Parallel arrays

[momotaro]

am trying to count the accurence and the position of dots in an array and fill another array with the locations found:
this is the code:

void get_dot_pos(char *IP, int *pos)
{
    int i = 0, j, cnt = 0;
    while(i < 15)
    {
        if(IP[i] != '.')
            i++;
        else
        {
            for(j = 0; j < 3; j++)
            {
                cnt++;
                pos[j] = i;
            }
        }
    }
}
 

[momotaro]

this one works better but still not that good!
am having the same value for pos[0] pos[1] pos[2]

void get_dot_pos(char *IP, int *pos)
{
    int i = 0, j = 0, cnt = 0;
    while(i < 15)
    {
        if(IP[i] != '.')
            i++;
        else
        {
            cnt++;
            pos[j++] = i;
        }
    }
}
 

[momotaro]

final version 100% working

void get_dot_pos(char *IP, int *pos)
{
    int i = 0, j = 0, cnt = 0;
    while(i < 15)
    {
        if(IP[i] != '.')
            i++;
        else
        {
            cnt++;
            pos[j++] = i;
            i ++;
        }
    }
}

[JosAH]

final version 100% working

void get_dot_pos(char *IP, int *pos)
{
    int i = 0, j = 0, cnt = 0;
    while(i < 15)
    {
        if(IP[i] != '.')
            i++;
        else
        {
            cnt++;
            pos[j++] = i;
            i ++;
        }
    }
}

Good! You did it all yourself with no help from here. You found your first solution
and if you have enough time you can notice that:

1) j and cnt increment simultaneously, i.e. both have the same values all the time;
2) i increments at every loop pass, no matter the if- or the else-part.

so, you can get rid of either j or cnt and you can simplify that loop somewhat:

int cnt, i;
for (i= cnt= 0; i < 15; i++)
   if (IP[i] == '.')
      pos[cnt++]= i;
 

This is called 'micro-optimization' and you should stay far from it or as late as
possible; only apply this when your program/algorithm works 100% as in your
case. Micro-optimizing it later is easy; coming up with a working solution is the
hard part.

kind regards,

Jos