Re: is it safe to create an object inside a function and pass out in std::vector?

From:

=?ISO-8859-1?Q?=D6=F6_Tiib?= <ootiib@hot.ee>

Newsgroups:

comp.lang.c++

Date:

Wed, 17 Nov 2010 16:28:52 -0800 (PST)

Message-ID:

<4d9f95e5-44ef-497b-98ce-890c0fdf631e@o14g2000yqe.googlegroups.com>

On Nov 16, 11:51 pm, Ruslan Mullakhmetov <tiaba...@gmail.com> wrote:

On 11/16/2010 1:52 AM, Ian Collins wrote:

On 11/16/10 11:45 AM, zl2k wrote:

hi, there,

Is it safe if I do like this

void PopulateVector(std::vector<Object> objs){
Object obj;
objs.push_back(obj);
}

Will the obj created inside of the function lost once I try to use it
outside of the function? I need the function to populate an empty
vector. Thanks for your comments.

Pass by reference:

void PopulateVector( std::vector<Object>& objs )

Objects are copied into containers, so yes it will be OK.

You could also use pointer instead of reference if you know that param
is going to be changed.
As for me i do not use non-const references at all except operators
overloading. Using pointers you definitely know that object could be
changed or specify 0 (NULL) if you do not want return value.

Passing NULL as output parameter when caller does not want that output
is the way of C interface. In C++ it is rude when interface demands
parameters that caller does not have; in C++ you can have overloads.

by the way, i've read about such rule in google code style guide.

Some other style guides suggest to minimize amount of raw pointers and
avoid usage of unary &. I like such rules better.