您可以调用此函数来获得黑暗(颤振)的精确程度。 Double eval -> Double想要转换的 Int I ->返回的小数点。

double doubleToPrecision(double eval, int i) {
double step1 = eval;//1/3
print(step1); // 0.3333333333333333

String step2 = step1.toStringAsFixed(2);
print(step2); // 0.33

double step3 = double.parse(step2);
print(step3); // 0.33
eval = step3;
return eval; }

您可以调用此函数来获得黑暗(颤振)的精确程度。 Double eval -> Double想要转换的 Int I ->返回的小数点。

double doubleToPrecision(double eval, int i) {
double step1 = eval;//1/3
print(step1); // 0.3333333333333333

String step2 = step1.toStringAsFixed(2);
print(step2); // 0.33

double step3 = double.parse(step2);
print(step3); // 0.33
eval = step3;
return eval; }

如果您想要一个双精度结果，将一个IEEE-754二进制浮点数舍入到特定的十进制位数本身就是有问题的。

In the same way that fractions such as 1/3 can't be exactly represented with a finite number of decimal digits, many (really infinitely many) decimal numbers can't be represented with a finite number of binary digits. For example, the decimal number 0.1 cannot be exactly represented in binary. While you could try to round 0.09999 to 0.1, as a double it would actually be "rounded" to 0.1000000000000000055511151231257827021181583404541015625. Most of the other answers that claim to round doubles with decimal precision actually return the nearest representable double.

你能做的就是使字符串表示形式看起来像一个漂亮的四舍五入的数字，这就是double.toStringAsFixed()所做的。这也是为什么当你打印0.100000000…，如果实现试图打印用户友好的值，则可能会看到0.1。然而，不要被愚弄了:double值实际上永远不会精确到0.1，如果你用这样不精确的值重复计算，就会累积错误。

请注意，以上所有内容都是浮点数工作方式所固有的，并不是Dart所特有的。还看到:

浮点数学坏了吗? 如果你不理解上面的解释，Tom Scott也做了一个很好的视频来解释浮点数是如何工作的。

底线:如果关心十进制精度，就不要使用二进制浮点类型。如果你和钱打交道，这一点尤其重要。

相反，你应该使用:

Integers. For example, if you are dealing with currency, instead of using double dollars = 1.23;, use int cents = 123;. Your calculations then always will be exact, and you can convert to the desired units only when displaying them to the user (and likewise can convert in the opposite direction when reading input from the user). A type designed to represent decimal numbers with arbitrary precision. For example, package:decimal provides a Decimal type. With such a type, some of the other answers (such as multiplying by 100, rounding, and then dividing by 100) then would be appropriate. (But really you should use Decimal.round directly.)

我在double上做了这个扩展

import 'dart:math';

extension DoubleExtension on double {

  /// rounds the double to a specific decimal place
  double roundedPrecision(int places) {
    double mod = pow(10.0, places) as double;
    return ((this * mod).round().toDouble() / mod);
  }

  /// good for string output because it can remove trailing zeros
  /// and sometimes periods. Or optionally display the exact number of trailing
  /// zeros
  String roundedPrecisionToString(
    int places, {
    bool trailingZeros = false,
  }) {
    double mod = pow(10.0, places) as double;
    double round = ((this * mod).round().toDouble() / mod);
    String doubleToString =
        trailingZeros ? round.toStringAsFixed(places) : round.toString();
    if (!trailingZeros) {
      RegExp trailingZeros = RegExp(r'^[0-9]+.0+$');
      if (trailingZeros.hasMatch(doubleToString)) {
        doubleToString = doubleToString.split('.')[0];
      }
    }
    return doubleToString;
  }

  String toStringNoTrailingZeros() {
    String doubleToString = toString();
    RegExp trailingZeros = RegExp(r'^[0-9]+.0+$');
    if (trailingZeros.hasMatch(doubleToString)) {
      doubleToString = doubleToString.split('.')[0];
    }
    return doubleToString;
  }
}

这是通过的测试。

import 'package:flutter_test/flutter_test.dart';
import 'package:project_name/utils/double_extension.dart';

void main() {
  group("rounded precision", () {
    test("rounding to 0 place results in an int", () {
      double num = 5.1234;
      double num2 = 5.8234;
      expect(num.roundedPrecision(0), 5);
      expect(num2.roundedPrecision(0), 6);
    });
    test("rounding to 1 place rounds correctly to 1 place", () {
      double num = 5.12;
      double num2 = 5.15;
      expect(num.roundedPrecision(1), 5.1);
      expect(num2.roundedPrecision(1), 5.2);
    });
    test(
        "rounding a number to a precision that is more accurate than the origional",
        () {
      double num = 5;
      expect(num.roundedPrecision(5), 5);
    });
  });

  group("rounded precision returns the correct string", () {
    test("rounding to 0 place results in an int", () {
      double num = 5.1234;
      double num2 = 5.8234;
      expect(num.roundedPrecisionToString(0), "5");
      expect(num2.roundedPrecisionToString(0), "6");
    });
    test("rounding to 1 place rounds correct", () {
      double num = 5.12;
      double num2 = 5.15;
      expect(num.roundedPrecisionToString(1), "5.1");
      expect(num2.roundedPrecisionToString(1), "5.2");
    });
    test("rounding to 2 places rounds correct", () {
      double num = 5.123;
      double num2 = 5.156;
      expect(num.roundedPrecisionToString(2), "5.12");
      expect(num2.roundedPrecisionToString(2), "5.16");
    });
    test("cut off all trailing zeros (and periods)", () {
      double num = 5;
      double num2 = 5.03000;
      expect(num.roundedPrecisionToString(5), "5");
      expect(num2.roundedPrecisionToString(5), "5.03");
    });
  });
}

上述解决方案没有适当地四舍五入数字。我使用:

double dp(double val, int places){ 
   num mod = pow(10.0, places); 
   return ((val * mod).round().toDouble() / mod); 
}

请参阅num.toStringAsFixed()的文档。

字符串磨损

返回this的小数字符串表示形式。

在计算字符串表示形式之前，将此转换为double。

如果this的绝对值大于或等于10^21，则该方法返回由this. tostringasexponential()计算的指数表示。

例子:

1000000000000000000000.toStringAsExponential(3); // 1.000e+21

否则，结果是与小数点后的fractionDigits位数最接近的字符串表示形式。如果fractionDigits等于0，则小数点将被省略。

参数fractionDigits必须是满足以下条件的整数:0 <= fractionDigits <= 20。

例子:

1.toStringAsFixed(3);  // 1.000
(4321.12345678).toStringAsFixed(3);  // 4321.123
(4321.12345678).toStringAsFixed(5);  // 4321.12346
123456789012345678901.toStringAsFixed(3);  // 123456789012345683968.000
1000000000000000000000.toStringAsFixed(3); // 1e+21
5.25.toStringAsFixed(0); // 5