Question about arrays

[ts4525]

I am reviewing some C++ code and ran across the following line of code:

class->method(label, array + integerValue)

the code in question is: array + integerValue

array is declared as: float array[850];
integerValue is declared as: int integerValue;

integerValue can be either 0 or 25.

So the question is what is array + integerValue actually doing?

[vlad from moscow]

The type of the expression is float *

So array + integerValue provided that integerValue is equal to 25 means

&array[25] that is equivalent to array + 25

[o3hC5Di1]

Hi there,

Let's say integerValue has value 25.
In that case what the code does is to pass the address of the 26th member of the array.

It works like this:

array is an array of floats, the single name array (without brackets) will hold the address in memory of the very first member of the array - array[0].
When we do arithmetic on that address, the compiler is smart enough to interpret that as such:

array (start address) + 25 times the size of a float (because the array consists of float values).

That lands you at the 26th member of the array member [code]array[25].
More info here: http://cplusplus.com/doc/tutorial/arrays/#accessing_values

Hope that's a bit clear, let us know if you need further clarification.

All the best,
NwN