16 is the 4th power of 2. So a hexadecimal number is converted to binary by replacing each hex digit by the 4-bit binary number having the same value. Conversely, in converting binary to hexadecimal, we group every 4 bits starting at the decimal (binary?) point and replace it with the equivalent hex digit. For example, the hexadecimal number 3F9 in binary is 1111111001, because 3 in binary is 11, F (decimal 15) is 1111, and 9 is 1001.
To store the hexadecimal number FF, we need to convert it to binary first. FF in hexadecimal is equivalent to 1111 1111 in binary, which requires 8 bits to represent. Each hexadecimal digit corresponds to 4 bits in binary, so two hexadecimal digits (FF) require 8 bits to store.
hexadecimal can express 16 bit binary in 4 place form, not 16.
0xc = 1100 Hexadecimal digits use exactly 4 binary digits (bits). The 0x0 to 0xf of hexadecimal map to 0000 to 1111 of binary. Thinking of the hexadecimal digits as decimal numbers, ie 0x0 to 0x9 are 0 to 9 and 0xa to 0xf are 10 to 15, helps with the conversion to binary: 0xc is 12 decimal which is 8 + 4 → 1100 in [4 bit] binary.
import java.util.Scanner; public class NumberSystem { public void displayConversion() { Scanner input = new Scanner(System.in); System.out.printf("%-20s%-20s%-20s%-20s\n", "Decimal", "Binary", "Octal", "Hexadecimal"); for ( int i = 1; i <= 256; i++ ) { String binary = Integer.toBinaryString(i); String octal = Integer.toOctalString(i); String hexadecimal = Integer.toHexString(i); System.out.format("%-20d%-20s%-20s%-20s\n", i, binary, octal, hexadecimal); } } // returns a string representation of the decimal number in binary public String toBinaryString( int dec ) { String binary = " "; while (dec >= 1 ) { int value = dec % 2; binary = value + binary; dec /= 2; } return binary; } //returns a string representation of the number in octal public String toOctalString( int dec ) { String octal = " "; while ( dec >= 1 ) { int value = dec % 8; octal = value + octal; dec /= 8; } return octal; } public String toHexString( int dec ) { String hexadecimal = " "; while ( dec >= 1 ) { int value = dec % 16; switch (value) { case 10: hexadecimal = "A" + hexadecimal; break; case 11: hexadecimal = "B" + hexadecimal; break; case 12: hexadecimal = "C" + hexadecimal; break; case 13: hexadecimal = "D" + hexadecimal; break; case 14: hexadecimal = "E" + hexadecimal; break; case 15: hexadecimal = "F" + hexadecimal; break; default: hexadecimal = value + hexadecimal; break; } dec /= 16; } return hexadecimal; } public static void main( String args[]) { NumberSystem apps = new NumberSystem(); apps.displayConversion(); } }
Assuming the original was in binary, the answer is 36.A
Okay, I'm pretty sure that 864 binary is 30 hexadecimal. - RG
The answer depends on what form you wish to convert binary and hex 2011 to.
Octal = 52746757 Binary = 101010111100110111101111
4F7B: Binary = 100111101111011 Decimal = 20347
The binary equivalent of the hexadecimal number EF16 is 1110111100010110.
01
ABCD1 = 10101011110011010001
It is used because it is easier to convert to and from binary to hexadecimal than decimal, and it uses less characters than binary. For instance: decimal: 65535 hex: FFFF binary: 1111111111111111
In order to convert binary to hexadecimal using assembly language, the programmer must possess an understanding on boolean algebra or binary system in other words. A compiler is also needed to complete the program.
The answer depends on what you are converting from: binary, ternary, octal, hexadecimal ...
hex =7