Add solutions for part 1

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2020-11-15 13:57:48 -05:00
parent e4f9fd2682
commit 742db6d102
479 changed files with 202980 additions and 13 deletions

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// This file is part of www.nand2tetris.org
// and the book "The Elements of Computing Systems"
// by Nisan and Schocken, MIT Press.
// File name: projects/12/Math.jack
/**
* A library of commonly used mathematical functions.
* Note: Jack compilers implement multiplication and division using OS method calls.
*/
class Math {
/** Initializes the library. */
function void init() {
return;
}
/** Returns the absolute value of x. */
function int abs(int x) {
if (x < 0) {
let x = -x;
}
return x;
}
/** Returns the product of x and y.
* When a Jack compiler detects the multiplication operator '*' in the
* program's code, it handles it by invoking this method. In other words,
* the Jack expressions x*y and multiply(x,y) return the same value.
*/
function int multiply(int x, int y) {
var int sum;
var int i;
var int mask;
var int xShifted;
let sum = 0;
let i = 0;
let mask = 1;
let xShifted = x;
if (x = 1) {
return y;
}
if (y = 1) {
return x;
}
while (i < 16) { // 16 because we got a 16 bit architecture
if (y & mask) {
let sum = sum + xShifted;
}
let xShifted = xShifted + xShifted; // xShifted << 2;
let mask = mask + mask; // mask << 2;
let i = i + 1;
}
return sum;
}
/** Returns the integer part of x/y.
* When a Jack compiler detects the multiplication operator '/' in the
* program's code, it handles it by invoking this method. In other words,
* the Jack expressions x/y and divide(x,y) return the same value.
*/
function int divide(int x, int y) {
var int q;
var int sign;
let sign = 1;
if (x < 0) {
let x = -x;
let sign = -sign;
}
if (y < 0) {
let y = -y;
let sign = -sign;
}
if (y > x) {
return 0;
}
let q = x / (y + y);
if ((x - (2 * q * y)) < y) {
return (q + q) * sign;
} else {
return (q + q + 1) * sign;
}
}
/** Returns the integer part of the square root of x. */
function int sqrt(int x) {
var int y;
var int p;
var int s;
var int s2;
let y = 0;
let p = 128; // 2^7
while (p > 0) {
let s = y + p;
let s2 = s * s;
if ((s2 - 1) < x) {
if (s2 > 0) {
let y = s;
}
}
let p = p / 2; // p >> 1;
}
return y;
}
/** Returns the greater number. */
function int max(int a, int b) {
if (a > b) {
return a;
}
return b;
}
/** Returns the smaller number. */
function int min(int a, int b) {
if (a < b) {
return a;
}
return b;
}
}