I was talking about passing a parameter using a pointer is the right solution

‘ void function( struct mytype mystruct[] ) { // access it with mystruct[ index ] } ‘ this how to pass it as a parameter

this one is not

but please note that the memset version is valid for specific values like 0, other values are possibly invalid, std::fill works for any values you want to assign.

Really ? I said array parameters will turn into pointers

according to the intel x86 calling conventions: when you pass parameters to a function, the register rax ( eax on 32bit machines) will contain the first parameter then the next parameter will be in rcx register then the third is stored in rdx register and so on so if an array is passed to a function then each element in the array will be pushed on the stack and the rax register will point to it (in case that the array was the first parameter). but if you use a pointer then the rax register will point to the address of the original array in memory whether it is on the stack or the heap or even on a different memory page if you are sharing memory between multiple programs

according to the intel x86 calling conventions: when you pass parameters to a function, the register rax ( eax on 32bit machines) will contain the first parameter then the next parameter will be in rcx register then the third is stored in rdx register and so on so if an array is passed to a function then each element in the array will be pushed on the stack and the rax register will point to it (in case that the array was the first parameter). but if you use a pointer then the rax register will point to the address of the original array in memory whether it is on the stack or the heap or even on a different memory page if you are sharing memory between multiple programs

which x86 calling convention is this? where is that written?

because when you pass a parameter to a function you are copying it’s content somewhere on the memory (usually the stack but sometimes cpu registers if the value is small enough) so if you have a big array then you pass it as a parameter to a function. you will end up with two instances on memory of the same array, this is why more memory will be required not mentioning that copying the content of a huge array might take sometime and will slow your program

There is no such a thing. Array arguments in C don't exist. Compilers accept it for various bad historical reasons and since array arguments don't actually exist in C, it is the size of the pointer, not the array. This is just a pointer copy, you don't copy the array elements via arguments. For ex, this will print the size of a pointer, not int [100]. void f(int arr[100]) { printf("%zu\n", sizeof(arr)); }

There is no such a thing. Array arguments in C don't exist. Compilers accept it for various bad historical reasons and since array arguments don't actually exist in C, it is the size of the pointer, not the array. This is just a pointer copy, you don't copy the array elements via arguments. For ex, this will print the size of a pointer, not int [100]. void f(int arr[100]) { printf("%zu\n", sizeof(arr)); }

I understand that arrays are an abstraction of blocks of memory pointed accessed by a pointer to its address and you where right they are not pushed on the stack which make a lot more sence but what do you mean by historical reasons ? isn’t it considered as one of the features of high level languages to provide such abstraction?

i have some stupid question. I want to increase a string by multiplying by a number, but it seems that in C is not possible to do this printf("some text" * 3) in order to myltiply this line 3 times. But I'm getting an error. Is there any way to do this. PS. Im started learn C, before i learn python and i can make it with this lines of code print("text"*3)

you can't do that in C, you need a loop

i have some stupid question. I want to increase a string by multiplying by a number, but it seems that in C is not possible to do this printf("some text" * 3) in order to myltiply this line 3 times. But I'm getting an error. Is there any way to do this. PS. Im started learn C, before i learn python and i can make it with this lines of code print("text"*3)

i have some stupid question. I want to increase a string by multiplying by a number, but it seems that in C is not possible to do this printf("some text" * 3) in order to myltiply this line 3 times. But I'm getting an error. Is there any way to do this. PS. Im started learn C, before i learn python and i can make it with this lines of code print("text"*3)

How can i access an element in an array. C program need help plz

plz give some effort to learn it before dropping your question to the group!! it is very very very basic programming in C there are a lot of harder things outside there and you DO NOT make questions for all of them, you can't make a question for each line in your code, so learn, this trivial thing shouldn't be missed when you start learning

those get pushed on the stack

Hello, can anyone help with a C++ problem? 🥺

Can any one tell me main difference between this two functions prototypes void printSales (const int theArray[REGIONS] [PRODUCTS]); void printSales (int theArray[REGIONS] [PRODUCTS]);

Please use an formatting page like pastebin

Can anyone help me make this code functional. I want the user to be able to access all the choices like weekly menu, contact details... Here's the link to the code https://pastebin.com/fmRn1ZY6 Thank you in advance

Why invalid ?

