Converting Hexadecimal string to int - c

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.

Answer 1

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;
}