Number of divisors with the function getopt

Question

I'm doing a program which receives a number and gives you the number of divisors. e.g. in the cmd :

practica -n 30

Expected output:

1,2,3,5,6,10,15,30

I have this code:

void divisor(char *temp1);
int main(int argc,char **argv){
int option;
char *n;
 while((option =getopt(argc,argv,"n")) !=-1){
  switch(option){
  case 'n':
  n=optarg;
  break;
 }
}

divisor(n);
}

void divisor(char *temp1){
char f=*temp1;
int n=f-'0';
int a;
a=1;
while (a<n)
{
 if(n%a==0)
 printf("%d,",a);
 a++;
}
a=n;
if(n%a==0)
printf("%d",a);
printf("\b ");
}

I'm using the command line and the program closes.

Answer 1

I think, your getopt() call should look like

while((option =getopt(argc,argv,"n:")) !=-1){  //notice the :

as you'll be supplying an argument (value) for that option.

That said, in your code, inside divisor() function,

char f=*temp1;
int n=f-'0';

looks wrong. temp is a char pointer an dereferencing the pointer only gives you the value of the first char stored, and the expected value , 30 is not stored as a single char, rather it's stored as lexicographical format.

I think, you can use strtol() to convert the lexicographical representation to int value, like

int x = strtol(temp, NULL, 0);