pset3 duration itoa - with "&"

Question

For 'duration' I used the following which works (at least within a testing file), using atoi on individual parts of the inputted string fraction s:

double n = atoi(&s[0]); double d = atoi(&s[2]); int duration = ((n/d))/(.125); return duration;

I added the & after trial and error, and as a result I have a few questions on how atoi works:

1) In Caesar I used: int key = atoi(argv[1]); which worked well, and converted the whole string in one go. In the case of the fraction, does atoi NOT work because it returns a NULL value when it hits the / character?

2) What exactly is the & doing in the code above? From reading, it looks like & is an "Address of Operator" - meaning the & pointing to the value that is stored in the s[0] location...but isn't that what writing s[0] is already doing?

Answer 1

First, understand what strings in C are.

Strings in C are pointers to a character, or char* (if you use type string, that's a custom alias for the same thing). The convention then says that the pointer points the first byte of a string (this time in the sense of the actual characters), and the string ends at the first character that's zero, often called "null terminator".

This division between the pointer and the memory where the characters are stored might be confusing, but gives extra flexibility (if you assign to a string variable, you copy the pointer, not the whole string, and that string can be of any length, the pointer doesn't care and always has the same size).

Now &s[0]. s[0] is the character the pointer points to. &s[0] is the address of that character (same as s).

atoi(&s[0]) will now try to convert for example "3/8" to an integer. It reads '3', fine. It reads '/', ouch. That's not part of an integer, so return what we got so far.

atoi(&s[2]) for the same string will read '8', fine. Then it reads '\0', null terminator means string end, done, return what we have so far.

As we here just want to convert single digits to their numeric value, you could just subtract '0' (or its ASCII value 48) like you did subtract 'A' or 'a' (or their counterparts 65 and 97) in caesar and vigenere.

And if you multiplied by 8 first, and only then divided, you wouldn't need any floating point numbers.