How to pass arrays to functions in C

Passing arrays to functions is a simple task can get really confusing when the compiler tries to change the code for you or just doesn't behave the way you expect.
Check out our Discord server: / discord

[@pointles2221]

OMG I FINALLY UNDERSTAND, THIS CONCEPT CAN BE SO HARD AND YOU MADE IT SO MUCH EASIER

[@kipster-ll6po]

"You, my friend, are a POINTER." Love it!!!

[@user-vg1qo5gi3l]

Thanks for explaining! Was looking for this for a while

[@sufiyansadiq9962]

this was exactly what i was looking for, thankyou so much for explaining this to me

[@wahabfiles6260]

Beautifully explained.

[@pokemaster1359]

You are amazing! Thank you for the understanding!

[@naboulsikhalid7763]

you just point(put) your finger on the things that mostly matter. ARRAY POINTER. thank you for your effort

[@juggles5474]

Awesome video, thank you!

[@mehdirexon]

it was helpful thank you.please upload more video about pointers and hard concepts in c/c++.

[@MuksEmmaN]

I love your videos please release more sir🙏

[@We.viriya]

Thx God. You help me! Love 🙏🏻❤

[@rechardnabo7751]

from the function the receive array as a pointer, how do u get the size or length of the array since it returns the 8 bytes

[@CodeVault]

You can't. You have to store it separately.

[@vectiphy]

How do we create an Array of functions i.e an arrray consisting of functions parameters.
is this possible?

[@CodeVault]

You mean a function pointer array?:
void fun1() { }
void fun2() { }
void fun3() { }
void (*function_array[3])() = {fun1, fun2, fun3};

[@numairdubas2688]

What about the argv array passed in the main function? Where does that point?

[@CodeVault]

That's taken care of by the OS. It is allocated in more or less the same way you'd do it by calling a function yourself. As for where in memory it's stored... it depends, here's an answer from others that already did this research: stackoverflow.com/questions/26416678/where-does-the-os-store-argv-and-argc-when-a-child-process-is-executed

[@numairdubas2688]

Thanks! Great video btw

[@hersona6560]

thanks

[@gonfreecs8174]

and if i don't want to update the array values what should i do??

[@CodeVault]

You could pass a copy then

[@gonfreecs8174]

@@CodeVault i pass the array, and the inside the function i copy it to a new array. But i mean thats the unique solution?

[@CodeVault]

Yea, because if you don't want to modify you have to make a copy of it. If you pass an array to a function:
int testFunction(int arr[]) { ... }
That arr identifier would be treated as a pointer. So it won't ever automatically make a copy of your array on the stack.
MAYBE, if you want it to be automatically copied you could have a struct that stores the array and pass that struct directly to that function:
typedef structArray { int arr[50]; } structArray;
int testFunction(structArray arr) { ... }
That should copy the array automatically I think.

[@gonfreecs8174]

@@CodeVault that could be a good solution. i will try, thanks

[@vytautaskuklys3829]

How to change values of arrays in a function when they are passed like this - int** arr?

[@CodeVault]

If it's a single dimensions array, you should pass it as int* and change it like any other array, if it's a multi-dimensional array then it's more complicated. There's this video on the topic: code-vault.net/lesson/a4mqix89a0:1610303947019

[@loonathefoxgirl6375]

Pointers in disguise
arr[n] is actually just a way to write *(arr +n). Thats why you can be cursed and do stuff like 3[arr] and it still works. Its just pointer math

[@cristiandragusin3814]

I have an exam tomorrow, so best spent 5 minutes!

[@QuimChaos]

what about multidimensional arrays? 8)

[@CodeVault]

There is this video on the topic:
code-vault.net/lesson/zgpd1zov1t:1603733527939