The use of &.

what does it mean when you place a & after something opposed to before it?

Code:

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&thing
thing&

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I understand the one first one would give me the address of the thing...and the 2nd line's form is used when i want to accept something as a reference (??)

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what is the difference between
(assuming the & one is even valid..i mean is it?)

Code:

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int func(int *p)

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Code:

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int func2(int p&)

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?

You might want to read this FAQ on references.

read the article,
hmm so is placing the ampersand AFTER incorrect syntax?

I am reading a book and they talk about methods such as

Quote:

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void process_array( IntArray& );
void process_array( IntArrayRC& );

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(c++, primer) not sure why the author would write it like this.

Quote:

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Originally Posted by rodrigorules

hmm so is placing the ampersand AFTER incorrect syntax?

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In your case, yes, in your book's case, no ;)

Identify the type and the variable. For reference declarations, the ampersand comes after the type name and before the variable name, if it is present.

Its important to note that Arrays are always passed by reference therefore it is not necessary to use the ampersand when passing an array through a function

Quote:

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Originally Posted by bleuz

Its important to note that Arrays are always passed by reference therefore it is not necessary to use the ampersand when passing an array through a function

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It is not true that arrays are always passed by reference. Rather, they are converted to pointers to their first elements, hence one can modify the contents of an array without passing it by reference.

To pass an array by reference looks like this:

Code:

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#include <iostream>

void foo(const int (&arr_ref)[10]) //by reference
{
    std::cout << sizeof(arr_ref) << '\n'; //size of the array: int[10]
}

void bar(const int arr_ptr[10]) //usual way, array decays to pointer (10 is ignored)
{
    std::cout << sizeof(arr_ptr) << '\n'; //size of the pointer: int*
}


int main()
{
    int arr[10];
    foo(arr);
    bar(arr);
}

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Note that no decay to pointer occurs and sizeof still reports the size of the array (10 * sizeof(int)).