What does someArray[--n] mean?

Question

I tried googling this but Google doesn't handle "--n" well. I saw this in my professor's code:

f[--n];
f[n++];

where f is an array of double values.

My guess is that it returns the value of f[n] before reducing (or adding) to n.

Answer 1

f[--n]; means :

n = n -1;
f[n];

f[n++]; means :

f[n];
n = n + 1;

Answer 2

It's actually a type of operator called a pre-decrement, and it's part of a family of 4 operators (see table of java operators)

For an integer-type variable called n:

1. post-increment n++ is the equivalent of n = n + 1, the 'post' part means that if you see it in a line of code (ex. foo(n++);) then the line of code will be called Before n is incremented.

2. pre-increment ++n is also the same as n = n + 1 but it occurs Before the line of code it belongs in has been run.

3. post-decrement n-- is the equivalent of n = n - 1 and occurs After the current line of code has been run

4. pre-decrement --n is the equivalent of n = n - 1 and occurs Before the current line of code has been run

Example of post vs pre decrement:

int n = 5;

System.out.println(n--); //This prints 5
System.out.println(n); //This prints 4

System.out.println(--n); //This prints 3
System.out.println(n); //this prints 3

Answer 3

you can look it up under predecrement (--n) or postincrement (n++). It works like this:

• f[--n]: first n is reduced by 1 then the value of f[n] (this is already the reduced n) is returned
• f[n++]: first the value of f[n] is returned then n is increased by 1

Example:

f {1,3,5} n=1;

• f[--n] returns 1
• f[n++] returns 3

Answer 4

The code

f[--n];
f[n++];

Is the same as

n--;
f[n];
f[n];
n++;