Splitting string sequence into integers [closed] - c

Question

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 6 years ago.
Improve this question
How do you split a string:
char *mystring = "12345"
into an integer array which looks like this:
[1, 2, 3, 4, 5]
I have tried something like the code below, but I'm not entirely sure if it's reliable, and I think it will be easy to break. This is the code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int
main(void) {
char *mystring = "12345";
int string_size, i, length;
string_size = strlen(mystring);
int values[string_size];
for (i = 0; mystring[i] != '\0'; i++) {
values[i] = mystring[i] - 48;
}
length = sizeof(values)/sizeof(*values);
for (i = 0; i < length; i++) {
printf("%d ", values[i]);
}
return 0;
}
Which outputs:
1 2 3 4 5
Is there a more C like way I can do this?

Answer 1

The odd thing I see, which isn't itself a problem, is that you calculate the length of the string/array three different ways:
string_size = strlen(mystring);
for (i = 0; mystring[i] != '\0'; i++) {
length = sizeof(values)/sizeof(*values);
where just one method is sufficient:
#include <stdio.h>
#include <string.h>
int main(void) {
char *mystring = "12345";
size_t length = strlen(mystring);
int values[length];
for (int i = 0; i < length; i++) {
values[i] = mystring[i] - '0';
}
for (int i = 0; i < length; i++) {
printf("%d ", values[i]);
}
printf("\n");
return 0;
}

Answer 2

You can replace 48 with '0' for readability.
You can change all loops to loop until string_size like the first one, no need to change the method for each loop.
And finally if you're going to return that array anywhere outside of local function, you should probably malloc() it rather than use a local/stack variable.
But otherwise, it's pretty simple and it works.