Count the number of digits in a hexadecimal number

Question

I am trying to count the number of digits that are available in a hexadecimal number. For example: 0x00000001 should return the count as 8 .

Could anyone suggest as to what is the most efficient way of doing this? I have tried converting it to a CString and get the length using ' GetLength() ' but that doesn't seem to work here.

Edit: Sorry if I forgot to mentioned, my variable that stores the hexadecimal number is an unsigned short.

Tried this:

unsigned short number;
CString HexValue;
HexValue.Format("%.8x",number);     // number = 0000000000000000, 16 0's
HexValue = "0x" + HexValue;
int length = HexValue.GetLength() - 2; // returns an 8 here

Answer 1

由于每个字节包含两个十六进制数字，请尝试使用整数类型的变量n

sizeof(n)*CHAR_BIT/4

Answer 2

如果0x00000001是一个字符串，则GetLength() - 2应该为8。如果它是存储在不同类型中的值（假设为int），则只需记住十六进制编码需要每8位2个字符来对整个整数进行编码值范围（ sizeof(int) * 2 ）

Answer 3

There is one hex digit per 4 bits.

std::cout << sizeof(unsigned short) * CHAR_BIT / 4 << "\n";
std::cout << sizeof(unsigned short) * CHAR_BIS % 4 << " LEFTOVER PARTIAL DIGITS\n";

Answer 4

The number of digits in a hexadecimal number n is max(1, ceil(log2( n ))). See Wikipedia, Binary logarithm for a fast algorithm to compute log2.

That won't give you the value 8 for 0x00000001 , though, since there's actually only digit needed to represent that number. The difference between 0x00000001 and 0x1 is a display issue only.