JavaScript的计算精度问题
Posted 无梦南柯
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直接使用代码吧
// 数字相加
xiangjia (arg1, arg2) {
let r1, r2, m, c;
try {
r1 = arg1.toString().split(‘.‘)[1].length;
} catch (e) {
r1 = 0;
}
try {
r2 = arg2.toString().split(‘.‘)[1].length;
} catch (e) {
r2 = 0;
}
c = Math.abs(r1 - r2);
m = Math.pow(10, Math.max(r1, r2));
if (c > 0) {
let cm = Math.pow(10, c);
if (r1 > r2) {
arg1 = Number(arg1.toString().replace(‘.‘, ‘‘));
arg2 = Number(arg2.toString().replace(‘.‘, ‘‘)) * cm;
} else {
arg1 = Number(arg1.toString().replace(‘.‘, ‘‘)) * cm;
arg2 = Number(arg2.toString().replace(‘.‘, ‘‘));
}
} else {
arg1 = Number(arg1.toString().replace(‘.‘, ‘‘));
arg2 = Number(arg2.toString().replace(‘.‘, ‘‘));
}
return (arg1 + arg2) / m;
}
// 数字相减
xiangjian (arg1, arg2) {
let r1, r2, m, n;
try {
r1 = arg1.toString().split(‘.‘)[1].length;
} catch (e) {
r1 = 0;
}
try {
r2 = arg2.toString().split(‘.‘)[1].length;
} catch (e) {
r2 = 0;
}
m = Math.pow(10, Math.max(r1, r2)); // last modify by deeka //动态控制精度长度
n = (r1 >= r2) ? r1 : r2;
return Number(((arg1 * m - arg2 * m) / m).toFixed(n));
}
// 数字相乘
xiangcheng (arg1, arg2) {
let m = 0, s1 = arg1.toString(), s2 = arg2.toString();
try {
m += s1.split(‘.‘)[1].length;
} catch (e) {
}
try {
m += s2.split(‘.‘)[1].length;
} catch (e) {
}
return Number(s1.replace(‘.‘, ‘‘)) * Number(s2.replace(‘.‘, ‘‘)) / Math.pow(10, m);
}
// 数字相除
xiangchu(num1 , num2) {
let t1 , t2 , r1 , r2
try {
t1 = `${num1}`.split(‘.‘)[1].length;
} catch (e) {
t1 = 0;
}
try {
t2 = `${num2}`.toString().split(‘.‘)[1].length;
} catch (e) {
t2 = 0;
}
r1 = Number(`${num1}`.replace(‘.‘ , ‘‘));
r2 = Number(`${num2}`.toString().replace(‘.‘ , ‘‘));
return(r1 / r2) * Math.pow(10 , t2 - t1);
}
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