Nested function call with array default parameter

Question

What I want to achieve: when test() is called without any argument, I want that nested() function is called with default arguments.

UPD: Requirements:

no copy-pasting default argument value from nested() function
possibility to override $arr argument value
UPD: function test() have more than one argument (in this case, ...func_get_args() won't work)
actually, this is simplified problem. Imagine that there are several classes that have something like test() method with default $arr parameter that calls nested() method of other class. I want to achieve that nested() method is called with default parameters when test() method is called with default parameters.

Is that possible?

<?php

function test(int $i = 0, array $arr = null)
{
  nested($arr);
}

function nested(array $arr = [1,2,3])
{
   var_dump($arr);
}

test();
test(5, [4,5,6]);

Check if $arr is NULL, and if so, call nested(); …?

04FS
– 04FS

2019-12-03 13:19:29 +00:00
Commented Dec 3, 2019 at 13:19 — 04FS
– 04FS, Commented Dec 3, 2019 at 13:19
Agreed with @04FS

Udit Trivedi
– Udit Trivedi

2019-12-03 14:07:45 +00:00
Commented Dec 3, 2019 at 14:07 — Udit Trivedi
– Udit Trivedi, Commented Dec 3, 2019 at 14:07

Nigel Ren · Accepted Answer · 2019-12-03 14:33:03Z

2

Instead of passing $arr into the nested function, you could instead pass in the actual parameters passed in using func_get_args() with argument unpacking (...)

function test(array $arr = null)
{
    nested(...func_get_args());
}

function nested(array $arr = [1,2,3])
{
    var_dump($arr);
}

test();

If you want it to work with a more complex argument set-up, then you can look at the number of arguments passed (func_num_args()), in this case, check if there are less than 2 arguments and if there are, then force a call without any parameters. You could check if $arr is null if there are more complex patterns, but this doesn't work if the users passes null...

function test(int $i = 0, array $arr = null)
{
    if ( func_num_args() < 2 )  {
        nested();
    }
    else    {
        nested($arr);
    }
}

edited Dec 3, 2019 at 14:33

answered Dec 3, 2019 at 13:20

Nigel Ren

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4 Comments

Serhii Smirnov Over a year ago

That wouldn't work if test() have more than one argument

M. Eriksson Over a year ago

@SergiiSmirnov - There's nothing in the question about multiple arguments. If there are more states, then you need to edit your question to include them. We can't guess what you want.

AbraCadaver Over a year ago

@SergiiSmirnov Sure it will. More will be passed to nested but ignored.

Nigel Ren Over a year ago

Updated the answer.

Trupti Barad · Accepted Answer · 2019-12-03 13:31:11Z

1

Check an array is empty or not

function test(array $arr = null)
{
  if (empty($arr)) {nested();}else{nested($arr);}
}

function nested(array $arr = [1,2,3])
{
   var_dump($arr);
}

test();

edited Dec 3, 2019 at 13:31

answered Dec 3, 2019 at 13:23

Trupti Barad

1691 silver badge8 bronze badges

4 Comments

mega6382 Over a year ago

Then what about when the $arr in test() has values and he needs that value inside nested()?

Trupti Barad Over a year ago

I have given condition

M. Eriksson Over a year ago

This won't do what they are asking for in their question though. It says "when test() is called without any argument" while you're checking if the argument is falsy. An empty array will evaluate as false as well, even though it would be an argument.

Serhii Smirnov Over a year ago

Thanks for the answer. But I need more, I've just updated the question's description

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