Question about Exercise 8 (@implementation issue)

[sugarboyloveOC]

Dear Steve,

In my work, I write the (add:, subtract:, multiply:, and divide:) method is :


-(double) accumulator
{
    return accumulator;
}

-(double) add: (double) value
{
    return accumulator += value;
}

-(double) subtract: (double) value
{
    return accumulator -= value;
}

-(double) multiply: (double) value
{
    return accumulator *= value;
}

-(double) divide: (double) value
{
    return accumulator /= value;
}

ANd it is work.

But when I read the other post from other people, some people  write this :

-(double) divide: (double) value
{
    accumulator /= value;
    return accumulator;
}

I try it and it is work too. SO I want to know is there any difference between with

-(double) divide: (double) value
{
    return accumulator /= value;
}

and

-(double) divide: (double) value
{
    accumulator /= value;
    return accumulator;
}


?

Looking for your reply !!!

[fujilla]

With
return accumulator /= value;
you are basically combining
accumulator /= value;
return accumulator;
into a single statement.

Nick
http://myfirstiphoneapp.co.uk