So I have a string like this:
char numbers[] = "123,125,10000000,22222222222]"
This is an example, there can be lots more of numbers in the array but it will certainly end with a ].
So now I need to convert it to an array of unsigned long longs. I know I could use strtoull() but it takes 3 arguments and I don't know how to use that second argument. Also I would like to know how I can make my array have the right length. I would like to have my code look like this, but not in pseudocode but in C:
char numbers[] // string of numbers seperated by , and at the end ]
unsigned long long arr[length] // get the correct length
for(int i = 0; i < length; i++){
arr[i]=strtoull(numbers,???,10)// pass correct arguments
}
Is this possible in C to do it like that?
The second argument to strtoull
is a pointer to a char *
that will receive a pointer to the first character after the number in the string argument. The third argument is the base to use for conversion. Base 0 allows for a 0x
prefix to specify hexadecimal conversion and a 0
prefix to specify octal, just like the C integer literals.
You can parse your line this way:
extern char numbers[]; // string of numbers separated by , and at the end ]
unsigned long long arr[length] // get the correct length
char *p = numbers;
int i;
for (i = 0; i < length; i++) {
char *endp;
if (*p == ']') {
/* end of the list */
break;
}
errno = 0; // clear errno
arr[i] = strtoull(p, &endp, 10);
if (endp == p) {
/* number cannot be converted.
return value was zero
you might want to report this error
*/
break;
}
if (errno != 0) {
/* overflow detected during conversion.
value was limited to ULLONG_MAX.
you could report this as well.
*/
break;
}
if (*p == ',') {
/* skip the delimiter */
p++;
}
}
// i is the count of numbers that were successfully parsed,
// which can be less than len