JS Subtraction Loss Accuracy problem Solving method _javascript Skills

In JavaScript, when you use decimal to perform a subtraction operation, you will find that the resulting results are sometimes followed by a long fractional fraction, which complicates the operation and affects the results of the calculation. The Internet query for the reasons, roughly as follows: in JavaScript, with decimal data operations, there will always be a number of decimal places. This is because the calculation of floating-point numbers in JavaScript is calculated in 2.

Copy Code code as follows:

/**
* Addition operation, to avoid the data added decimal point after the number of digits and computational precision loss.
*
* @param num1 addends 1 | Num2 Addends 2
*/
function Numadd (NUM1, num2) {
var basenum, BaseNum1, baseNum2;
try {
BASENUM1 = Num1.tostring (). Split (".") [1].length;
catch (e) {
BASENUM1 = 0;
}
try {
baseNum2 = Num2.tostring (). Split (".") [1].length;
catch (e) {
baseNum2 = 0;
}
Basenum = Math.pow (Math.max (BASENUM1, baseNum2));
Return (NUM1 * basenum + num2 * basenum)/basenum;
};
/**
* Addition operation, to avoid the data phase reduction points after the number of digits and computational precision loss.
*
* @param num1 Bing | Num2 meiosis
*/
function Numsub (NUM1, num2) {
var basenum, BaseNum1, baseNum2;
var precision;//precision
try {
BASENUM1 = Num1.tostring (). Split (".") [1].length;
catch (e) {
BASENUM1 = 0;
}
try {
baseNum2 = Num2.tostring (). Split (".") [1].length;
catch (e) {
baseNum2 = 0;
}
Basenum = Math.pow (Math.max (BASENUM1, baseNum2));
Precision = (baseNum1 >= baseNum2)? basenum1:basenum2;
Return ((NUM1 * basenum-num2 * basenum)/basenum). toFixed (precision);
};
/**
* Multiplication operations, to avoid data multiplication decimal point after the number of digits and computational precision loss.
*
* @param num1 by multiplier | Num2 Multiplier
*/
function Nummulti (NUM1, num2) {
var basenum = 0;
try {
Basenum + = Num1.tostring (). Split (".") [1].length;
catch (e) {
}
try {
Basenum + = Num2.tostring (). Split (".") [1].length;
catch (e) {
}
Return number (num1.tostring (). Replace (".", "")) * Number (num2.tostring (). Replace (".", ""))/Math.pow (basenum);
};
/**
* Division operations, to avoid dividing the decimal point after the number of digits and computational precision loss.
*
* @param NUM1 Dividend | NUM2 Divisor
*/
function Numdiv (NUM1, num2) {
var baseNum1 = 0, baseNum2 = 0;
var baseNum3, BaseNum4;
try {
BASENUM1 = Num1.tostring (). Split (".") [1].length;
catch (e) {
BASENUM1 = 0;
}
try {
baseNum2 = Num2.tostring (). Split (".") [1].length;
catch (e) {
baseNum2 = 0;
}
With (Math) {
BASENUM3 = number (num1.tostring (). Replace (".", ""));
BASENUM4 = number (num2.tostring (). Replace (".", ""));
Return (BASENUM3/BASENUM4) * POW (BASENUM2-BASENUM1);
}
};