multiplication

I'm writing a Fixedpoint class, but have ran into bit of a snag... The multiplication, division portions, I am not sure how to emulate. I took a very rough stab at the division operator but I am sure it's wrong. Here's what it looks like so far: class Fixed { Fixed(short int _value, short int _part) : value(long(_value + (_part >> 8))), part(long(_part & 0x0000FFFF)) {}; ... inline Fixed operator -() const // example of some of the bitwise it's doing { return Fixed(-value - 1, (~part)&0x0000FFFF); }; ... inline Fixed operator / (const Fixed & arg) const // example of how I'm probably doing it

Can someone please give me pointers on how I can go about making a code that multiplies using shifts in MIPS assembly? I don't understand how having a number 2^n can help me multiply using an odd multiplicand I currently have this code, I'm trying to make a calculator .text li $v0, 4 la $a0, ask_1 syscall li $v0,5 syscall move $s1, $v0 li $v0, 4 la $a0, ask_2 syscall li $v0,5 syscall move $s2, $v0 #sll $s2, $s2, 3 #$s2 * $s2^3 = result srl $s2, $s2, 1 li $v0, 1 la $a0, ($s2) syscall .data ask_1: .asciiz "Enter Multiplier\n" ask_2: .asciiz "Enter Multiplicand\n" result: .asciiz "The Answer is:

This question already has answers here : Is floating point math broken? (31 answers) Closed 4 years ago . I am executing the following code in java but i got two different answers for what should be the same number mathematically. public class TestClass { public static void main(String[] args) { double a=0.01; double b=4.5; double c=789; System.out.println("Value1---->"+(a*b*c)); System.out.println("Value2---->"+(b*c*a)); } } Output: Value1---->35.504999999999995 Value2---->35.505 Please tell me why I get this result. Floating point numbers have a certain precision. Some fractions can not be

I read in a paper that division and multiplication of a number by a power of 2 is a trivial process. I have searched a lot of internet for the explanation but doesn't get it. Can any one explain in easy words what does this actually meant to say. Vlad It is trivial from the bit operations perspective. Multiplying by 2 is equivalent to a shift left by 1 bit, division is a right shift. similarly it is the same trivial to multiply and divide by any power of 2. int a = 123; // or in binary format: a = 0b01111011; assert (a * 2) == (a << 1); // 2 = 2^1, (a << 1) = 11110110 assert (a / 2) == (a >> 1

If I want to take the product of a list of floating point numbers, what's the worst-case/average-case precision lost by adding their logs and then taking exp of the sum as opposed to just multiplying them. Is there ever a case when this is actually more precise? tmyklebu Absent any overflow or underflow shenanigans, if a and b are floating-point numbers, then the product a*b will be computed to within a relative error of 1/2 ulp. A crude bound on the relative error after multiplying a chain of N double s therefore results in answer off by a factor of at most (1 - epsilon/2) -N , which is about

I want to be able to compute g^x = g * g * g * ... * g (x times) where g is in a finite field GF(2^m). Here m is rather large, m = 256, 384, 512, etc. so lookup tables are not the solution. I know that there are really fast algorithms for a similar idea, modpow for Z/nZ (see page 619-620 of HAC ). What is a fast, non-table based way to compute cycles (i.e. g^x)? This is definitely a wishful question but here it comes: Can the idea of montgomery multiplication/exponentiation be 'recycled' to Galois fields? I would like to think so because of the isomorphic properties but I really don't know.

As much as I love Python, the reference and deepcopy stuff sometimes freaks me out. Why does deepcopy not work here: >>> import copy >>> a = 2*[2*[0]] >>> a [[0, 0], [0, 0]] >>> b = copy.deepcopy(a) >>> b[0][0] = 1 >>> b [[1, 0], [1, 0]] #should be: [[1, 0], [0, 1]] >>> I am using a numpy array as a workarround which I need later on anyway. But I really had hoped that if I used deepcopy I would not have to chase any unintended references any more. Are there any more traps where it does not work? It doesn't work because you are creating an array with two references to the same array. An