在.NET什么是找到字符的整数的长度最好的方式，如果它再presented作为一个字符串？

In .NET what is the best way to find the length of an integer in characters if it was represented as a string?

例如。

1 = 1个字符 10 = 2个字符 99 = 2个字符 100 = 3个字符 1000 = 4个字符

1 = 1 character 10 = 2 characters 99 = 2 characters 100 = 3 characters 1000 = 4 characters

答案显然是为int转换为字符串，并得到它的长度，但我想最佳的性能，而无需创建一个新的字符串的开销。

The obvious answer is to convert the int to a string and get its length but I want the best performance possible without the overhead of creating a new string.

推荐答案

您可以使用logartihms计算INT的长度：

you can use logartihms to calculate the length of the int:

public static int IntLength(int i) {
  if (i <= 0) throw new ArgumentOutOfRangeException();

  return (int)Math.Floor(Math.Log10(i)) + 1;
}

测试通过：

[Test]
public void TestIntLength() {
  Assert.AreEqual(1, IntLength(1));
  Assert.AreEqual(1, IntLength(9));
  Assert.AreEqual(2, IntLength(10));
  Assert.AreEqual(2, IntLength(99));
  Assert.AreEqual(3, IntLength(100));
  Assert.AreEqual(3, IntLength(999));
  Assert.AreEqual(4, IntLength(1000));
  Assert.AreEqual(10, IntLength(int.MaxValue));
}

快速测试已经表明，对数方法比int.ToString（）。持续方法快4倍..

a quick test has shown that the log-method is 4 times faster than the int.ToString().Length method..

由GVS以下所示的方法（使用if语句）是另一个6倍以上日志方法快（！）：

the method shown by GvS below (using if-statements) is another 6 times (!) faster than the log method:

public static int IntLengthIf(int i) {
  if (i < 10) return 1;
  if (i < 100) return 2;
  if (i < 1000) return 3;
  if (i < 10000) return 4;
  if (i < 100000) return 5;
  if (i < 1000000) return 6;
  if (i < 10000000) return 7;
  if (i < 100000000) return 8;
  if (i < 1000000000) return 9;
  throw new ArgumentOutOfRangeException();
}

下面是详细的时序为数字1至10000000：

here are the exact timings for the numbers 1 to 10000000:

IntLengthToString: 4205ms
IntLengthLog10: 1122ms
IntLengthIf: 201ms