How the binary multiplication of floats can be represented in C? - c

Question

The binary multiplication algorithm with ints can be represented as follows:
unsigned int multiply(unsigned int multiplier, unsigned int multiplicand) {
unsigned int product = 0;
while (multiplier != 0) {
if ((multiplier & 1) != 0) {
product = product + multiplicand;
}
multiplier = multiplier >> 1;
multiplicand = multiplicand << 1;
}
return product;
}
This function performs the multiplication of two unsigned ints without the operator "*".
However it does not work with floats because a float is composed of three parts:
IEEE 754 single-precision binary floating-point format
These parts can be isolated as follows:
#include <stdio.h>
typedef union {
float f;
struct {
unsigned int mantisa : 23;
unsigned int exponent : 8;
unsigned int sign : 1;
} parts;
} float_cast;
int main() {
float_cast d1;
d1.f = 0.15625;
printf("sign = %x\n",d1.parts.sign);
printf("exponent = %x\n",d1.parts.exponent);
printf("mantisa = %x\n",d1.parts.mantisa);
return 0;
}
With the parts separated as ints I can manipulate the bits. But how to make a function that multiplies the parts of a float?
Thanks in advance

Answer 1

Multiply the mantissas as integers.
Add the exponents
Xor the signs.
There are some details.
The mantissas should both be normalized, meaning that either the high-order bit is 1 or the mantissa is 0. For full compliance you need to deal with denorms and other special cases -- infinities, NaNs, zeros -- and you may need to normalize, denorm, or overflow (set to infinity).
The product is in the range [1, 4), assuming the values were in [1, 2). If the product of the mantissas is greater than 2, some fixups are necessary: increment the exponents by 1; shift both mantissas right one.
Exponents are normally stored with an offset. Suppose the the real value of the exponent is e + m, where m is the constant offset. The m needs to be subtracted from the sum of the representations of the two exponents, in order to get the exponent of the product.

Answer 2

Here is my solution and Answer:
#include <stdio.h>
float multiplyfloat(float multiplier, float multiplicand) {
typedef union {
float f;
unsigned int i;
struct {
unsigned int mantissa : 23;
unsigned int exponent : 8;
unsigned int sign : 1;
} parts;
struct {
unsigned int mantissa : 23;
unsigned int b23 : 1;
unsigned int b31_24 : 8;
} parts2;
} float_cast;
float_cast product, f1, f2, m1, m2;
product.f = 0.f;
f1.f = multiplier;
f2.f = multiplicand;
m1 = f1;
m2 = f2;
m1.parts2.b23 = m2.parts2.b23 = 1;
m1.parts2.b31_24 = m2.parts2.b31_24 = 0;
while (m1.parts.mantissa) {
if (m1.parts2.b23) {
product.i += m2.i;
}
m2.i >>= 1;
m1.i <<= 1;
}
if (product.parts.exponent > 1) {
product.parts.mantissa >>= product.parts.exponent - 1;
}
product.parts.exponent += f1.parts.exponent + f2.parts.exponent - 128;
product.parts.sign = f1.parts.sign != f2.parts.sign;
return product.f;
}
int main() {
float a = 134.337368;
float b = 151.23000000001;
float res = multiplyfloat(a, b);
printf("result = %f\n", res);
printf("compare = %f\n", a * b);
system("pause");
return 1;
}
Any questions, just comment below. Thanks