Arrays as function parameters in C

Source code can be found here:

===== Support us through our store =====

===== Check out our website =====

===== Check out our Discord server =====

[MsHofmannsJut]

Just remember, that when you're accessing array elements, all array subscripts are sugar for pointer arithmetics.
a[2] is sugar for *(a+2).
b[size][2] is sugar for *( *(b+size) + 2 ).

Then, of course, it makes sense why `arrParam[2][2]` does not work for pointer-decayed arrays:
it's *( *(arrParam + 2) + 2 ), which in this case *( 3 + 2 ).
and explains why arrParam[10] is ok: *( arrParam + 10 )

The additional subscript doesn't add a dimension to the pointer. This would make no sense.
[x][y] multiplies the x dimension y times. Just like you would if you were drawing a table.
You can initialise a multidimensional array like this:
`arr[2][2] = { 1, 2, 3, 4 };`
and it will be the same as
`arr[2][2] = {{ 1, 2 }, { 3, 4 }}`

arr[0][2] == arr[1][0] == *( arr[0] + 2 ) == *( arr[1] ) == *( *arr + 2 )

[nicolasedome]

This is a fancy channel of C programming. I love this. Thanks for your effort.

[thosewhowishb]

Wow, I never realized it was like so! I could swear using arrParam[0][0] inside the function would work. Thanks for this!

[fifaham]

Excellent presentation of dealing with multi arrays outside the calling fucntion.

[divyanshuverma]

After many unsuccessful attempts, I finally understood the logic behind arrays in c and a good way of passing a 2d array in a function.
Wow, amazing work dude.

[iAmTheWagon]

Best teacher. My pain and anguish have been extinguished!

[KangJangkrik]

Thanks sir, you just saved me from another segmentation fault pain

[tomaszstanislawski]

The cast to `(int*)` can easily invoke UB since accessing 1-D array is allowed only for indices 0 to 5. The proper declaration of `printSize` would be `void printSize(int (*arrParam)[5])` or `void printSize(int arrParam[][5])`.

[SuperSamsosa]

Top explanation. Seen others but yours is concise and good to understand. Thx

[zltn_brkl]

You are the best always telling the gist.

[zimablue]

what do you think about this way:

double a1[] = { 0.5, -1.0, -1.0 };
double a2[] = {-1.0, 1.0, 2.0 };
double a3[] = {-1.0, 2.0, 1.0 };

double *A[3];

A[0] = a1;
A[1] = a2;
A[2] = a3;

if you pass A to a function as a double**, you have no problem at all

[jettgoldberg]

0:13 me whenever I learn anything related to coding LMAOOO
Thank you for making this video! I didn't understand this concept at all in class and I need to learn it to do homework tonight

[MesbahSalekeen]

This channel is not for beginner level c tutorial. It's more like you learnt the basic c terms and trying to make a project and getting your concepts more clear.

[escapefelicity]

It might be a REAL GOOD IDEA to post the code

[tomerfata]

Can you also explain about dynamically allocated arrays as parameters?

[itsmespiazzy]

Hey, CodeVault. I'm having a hard time understanding this topic.
In 11:37 you say if I dereference a 2D Array I'll get just the first integer of the matrix, but when I done that in my main() function actually what I get is an error, because it expects an variable of type array, with is true because matrix[0] have an array.
I really don't get that part, why *matrix works differentelly in a function parameter and main.

[yuui]

Thank you for the video! I just have one question, why do you have to convert the two dimensional array to an integer pointer in the main function before passing it to the function, I thought that an array becomes an integer pointer anyway when getting passed to a function?

[Hevletica]

What about passing the length of the second dimension as an argument in a function definition?

[muaadakmal]

Sir, an array declared in heap using malloc is global or not, as it retains in memory even after function returns, so can we edit it in any other function without passing as argument instead simply calling inside the fun by the ptr name

[stack.]

I think you can just do something like... without the cast

void printArr(int *arrArg) {
printf(" arg[0][2] : %d", (arrArg[0]+2));
}

int main(void) {
int arr[][5] = {
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},


};
printArr(arr);
}