I need to perform a simple multiplication of 400 * 256.3. The result is 102520. Straight forward and simple. But to implement this multiplication in C++ (or C) is a little tricky and confusing to me.
I understand floating point number is not represented as it is in computer. I wrote the code to illustrate the situation. Output is attached too.
So, if I do the multiplication using float type variable, I am subjected to rounding error. Using double type variable would have avoided the problem. But let's say I have a very limited resource on the embedded system and I have to optimize the variable type to the very best I could, how can I perform the multiplication using float type variable and not susceptible to rounding error?
I knew the floating point math done by computer is not broken at all. But I am curious for best practice to perform floating point math. 256.3 is just a value for illustration. I would not know what floating point value I will get during runtime. But it is for sure, a floating point value.
int main()
{
//perform 400 * 256.3
//result should be 102520
float floatResult = 0.00f;
int intResult = 0;
double doubleResult = 0.00;
//float = int * float
floatResult = 400 * 256.3f;
printf("400 * 256.3f = (float)->%f\n", floatResult);
//float = float * float
floatResult = 400.00f * 256.3f;
printf("400.00f * 256.3f = (float)->%f\n", floatResult);
printf("\n");
//int = int * float
intResult = 400 * 256.3f;
printf("400 * 256.3f = (int)->%d\n", intResult);
//int = float * float;
intResult = 400.00f * 256.3f;
printf("400.00f * 256.3f = (int)->%d\n", intResult);
printf("\n");
//double = double * double
doubleResult = 400.00 * 256.3;
printf("400.00 * 256.3 = (double)->%f\n", doubleResult);
//int = double * double;
intResult = 400.00 * 256.3;
printf("400.00 * 256.3 = (int)->%d\n", intResult);
printf("\n");
//double = int * double
doubleResult = 400 * 256.3;
printf("400 * 256.3 = (double)->%f\n", doubleResult);
//int = int * double
intResult = 400 * 256.3;
printf("400 * 256.3 = (int)->%d\n", intResult);
printf("\n");
//will double give me rounding error?
if (((400.00 * 256.3) - 102520) != 0) {
printf("Double give me rounding error!\n");
}
//will float give me rounding error?
if (((400.00f * 256.3f) - 102520) != 0) {
printf("Float give me rounding error!\n");
}
return 0;
}
floatunless you know you really need it. Use double instead – Ingate