalculated to 0-100?. If so then you might want...

float valueCalculated = fmaxf(fminf((val1 / val2) * 100, 100), 0);

Answer 5

那这个呢：

float valueCalculated = ((val1 / val2) * 100 < 100) ? (val1 / val2) * 100 : 100

Answer 6

I'm not an iphone dev, but isn't modulo what you're looking for?

float valueCalculated = ((val1 / val2) * 100) % 101;

?

Answer 7

如果只想通过“加权”将计算的值限制为0到100，则可以将该值除以可能的最大值，然后乘以100。这假定该值不是负数。