How to cast a string to a integer - c

Question

I have a 40 digit hex number stored in a string, I have to store it inside a struct called Int40 that only contains a pointer to an int.
typedef struct Int40
{
// a dynamically allocated array to hold a 40
// digit integer, stored in reverse order
int *digits;
} Int40;
this is what I have tried
Int40 *parseString(char *str)
{
Int40 *value = malloc(sizeof(Int40) * MAX40);
for (int i = 0; i < MAX40; i++)
{
value[i] = (int)str[i];
}
return value;
}
int main()
{
Int40 *p;
p = parseString("0123456789abcdef0123456789abcdef01234567");
printf("-> %d\n", *p);
}
I know that an Int cant contain 40 digits thats why I tried to store each number from the string in an array of integers but my code doesnt seem to work.
Edit: Also the number contains letters because is a hex number, will I have to the get the ascii value of the hex number to be able to store it in the array of int, how do i do that?

Answer 1

You might do something like the following (note that I omit verification of the argument char* and also assume that hex characters are lowercase)
// with if statements:
Int40 *parseString(char *str)
{
Int40 *value = malloc(sizeof(Int40) * MAX40);
// save the digits array locally (same memory address as value)
int* digits = value->digits;
for (int i = 0; i < MAX40; i++)
{
char c = str[i];
// decimal digits case
if (c >= '0' && c <= '9') {
digits[i] = c - '0'; // subtract '0' to get the numberical value of c
} else { // hex case
digits[i] = (c - 'a') + 10; // subtract 'a' to get the numerical value of c as 0 + 10 for hex characters A - F
}
}
return value;
}
An alternative:
// with a switch statements:
Int40 *parseString(char *str)
{
Int40 *value = malloc(sizeof(Int40) * MAX40);
// save the digits array locally (same memory address as value)
int* digits = value->digits;
for (int i = 0; i < MAX40; i++)
{
char c = str[i];
switch (c) {
// hex 10 - 15
case 'a': case 'b': case 'c':
case 'd': case 'e': case 'f':
digits[i] = (c - 'a') + 10;
break;
// hex 0 - 9
default:
digits[i] = c - '0';
}
}
return value;
}

Answer 2

#include <stdio.h>
#include <stdlib.h>
typedef int* hex;
hex parseString(char *str)
{
hex value = (hex)malloc(sizeof(int)*40);
for (int i = 0; i < 40; i++)
{
value[i] = str[i];
}
return value;
}
int main()
{
hex p;
p = parseString("0123456789abcdef0123456789abcdef01234567");
printf("-> %d\n", p[0]);
}
...
#include <stdio.h>
#include <stdlib.h>
typedef struct Int40
{
int* hex;
}Int40;
Int40 parseString(char *str)
{
Int40 value;
value.hex = (int*)malloc(sizeof(int)*40);
for (int i = 0; i < 40; i++)
{
value.hex[i] = str[i];
}
return value;
}
int main()
{
Int40 p;
p = parseString("0123456789abcdef0123456789abcdef01234567");
printf("-> %d\n", p.hex[0]);
}