Variable P Passed by Reference Before Being Initialized

Variable p passed by reference before being initialized

Your declaration of p is just that, a declaration. You haven't initialised it. You need to change it to

var p = [Human]()

Or, as @MartinR points out,

var p: [Human] = []

There are other equivalent constructs, too, but the important thing is you have to assign something to the declared variable (in both cases here, an empty array that will accept Human members).

Update
For completeness, you could also use:

var p: Array<Human> = []

or

var p = Array<Human>()

variable used before being initialized

This is because of the way of initialising variables in Swift, In Swift Class each property need some default value before being used.

Class initialisation in Swift is a two-phase process. In the first
phase, each stored property is assigned an initial value by the class
that introduced it. Once the initial state for every stored property
has been determined, the second phase begins, and each class is given
the opportunity to customize its stored properties further before the
new instance is considered ready for use.

Try changing your code as below:

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
  let destViewController = segue.destinationViewController as! SecondTableViewController
  var secondArrays :  SecondTableData?

  if let indexPath = self.tableView.indexPathForSelectedRow {
    secondArrays = secondArrayData[indexPath.row]

    //Here you're making sure that the secondArrays would must have a value
    destViewController.secondArray = secondArrays.secondTitle
  }

}

C intro - How to pass a parameter by reference in function?

1. It is perfectly valid. You can initialize and pass any number of pointer variables with their reference.

2. This is also valid..when you pass the variable address, you should store it into a pointers

you have to do some changes in your code,
You can assign directly a/b and a*b pointer variables *c & *d
Then you have to read double number with %lf format argument.

#include <stdio.h>
#include <string.h>

void myFunction(double a, double b, double *c, double *d)
{
*c = a/b;   //change
*d = a*b;   //change
printf("%lf %lf",*c,*d);
return;
//printf statements
}

int main()
{
//first and second double hold the scanf inputs
double first;
double second;

//unsure here - to reference c and d as parameters in the function, do I simply declare unfilled double variables here?
double *c;
double *d;

printf("Enter your first number\n");
scanf("%lf", &first);   //change
printf("Enter your second number\n");
scanf("%lf", &second);  //change

//call the function, first and second by value, &c / &d by reference - correct?
myFunction(first, second, &c,&d);
}

C++ Using a reference to the variable being defined

Syntactically it is, however if you try this

#include <iostream>
using namespace std;

typedef int T;
bool f(T& x)
{
    return true;
}
int main()
{
    T x = (f(x) ? x : T());
    cout << x;
}

it outputs some random junk.
However, if you modify

bool f(T& x)
{
    x = 10;
    return true;
}

then it outputs 10.
In the first case, the object x is declared, and the compiler assigns some pseudo-arbitrary value (so you do not initialize it), whereas in the second you specifically assign a value (T(), i.e. 0) after the declaration, i.e. you initialize it.

I think your question is similar to this one:
Using newly declared variable in initialization (int x = x+1)?

How do I pass a variable by reference?

Arguments are passed by assignment. The rationale behind this is twofold:

1. the parameter passed in is actually a reference to an object (but the reference is passed by value)
2. some data types are mutable, but others aren't

So:

• If you pass a mutable object into a method, the method gets a reference to that same object and you can mutate it to your heart's delight, but if you rebind the reference in the method, the outer scope will know nothing about it, and after you're done, the outer reference will still point at the original object.

• If you pass an immutable object to a method, you still can't rebind the outer reference, and you can't even mutate the object.

To make it even more clear, let's have some examples.

List - a mutable type

Let's try to modify the list that was passed to a method:

def try_to_change_list_contents(the_list):
    print('got', the_list)
    the_list.append('four')
    print('changed to', the_list)

outer_list = ['one', 'two', 'three']

print('before, outer_list =', outer_list)
try_to_change_list_contents(outer_list)
print('after, outer_list =', outer_list)

Output:

before, outer_list = ['one', 'two', 'three']
got ['one', 'two', 'three']
changed to ['one', 'two', 'three', 'four']
after, outer_list = ['one', 'two', 'three', 'four']

Since the parameter passed in is a reference to outer_list, not a copy of it, we can use the mutating list methods to change it and have the changes reflected in the outer scope.

Now let's see what happens when we try to change the reference that was passed in as a parameter:

def try_to_change_list_reference(the_list):
    print('got', the_list)
    the_list = ['and', 'we', 'can', 'not', 'lie']
    print('set to', the_list)

outer_list = ['we', 'like', 'proper', 'English']

print('before, outer_list =', outer_list)
try_to_change_list_reference(outer_list)
print('after, outer_list =', outer_list)

Output:

before, outer_list = ['we', 'like', 'proper', 'English']
got ['we', 'like', 'proper', 'English']
set to ['and', 'we', 'can', 'not', 'lie']
after, outer_list = ['we', 'like', 'proper', 'English']

Since the the_list parameter was passed by value, assigning a new list to it had no effect that the code outside the method could see. The the_list was a copy of the outer_list reference, and we had the_list point to a new list, but there was no way to change where outer_list pointed.

String - an immutable type

It's immutable, so there's nothing we can do to change the contents of the string

Now, let's try to change the reference

def try_to_change_string_reference(the_string):
    print('got', the_string)
    the_string = 'In a kingdom by the sea'
    print('set to', the_string)

outer_string = 'It was many and many a year ago'

print('before, outer_string =', outer_string)
try_to_change_string_reference(outer_string)
print('after, outer_string =', outer_string)

Output:

before, outer_string = It was many and many a year ago
got It was many and many a year ago
set to In a kingdom by the sea
after, outer_string = It was many and many a year ago

Again, since the the_string parameter was passed by value, assigning a new string to it had no effect that the code outside the method could see. The the_string was a copy of the outer_string reference, and we had the_string point to a new string, but there was no way to change where outer_string pointed.

I hope this clears things up a little.

EDIT: It's been noted that this doesn't answer the question that @David originally asked, "Is there something I can do to pass the variable by actual reference?". Let's work on that.

How do we get around this?

As @Andrea's answer shows, you could return the new value. This doesn't change the way things are passed in, but does let you get the information you want back out:

def return_a_whole_new_string(the_string):
    new_string = something_to_do_with_the_old_string(the_string)
    return new_string

# then you could call it like
my_string = return_a_whole_new_string(my_string)

If you really wanted to avoid using a return value, you could create a class to hold your value and pass it into the function or use an existing class, like a list:

def use_a_wrapper_to_simulate_pass_by_reference(stuff_to_change):
    new_string = something_to_do_with_the_old_string(stuff_to_change[0])
    stuff_to_change[0] = new_string

# then you could call it like
wrapper = [my_string]
use_a_wrapper_to_simulate_pass_by_reference(wrapper)

do_something_with(wrapper[0])

Although this seems a little cumbersome.

---

Related Topics