fangio fangio - 1 month ago 22
C Question

strtoull use properly in C

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 */
    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
    if (errno != 0) {
        /* overflow detected during conversion.
           value was limited to ULLONG_MAX.
           you could report this as well.
    if (*p == ',') {
        /* skip the delimiter */
// i is the count of numbers that were successfully parsed,
//   which can be less than len