How do I read hex numbers into an unsigned int in C - c

Question

I'm wanting to read hex numbers from a text file into an unsigned integer so that I can execute Machine instructions. It's just a simulation type thing that looks inside the text file and according to the values and its corresponding instruction outputs the new values in the registers.
For example, the instructions would be:
1RXY -> Save register R with value in
memory address XY
2RXY -> Save register R with value XY
BRXY -> Jump to register R if xy is
this and that etc..
ARXY -> AND register R with value at
memory address XY
The text file contains something like this each in a new line. (in hexidecimal)
120F
B007
290B
My problem is copying each individual instruction into an unsigned integer...how do I do this?
#include <stdio.h>
int main(){
FILE *f;
unsigned int num[80];
f=fopen("values.txt","r");
if (f==NULL){
printf("file doesnt exist?!");
}
int i=0;
while (fscanf(f,"%x",num[i]) != EOF){
fscanf(f,"%x",num[i]);
i++;
}
fclose(f);
printf("%x",num[0]);
}

Answer 1

You're on the right track. Here's the problems I saw:
You need to exit if fopen() return NULL - you're printing an error message but then continuing.
Your loop should terminate if i >= 80, so you don't read more integers than you have space for.
You need to pass the address of num[i], not the value, to fscanf.
You're calling fscanf() twice in the loop, which means you're throwing away half of your values without storing them.
Here's what it looks like with those issues fixed:
#include <stdio.h>
int main() {
FILE *f;
unsigned int num[80];
int i=0;
int rv;
int num_values;
f=fopen("values.txt","r");
if (f==NULL){
printf("file doesnt exist?!\n");
return 1;
}
while (i < 80) {
rv = fscanf(f, "%x", &num[i]);
if (rv != 1)
break;
i++;
}
fclose(f);
num_values = i;
if (i >= 80)
{
printf("Warning: Stopped reading input due to input too long.\n");
}
else if (rv != EOF)
{
printf("Warning: Stopped reading input due to bad value.\n");
}
else
{
printf("Reached end of input.\n");
}
printf("Successfully read %d values:\n", num_values);
for (i = 0; i < num_values; i++)
{
printf("\t%x\n", num[i]);
}
return 0
}

Answer 2

You can also use the function strtol(). If you use a base of 16 it will convert your hex string value to an int/long.
errno = 0;
my_int = strtol(my_str, NULL, 16);
/* check errno */
Edit: One other note, various static analysis tools may flag things like atoi() and scanf() as unsafe. atoi is obsolete due to the fact that it does not check for errors like strtol() does. scanf() on the other hand can do a buffer overflow of sorts since its not checking the type sent into scanf(). For instance you could give a pointer to a short to scanf where the read value is actually a long....and boom.

Answer 3

You're reading two numbers into each element of your array (so you lose half of them as you overwrite them. Try using just
while (i < 80 && fscanf(f,"%x",&num[i]) != EOF)
i++;
for your loop
edit
you're also missing the '&' to get the address of the array element, so you're passing a random garbage pointer to scanf and probably crashing. The -Wall option is your friend.

Answer 4

In this case, all of your input is upper case hex while you are trying to read lower case hex.
To fix it, change %x to %X.

Answer 5

Do you want each of the lines (each 4 characters long) separated in 4 different array elements? If you do, I'd try this:
/* read the line */
/* fgets(buf, ...) */
/* check it's correct, mind the '\n' */
/* ... strlen ... isxdigit ... */
/* put each separate input digit in a separate array member */
num[i++] = convert_xdigit_to_int(buf[j++]);
Where the function convert_xdigit_to_int() simply converts '0' (the character) to 0 (an int), '1' to 1, '2' to 2, ... '9' to 9, 'a' or 'A' to 10, ...
Of course that pseudo-code is inside a loop that executes until the file runs out or the array gets filled. Maybe putting the fgets() as the condition for a while(...)
while(/*there is space in the array && */ fgets(...)) {
}