How to implement computer arithmetic?

The computer is a generic computing means to achieve a CPU, a CPU inside the ALU

Wiki description

　　An arithmetic logic unit ( English: Arithmetic Logic Unit, the ALU ) is the central processor of an execution unit, a core component of all central processors, the AND gate and OR gate arithmetic logic units, the main function is binary in the arithmetic operation , such as by subtraction (not including the integer division). Essentially, all modern CPU architectures, are in binary complement form to represent.

Here are a few questions:

　　1. implemented without the use of arithmetic adder

　　2. Do not use arithmetic to realize subtraction

　　3. Save multiplication and division implemented without using multiplication

If you know the answer, there is no need Kanla down, on the contrary continue reading

　　Said first adder

Code

int _add(int v1,int v2) {
　　int nRet = 0;
　　do {
　　　　nRet = v1 ^ v2;
　　　　v2 =(v1 & v2) << 1;
　　　　v1 = nRet;
　　} while (v2);
　　return nRet;
}

　Manual testing

　　5+5=？

　　　　101　　　　 101

xor　　 101　　and　101

------------------------------------------

　　　 000　　　　 101

　　　　0000　　　　 0000

xor　　 1010　　and　1010

------------------------------------------

　　　 1010　　　　 0000

The results are: conversion bit binary 1010 (2 ^ 0 * 0) + (1 * 2 ^ 1) + (0 * 2 ^ 2) + (1 * 2 ^ 3) = 10

Ye who were said subtraction

　　Addition and subtraction same reason, as to why you need to know complement (complement design Niubi I think nobody should refute)

_bitSub int (int A, int B) { 
	B = -b; 
	do { 
		int nXorResult = A ^ B; 
		B = (A & B) <<. 1; // B is determined to exit the loop means 
		a = nXorResult; // a is results 
	} the while (B);	 
	return A; 
}

Manual testing (originally wanted to be lazy, do not continue to demonstrate ......)

　　 1-1　(invert the sign of the subtrahend and then using an addition operation mode)

　　 0001　　　　0001

xor　　1111　　and 1111

-------------------------------------

　　 1110　　　　0001

　　 1110　　　 1110

xor　　0010　　and 0010

--------------------------------------

　　 1100　　　　 0010

　　 1100　　　 1100

xor　　0100　　and 0100

--------------------------------------

　　 1000　　　　 0100

　　 1000　　　 1000

xor　　1000　　and 1000

--------------------------------------

　　 0000　　　　 1000

　　 00000　　　 00000

xor　　10000　　and 10000

--------------------------------------

　　 10000 00000

It should be noted, ye have done a four bit computing, direct cut off the excess, so the result is a binary 0000

　1--1 (invert the sign of the subtrahend and then using an addition operation mode)

　　　　0001　　　　0001

xor　　 0001　　　　0001
-------------------------------------

　　　　0000　　　　0001　　

　　　　0000　　　　0000

xor　　 0010　　　　0010
-------------------------------------

　　　　0010　　　　0000

Results: binary to decimal 0010 is 2

Open proved correct

Achieve multiplication first consider the question

Decimal: 100 * 10 = seconds you should know the answer, 100 * 100 it? 100 * 100 = 10000

Note: The suffix b represents the binary digit

Binary: 100b * 100b = Do you know how much it equal? In fact, here and decimal is the same reason 100b * 100b = 10000b (through this is actually the result of two multiplicand shift to the left, 2 = 100 100 << 00 )

int _mul(int v1,unsigned int v2) {
    int nRet = 0;
    int nLeftMove = 0;
    do {
        if (v2 & 1) {
            nRet += v1 << nLeftMove;
        }
        v2 = v2>>1;
        nLeftMove++;
    }
    while(v2); 
    return nRet;
}

　　5*2=101b*10b

First: Analyzing 1 0 B 1 is not 0, the condition is not satisfied is not moved multiplicand

Second: Analyzing 1 0B 1 is 1, a multiplicand move condition is satisfied, 101b << 1b = 1010b

Third: the multiplier has been judged over, ending

How to implement computer arithmetic?

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