Convert Decimal to Hex and store to array in C - c

Question

I've been trying to convert decimal from 1 to 99 to hexadecimal and store them into an array.
What I have is something like this:
int main(){
int i =1;
char dump_holder[3];
char hex_holder[100];
for (i=1;i<100;i++){
sprintf(dump_holder, "%02x",i);
hex_holder[i] = atoi(dump_holder);
printf("this: %02x\n", hex_holder[i]);
}
return 0;}
I'm getting correct value a certain number. This code returns:
this: 01
this: 02
this: 03
this: 04
this: 05
this: 06
this: 07
this: 08
this: 09
this: 00
this: 00
this: 00
this: 00
this: 00
this: 00
this: 0a
this: 0b
this: 0c
this: 0d
this: 0e
this: 0f
this: 10
this: 11
this: 12
this: 13
this: 01
this: 01
this: 01
this: 01
this: 01
this: 01
this: 14
this: 15
this: 16
this: 17
this: 18
this: 19
this: 1a
this: 1b
this: 1c
this: 1d
I think the stray values are the null terminator, but I'm not sure.

Answer 1

well, the basic thing is I cannot understand what you are actually trying to achive
but I will try my best:
int main(){
int i =1;
char dump_holder[3];
char hex_holder[100];
for (i=1;i<100;i++){
/* convert numerical value to string */
sprintf(dump_holder, "%02x",i);
/* convert string value back to numerical value */
//hex_holder[i] = atoi(dump_holder); //won't work
hex_holder[i] = strtol(dump_holder, NULL, 16); // this will
/* print the numerical value in hex representation */
printf("this: %02x\n", hex_holder[i]);
}
return 0;}
are you trying to create a string representation of the values in hex format?
if so, you are going about this the wrong way
right now you are not doing much except wasting proccessing power
even so i added a small code that will actuallt conver the code to a string representation of the values. Maybe that is what you actually intended to do
int main()
{
int i = 1;
char dump_holder[3];
/* the array should be an array of strings (restricted here to 2chars) */
char hex_holder[100][2];
for (i=1;i<100;i++){
/* convert numerical value to string representation */
sprintf(hex_holder[i], "%02x",i);
/* print the string produced */
printf("this: %s\n", hex_holder[i]);
}
return 0;
}

Answer 2

atoi assumes decimal representation, try strtol(dump_holder, NULL, 16); instead.

Answer 3

To understand the problem in your code, you first need to understand -
How atoi works?
Read about atoi and go through the example demonstrating its behavior
here.
In your program output, after 09 you are getting six times 00 because the atoi is returning 0 for the hex values 0a, 0b, 0c, 0d, 0e and 0f.
After this the program output is:
this: 0a
this: 0b
this: 0c
this: 0d
this: 0e
this: 0f
Reason is -
The hex value after 0f is 10. 10 is stored in dump_holder variable which is passed to atoi.
atoi returned integer value 10 which is stored in hex_holder[i].
Your program is printing the value stored in hex_holder[i], which is of type char, using %x format specifier. So, the value 10 is getting printed as 0a.
Using incorrect format specifier is undefined behavior which includes it may do exactly what the programmer intended or silently generating incorrect results or anything else.
Just to convert decimal from 1 to 99 to hexadecimal and store them into an array, you can do:
#include <stdio.h>
int main() {
char hex_holder[100][3];
for (int i = 1; i < 100; i++) {
snprintf (hex_holder[i], 3, "%02x", i);
printf("this: %s\n", hex_holder[i]);
}
return 0;
}