an English speaking Russain, with arabic profile. seems funny😊

Best superspy, wanna be my partner?

true, in the above example why argv is of type char **. and what that means?

char** is basically a pointer to memory where char* lies. It doesn't say if there are other pointers or how many. As for array with empty [] on low level it's the same pointer. Array with a size is created on stack, not in the heap, and threated differently by compiler.

true, in the above example why argv is of type char **. and what that means?

argv is char** because is an Array of string. String in C are Array. Array in C is * And now have char**

an English speaking Russain, with arabic profile. seems funny😊

By the way, why you started from pure C? Why not C++ with its std::array, std::vector and other pretty things? Is it some requirements from your university?

I have nothing against learning C, but it's much more low-level than modern C++

By the way, why you started from pure C? Why not C++ with its std::array, std::vector and other pretty things? Is it some requirements from your university?

Because C > C++

argv is char** because is an Array of string. String in C are Array. Array in C is * And now have char**

> Array in C is * No that's false. Arrays do decay into pointers, but they aren't pointers

> Array in C is * No that's false. Arrays do decay into pointers, but they aren't pointers

yes

But [] is equal to * char* v[] Is char** v

For this reason have write array is *

int main(int count, char* args[]) And int main(int count, char** args) they are the same thing

int main(int count, char* args[]) And int main(int count, char** args) they are the same thing

Actually they are not

Actually they are not

In this case are identical

https://stackoverflow.com/questions/10186765/what-is-the-difference-between-char-array-vs-char-pointer-in-c

Yes

int main(int count, char* args[]) And int main(int count, char** args) they are the same thing

[] Not create array

But some may think that generally char* and char [] are same

[N] create Array and isn't identical to *

Compiler not know symbol []

I have nothing against learning C, but it's much more low-level than modern C++

What is low level? U mean C is more stupid and retarded

Compiler translate [] to * And a[1] Is *(a+1) For demostrate this you can build this code without problem int a[10]; 1[a]=0;

Difference between Array and pointer is only in declaration. int a[10]; int* b; a is an Array than reserve space in stack. b is only a pointer.

an English speaking Russain, with arabic profile. seems funny😊

Call me the Mystery Man.

Difference between Array and pointer is only in declaration. int a[10]; int* b; a is an Array than reserve space in stack. b is only a pointer.

No, actually not. char* a = "1234"; char b[] = "1234"; sizeof(a) returns 4 or 8 depending on 32/64bits, sizeof(b) returns 5

Difference is not in declaration only

No, actually not. char* a = "1234"; char b[] = "1234"; sizeof(a) returns 4 or 8 depending on 32/64bits, sizeof(b) returns 5

In this case [] is different to [] in a function argument

Yes ofc

No, actually not. char* a = "1234"; char b[] = "1234"; sizeof(a) returns 4 or 8 depending on 32/64bits, sizeof(b) returns 5

[] get automatic size to inizializzation

You can't in a function arguments

void foo(char z[]) Z not reserve space for Array and is Always only a pointer

Because C > C++

U mean C != C++

U mean C != C++

Tru

Call me the Mystery Man.

Call you weirdo 😂

By the way, why you started from pure C? Why not C++ with its std::array, std::vector and other pretty things? Is it some requirements from your university?

I'm just into it

In this case are identical

In this context they might have the same effect, but that doesn't make them the same thing

Call you weirdo 😂

You are the one learning C tho

In this context they might have the same effect, but that doesn't make them the same thing

compiler not know []

You are the one learning C tho

Learning to later have to unlearn lol

compiler not know []

If it didn't that code won't compile lol

And it does

I have nothing against learning C, but it's much more low-level than modern C++

You seem to tell this like knowing a low-level programming language is something bad

If it didn't that code won't compile lol

can build because translate to *

Learning to later have to unlearn lol

I pity the young woman, by the time she's done, her kids are already writing C++29

can build because translate to *

can build because translate to * based on the context that [] is used in

can build because translate to * based on the context that [] is used in

int a[5]; 1[a]=0;

That I can agree with

But it pretty much makes your initial statement invalid

int a[5]; 1[a]=0;

So?

You seem to tell this like knowing a low-level programming language is something bad

C is bad-level programming.

So?

This works because compiler not know []

This works because compiler not know []

Yes it does lol

C is bad-level programming.

And you're gonna explain I guess?

If it didn't how would it compile it?

And you're gonna explain I guess?

Wrong question.

Is this bot

Is this bot

No

Yes it does lol

1[a] translate to *(1+a) a[1] to *(a+1)

1[a] translate to *(1+a) a[1] to *(a+1)

So what does int a[10]; translate to?

This works because compiler not know []

It only knows pointers huh

So what does int a[10]; translate to?

It's unique difference

Hint: it fucking doesn't

It's unique difference

Exactly

can build because translate to * based on the context that [] is used in

^

in this end this querrel is pointless

Exactly

But in this case not allocate memory

both syntaxes are valid

doesnt matter what the compiler sees in them

in the end, one will be evaluated to some space reserved on the stack or in the heap

on function argument you can't allocate array

the other will result in some pointer arithmetic

on function argument you can't allocate array

myFunc (new Object()); ?

myFunc (new Object()); ?

In C?

In C?

myFunc (malloc(sizeof(myObject)))

the other will result in some pointer arithmetic

And are those two things the same in your world? Because for me no they aren't

And are those two things the same in your world? Because for me no they aren't

nope, they aren't

Exactly

they are inherently different

void foo(int z[10]); Z is simple pointer, not allocate memory

Wrong question.

Not sure what you referred to, though. Bother to explain your statement?