C++ syntax for pointer and reference

Question

I'm doing a searching for swapping two nodes of linked list and found out a block of code as follow:

void swapNode(call * &head, call * &first, call * &second) {
    // swap values, assuming the payload is an int:
    int tempValue = first->value;
    first->value = second->value;
    second->value = tempValue;
}

My question is what the meaning for put the ampersand after asterisk?

As mentioned below that specifies a reference. But it looks like you don't need it since you are not updating any of the parameters. — Anon Mail
– Anon Mail, Commented Oct 17, 2017 at 0:44
@AnonMail actually I don't use the head parameter, just swap the values in memory addresses first and second point to — Phi Truong
– Phi Truong, Commented Oct 17, 2017 at 0:54

CS Pei · Accepted Answer · 2017-10-17 00:45:44Z

1

Ampersand (&) here denotes reference. it is not the address operator.

This is the C++ way of being able to change the value of a passed-in pointer because C/C++ pass parameters by value. In C, you would use a double pointer to achieve the same.

edited Oct 17, 2017 at 0:45

answered Oct 17, 2017 at 0:33

CS Pei

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4 Comments

Phi Truong Over a year ago

could you give me simpler examples about denote reference?

CS Pei Over a year ago

& denotes reference, not that it is a denote reference.

Phi Truong Over a year ago

so the & used for both reference and address operator ?

Anon Mail Over a year ago

@PhiTruong If you use int var as a parameter to a function, any changes to var are local to the function (i.e. they are not seen outside the function). If you use int & var as a parameter to a function any changes to var are seen inside and outside the function.

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