CODEWITHSUNDEEP
chexmd5scanf
Bakus123
Bakus123

Reputation: 1399

Convert hash to 4 hex numbers

I would like to convert my MD5 hash to 4 hex numbers. What's wrong in my code?

//hash = 8ce4b16b22b58894aa86c421e8759df3
char *hash = argv[1];
unsigned int parts[4];

sscanf(&hash[0], "%x", &parts[0]);
sscanf(&hash[8], "%x", &parts[1]);
sscanf(&hash[16], "%x", &parts[2]);
sscanf(&hash[24], "%x", &parts[3]);

printf("Part[0]: %x\n", parts[0]);
printf("Part[1]: %x\n", parts[1]);
printf("Part[2]: %x\n", parts[2]);
printf("Part[3]: %x\n", parts[3]);

Upvotes: 0

Views: 557

Answers (2)

ultrasonic
ultrasonic

Reputation: 21

In your code, you are trying to tell sscanf() where to start parsing the string (stored in the variable named "hash"), but you aren't telling sscanf() where it should stop parsing it. Try inserting whitespace characters into "hash" where you want the 4 numbers to be divided, first, and then your code may work. Also, I would write "&(hash[0])" instead of "&hash[0]" because I don't want to clutter my memory banks with trivia such as the operator precedence rules for C - easier just to use parentheses to force what I want.

Upvotes: 1

Sourav Ghosh
Sourav Ghosh

Reputation: 134356

In your code, you wanted but forgot to limit the input for the HEX value to 8 (in characters). You need to use the field width specifier with the format specifier. You will need something like

sscanf(&hash[0], "%8x", &parts[0]);
sscanf(&hash[8], "%8x", &parts[1]);
sscanf(&hash[16], "%8x", &parts[2]);
sscanf(&hash[24], "%8x", &parts[3]);

Upvotes: 1

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