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js运算精度

js运算精度

作者: smallTigerZ | 来源:发表于2019-09-29 08:53 被阅读0次

https://blog.csdn.net/helloxiaoliang/article/details/72723387

https://blog.csdn.net/DADADIE/article/details/50385146

// 两个浮点数求和

        accAdd(num1, num2) {

            var r1, r2, m;

            try {

                r1 = num1.toString().split('.')[1].length;

            } catch (e) {

                r1 = 0;

            }

            try {

                r2 = num2.toString().split(".")[1].length;

            } catch (e) {

                r2 = 0;

            }

            m = Math.pow(10, Math.max(r1, r2));

            // return (num1*m+num2*m)/m;

            return Math.round(num1 * m + num2 * m) / m;

        },

        // 两个浮点数相减

        accSub(num1, num2) {

            var r1, r2, m;

            try {

                r1 = num1.toString().split('.')[1].length;

            } catch (e) {

                r1 = 0;

            }

            try {

                r2 = num2.toString().split(".")[1].length;

            } catch (e) {

                r2 = 0;

            }

            m = Math.pow(10, Math.max(r1, r2));

            n = (r1 >= r2) ? r1 : r2;

            return (Math.round(num1 * m - num2 * m) / m).toFixed(n);

        },

        // 两数相除

        accDiv(num1, num2) {

            var t1, t2, r1, r2;

            try {

                t1 = num1.toString().split('.')[1].length;

            } catch (e) {

                t1 = 0;

            }

            try {

                t2 = num2.toString().split(".")[1].length;

            } catch (e) {

                t2 = 0;

            }

            r1 = Number(num1.toString().replace(".", ""));

            r2 = Number(num2.toString().replace(".", ""));

            return (r1 / r2) * Math.pow(10, t2 - t1);

        },

        //两数相乘

        accMul(num1, num2) {

            var m = 0,

                s1 = num1.toString(),

                s2 = num2.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);

        }

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