C#•2y ago

❔ Parameter Shenanigans

I'm working on a linear search method that will return the count (number of iterations it took to find the target), the parameters are obviously the int table, the target number "t", and the count "c". I'm wondering if the c is incremented inside the method's for-loop, will the returned "c" be the new altered "c" or will it be the one that was used as an input for the function to be called.

25 Replies

DarmaOP•2y ago

    public static int linearSearch(int[] nums, int t, int c)
    {
        bool isFound = false;

        for (int i = 0; i < nums.Length; i++)
        {
            c++;
            
            if (t == nums[i])
            {
                isFound = true;
                Console.WriteLine("Found!");
                break;
            }
        }

        return  c;
    }

    public static int linearSearch(int[] nums, int t, int c)
    {
        bool isFound = false;

        for (int i = 0; i < nums.Length; i++)
        {
            c++;
            
            if (t == nums[i])
            {
                isFound = true;
                Console.WriteLine("Found!");
                break;
            }
        }

        return  c;
    }

Thinker•2y ago

Nothing will happen with the parameter you call the method with.

Doombox•2y ago

in this case it's important to understand how value types actually work if you want to understand the mechanics of what's happening

Thinker•2y ago

this has nothing to do with value types just passing by value

Doombox•2y ago

sure, phrasing once c is passed to the method, only the value is actually passed, because int is a value type so whatever happens to be assigned to c by the return c; statement is what will be returned

DarmaOP•2y ago

Ah thanks

trunksvn•2y ago

😳

Doombox•2y ago

if the int parameter c was actually a reference type, just a class called Foo, then (assuming you didn't directly assign something new to that reference) you would actually be returning a reference to the original object we also have ref and out keywords for parameters which can make things a bit more confusing

DarmaOP•2y ago

yeah it is a bit confusing so from what I can understand is that since I'm declaring count as an int variable it won't have any actual changes unless I redefine it before the return but that would be hard to do since count is being altered inside a for loop and I won't be able to access it once I go outside the loop's scope

Doombox•2y ago

c++; this is essentially shorthand for c = c + 1; in this case the value of c is scoped to the method, so wherever you modify this value in the scope of the method, c will always represent whatever value you've assigned to it

DarmaOP•2y ago

ah so it would in fact return let's say "5" if the loop took 5 tries to get to the target number then

Doombox•2y ago

yes my question is why would you pass c into the method? surely that should always start at 0?

DarmaOP•2y ago

int count = 0;
int target = 15;
int[] array = new int[40];
linearSearch(array, target, count);

int count = 0;
int target = 15;
int[] array = new int[40];
linearSearch(array, target, count);

Doombox•2y ago

yeah if the count should always start at 0, then there's no real need for the method to accept it

public static int linearSearch(int[] nums, int t)
{
  int c = 0;
  c++; // etc...
  return c;
}

public static int linearSearch(int[] nums, int t)
{
  int c = 0;
  c++; // etc...
  return c;
}

DarmaOP•2y ago

OH Man

Doombox•2y ago

then what you want is int count = linearSearch(array, target); because your method returns the count

DarmaOP•2y ago

ok that makes so much sense lol

Doombox•2y ago

your way could work if you used ref or out but there's no reason to do that returning the number makes way more sense

DarmaOP•2y ago

btw what are ref and out

MODiX•2y ago

Doombox#1389

REPL Result: Success

void AddOne(ref int val) => val++;
int i = 1;
AddOne(ref i);
Console.WriteLine(i);

void AddOne(ref int val) => val++;
int i = 1;
AddOne(ref i);
Console.WriteLine(i);

Console Output

Compile: 602.973ms | Execution: 37.258ms | React with ❌ to remove this embed.

Doombox•2y ago

it allows you to actually treat value types as if they were references to that value instead of just the value itself out is kinda old school and is mainly used for the TryParse pattern stuff

MODiX•2y ago

Doombox#1389

REPL Result: Success

var num = "123";
if(int.TryParse(num, out var parsed))
{
Console.WriteLine(parsed);
}

var num = "123";
if(int.TryParse(num, out var parsed))
{
Console.WriteLine(parsed);
}

Console Output

Compile: 538.635ms | Execution: 36.933ms | React with ❌ to remove this embed.

Doombox•2y ago

out is important to understand because TryParse is very commonly used, but you shouldn't have to create methods of your own that use them really, there are generally better and simpler ways to achieve similar results they have uses, but you'll get to that in the future

DarmaOP•2y ago

alright, thank you very much

Accord•2y ago

Was this issue resolved? If so, run /close - otherwise I will mark this as stale and this post will be archived until there is new activity.

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❔ Parameter Shenanigans