Convert a string of hexidecimals to an array of integers

[zeusprog]

Please how do I convert a string of hexidecimals “10B71356C845FB08E04951AAA2FF9F49” to integer? I tried using getBytes then Integer.valueOf but I got 32 integers instead of 16.

Any help would be greatly appreciated.

Thank you.

[NormR1]

Can you give an example of a hex String and what int value you want to get from it?
An int holds 32 bits which can be represented with 8 hex digits. The example hex String you posted is longer than 8 digits and would not fit in an int.
The Integer class's parseInt() method can be used for conversion using the proper base.

[JamesCherrill]

getBytes will extract the text bytes from the string - it doesn't know or care about hex. It will give you {'1', '0', 'B' ...
Then you convert each of those and you get 32 integer values each in the range 0-15.
To decode that as 16 integer values each in the range 0-255 you need to use substring to split the string into 16 strings each 2 chars long, then parse those.

[Starstreak]

Are these pairs of hex numbers, like 10 B7 etc..?

[RainbowMatrix]

C program for hexadecimal to binary conversion

#include<stdio.h>
#define MAX 1000

int main(){
    char binaryNumber[MAX],hexaDecimal[MAX];
    long int i=0;

    printf("Enter any hexadecimal number: ");
    scanf("%s",hexaDecimal);

    printf("\nEquivalent binary value: ");
    while(hexaDecimal[i]){
         switch(hexaDecimal[i]){
             case '0': printf("0000"); break;
             case '1': printf("0001"); break;
             case '2': printf("0010"); break;
             case '3': printf("0011"); break;
             case '4': printf("0100"); break;
             case '5': printf("0101"); break;
             case '6': printf("0110"); break;
             case '7': printf("0111"); break;
             case '8': printf("1000"); break;
             case '9': printf("1001"); break;
             case 'A': printf("1010"); break;
             case 'B': printf("1011"); break;
             case 'C': printf("1100"); break;
             case 'D': printf("1101"); break;
             case 'E': printf("1110"); break;
             case 'F': printf("1111"); break;
             case 'a': printf("1010"); break;
             case 'b': printf("1011"); break;
             case 'c': printf("1100"); break;
             case 'd': printf("1101"); break;
             case 'e': printf("1110"); break;
             case 'f': printf("1111"); break;
             default:  printf("\nInvalid hexadecimal digit %c ",hexaDecimal[i]); return 0;
         }
         i++;
    }

    return 0;
}

Sample output:

Enter any hexadecimal number: 2AD5
Equivalent binary value: 0010101011010101

[JamesCherrill]

You do realise that this is the Java forum? And that Java already comes with APIs that include parsing/converting numbers from/to any arbitrary base?

[zeusprog]

Thanks guys