C如何在没有浮点精度的情况下计算百分比(千) [英] C How to calculate a percentage(perthousands) without floating point precision

问题描述

How do you calculate a percentage from 2 int values into a int value that represents a percentage(perthousands for more accuracy)?

Background/purpose: using a processor that doesn't have a FPU, floating point computations take 100's of times longer.

int x = 25;
int y = 75;
int resultPercentage; // desire is 250 which would mean 25.0 percent

resultPercentage = (x/(x+y))*1000; // used 1000 instead of 100 for accuracy
printf("Result= ");
printf(resultPercentage);

output:

Result= 0

When really what I need is 250. and I can't use ANY Floating point computation.

Example of normal fpu computation:

int x = 25;
int y = 75;
int resultPercentage; // desire is 250 which would mean 25.0 percent

resultPercentage = (int)( ( ((double)x)/(double(x+y)) ) *1000); //Uses FPU slow

printf("Result= ");
printf(resultPercentage);

output:

Result= 250

But the output came at the cost of using floating point computations.

解决方案

resultPercentage = (x/(x+y))*1000; does not work as (x/(x+y)) is likely 0 or 1 before the multiplcation *1000 occurs. Instead:

For a rounded unsigned integer calculation of x/(x+y), let a = x and b = x+y then to find a/b use:

result = (a + b/2)/b;

For a rounded unsigned integer percent % calculation of a/b use

result = (100*a + b/2)/b;

For a rounded unsigned integer permil ‰ calculation of a/b use

result = (1000*a + b/2)/b;

For a rounded unsigned integer permyriad ‱ calculation of a/b use

result = (10000*a + b/2)/b;

@H2CO3 wells points out the concerns about eating up the integer range so using wider integers (long, long long) are needed for the multiplication and maybe x+y.

result = (100L*a + b/2)/b;

---

Of course, replace

// printf(resultPercentage);
printf("%d
", resultPercentage);

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