Converting Hexadecimal string to int

Question

Write a C function, that accepts a null-terminated string, containing a hexadecimal string, and returns the integer value. You cannot call any C library function, except strlen() to code the function. The decimal string will only contain 0-4 ASCII characters from ‘0’ through ‘9’ and ‘A’ through ‘F’. No error handling is required. If the string is empty, then return a value of 0.

I've constantly tried fixing my errors, but once I fix them, new errors pop up, resulting in me being confused.

#include <stdlib.h>  /*used for EXIT_SUCCESS */
#include <stdio.h>   /*used for printf */
#include <string.h>  /* used for strlen */
#include <stdbool.h> /* used for bool */
#include <math.h>


unsigned int hexStringTouint(const char str[], int length, int n[])
{
    int i, j;
    int intvalue = 0;
    int digit;
    for(i = (length-1), j = 0; i --, j++)
    {
        if(n[i]>='0' && n[i] <='9')
        {
            digit = n[i] - 0x30;
        }
        else if(n[i]>= 'A' && n[i] <= 'F')
        {
            switch(n[i])
            {
                case 'A': digit = 10; break;
                case 'B': digit = 11; break;
                case 'C': digit = 12; break;
                case 'D': digit = 13; break;
                case 'E': digit = 14; break;
                case 'F': digit = 15; break;
            }
        }
        intvalue += digit*pow(16,j);
    }
    printf("int value is %d\n", intvalue);
    return 0;
}

int main(void)
{
    int i, length, intvalue;
    unsigned char n[] = "";
    printf("Enter your hexadecimal string: ");
    scanf("%c\n", n);
    intvalue = 0;
    length = strlen(n);
    return EXIT_SUCCESS;
}

I am getting error messages saying expected ';' in 'for' statement specifier and how const char* converts between pointers and integers.

for needs 3 statements. you missed the condition to break the loop. it should be for(i = (length-1), j = 0; i >= 0; i --, j++) in your case. — seleciii44
– seleciii44, Commented Jun 22, 2019 at 5:41
A hint: If char c is the variable with current hexdigit character: The values of digits '0' to '9' are c - '0'. Similar, the values of digits 'A' to 'F' are c - 'A' + 0xa. Usually, upper case letters are accepted as well as lower case letters. In this case, a toupper(c) wouldn't hurt. (It doesn't change any non-alpha character.) Altogether: if (isxdigit(c)) digit = isdigit(c) ? c - '0' : toupper(c) - 'A' + 0xa; - a one liner... ;-) — Scheff's Cat
– Scheff's Cat, Commented Jun 22, 2019 at 7:11

chux · Accepted Answer · 2019-06-22 13:19:15Z

Much of OP's code is the right track, yet various coding errors exist.

Add test condition

// for(i = (length-1), j = 0; i --, j++)
//                         v 
for(i = (length-1), j = 0; i >= 0; i --, j++)

cannot call any C library function, except strlen()

// intvalue += digit*pow(16,j);
intvalue += digit*(1u << (4*j));

Return the value

Requirement "... and returns the integer value"

//int intvalue = 0;
//...
//printf("int value is %d\n", intvalue);
//return 0;

unsigned intvalue = 0;
...
printf("integer value is %u\n", intvalue);
return intvalue;

Buffer too small

// unsigned char n[] = "";
// scanf("%c\n", n);

char n[100] = "";
scanf("%99s", n);

hexStringTouint()

Function not called in main().

Other issues exist

A simpler approach

unsigned hexStringTouint2(const char *str) {
  unsigned value = 0;
  while (*str) {
    unsigned digit = hexCharTouint(*str);  // tbd code
    value = value*16 + digit;
    str++;
  }
  return value;
}

Collectives™ on Stack Overflow

Converting Hexadecimal string to int

1 Answer 1

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related