Js addition subtraction multiplication and division lost precision problem solution

In javascript, when you use decimals for addition, subtraction, multiplication, and division, you will find that the result is sometimes followed by a long decimal, which can complicate the operation and affect the result. The reason is as follows: in javascript, there are always many decimal places in the calculation of data with decimal.

 
 
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(10, Math.max(baseNum1, baseNum2)); 
return (num1 * baseNum + num2 * baseNum) / baseNum; 
}; 
 
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(10, Math.max(baseNum1, baseNum2)); 
precision = (baseNum1 >= baseNum2) ? baseNum1 : baseNum2; 
return ((num1 * baseNum - num2 * baseNum) / baseNum).toFixed(precision); 
}; 
 
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(10, baseNum); 
}; 
 
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(10, baseNum2 - baseNum1); 
} 
};