C prog - COUNTER to find the number of alphanumeric's character in a string?

Question

I'm trying to write a function to count the number of alphanumeric character(a,b,c,d,..,z), in a string. My code looks like it doesn't have error..., but when I run it, function stops to count characters before the end of the string. Someone can find why is that? Thanks for your times.

int counter(char text[])
{
    int length = strlen(text), counter = 0;

    for (int i = 0; i < length; i++)
    {
        if (isalpha(text[i]))
        {
            counter++;
            i++;
        }
        else i++;
    }
    return counter;
}

Answer 1

Your problem is pretty simple. In a for loop, after the end of each loop, the third parameter (<increment command>) of the loop is executed, every time.

for (<initialization>; <condition>; <increment command>)

In your case your <increment command> is i++, so i gets incremented by 1 in each iteration. But, in your loop, after each iteration, you also increment i yourself. So i gets incremented by 2 in every loop instead of just 1. To solve that just remove the i++s from inside the loop. The increment command will take care of it itself.

So your function should be:

int counter(char text[])
{
    int length = strlen(text), counter = 0;

    for (int i = 0; i < length; i++)
    {
        if (isalpha(text[i]))
        {
            counter++;
        }
    }
    return counter;
}

You don't even need the else clause ;).

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