Can anyone help me make this code functional. I want the user to be able to access all the choices like weekly menu, contact details... Here's the link to the code https://pastebin.com/fmRn1ZY6 Thank you in advance

You can use switch case, and display the desired output, required for the user.

Can anyone help me make this code functional. I want the user to be able to access all the choices like weekly menu, contact details... Here's the link to the code https://pastebin.com/fmRn1ZY6 Thank you in advance

Not all program use cdecl or stdcall calling convention

How can i access an element in an array. C program need help plz

I have done that my tests are still failing 😞

You do it with: array[ index ] or you can use a pointer: int array[3] = { 1 , 2 , 3 }; int * array_ptr = array; Then access it with: *( array_ptr ) //will return 1 *( array_ptr + 1 * sizeof(int) ) //will return 2 *( array_ptr + 2 * sizeof(int) ) //will return 3

You do it with: array[ index ] or you can use a pointer: int array[3] = { 1 , 2 , 3 }; int * array_ptr = array; Then access it with: *( array_ptr ) //will return 1 *( array_ptr + 1 * sizeof(int) ) //will return 2 *( array_ptr + 2 * sizeof(int) ) //will return 3

This is wrong. This is Undefined Behavior because you are accessing elements beyond the array's limits

He assumes arays are 3 but 4 indices.Is that what you mean?

No. Assuming int is 4 bytes long, he is accessing array elements at position 4 and 8 which is Undefined Behavior as the original array is just 3 elements long (meaning only positions 0,1 and 2 are valid).

Isnt that a defined Array size he declared there? Its allowed

What are you talking about? His array size is 3. So you can access only elements arr[0], arr[1] and arr[2]. In his code he is doing something like *(array_ptr+1*sizeof(int)) which is equivalent to arr[4] assuming integers are 4 bytes long. This is wrong. *(array_ptr+2*sizeof(int)) accesses the element arr[8] which is also wrong.

No. Assuming int is 4 bytes long, he is accessing array elements at position 4 and 8 which is Undefined Behavior as the original array is just 3 elements long (meaning only positions 0,1 and 2 are valid).

The number of elements of the array is 3 multiplied by the the size of int for each element Which makes the total 12 bytes not 3 bytes(like strings) and I'm accessing at offset of 8 bytes which put me at the start of the 9th byte and my array is 12 bytes in size

It's impossible to shrink the size of an array, so you can't erase element in an array, if you want to do this, use variable length containers like std::vector and then use their erase_if method instead. However, there is a workaround for arrays. Those who like functional programming styles may tend to create new sequences instead of modifying the old one. For example, your_array | std::views::filter(pred) will create a new range containing all elements in your_array satisfying the predicate you offer. You can also use the old school std::remove_if and then use std::span or something to create a new non-owning wrapper.

No you are wrong. Try this code: #include <iostream> int main(){ int a[3]={99,999,9999}; std::cout << *(a+2*sizeof(int)); } See if it prints 9999. Now try this instead #include <iostream> int main(){ int a[3]={99,999,9999}; std::cout << *(a+2); } See if this prints 9999 And then understand why you are wrong.

Alright.. you were right But I'm still confused if you increment an address by 1 why would it access the forth byte instead of the second byte in memory I've written a fair amount of assembly to know that in fact we are accessing the 5th byte in memory

It is because when you do a+2, the compiler knows that a is a pointer to int and that you are trying to access an integer two places away from the current int. So it automagically increments the pointer to point to the right integer. This saves us from using sizeof calculations. So basically if you have something like this: int* ptr = <...>; *(ptr+2) = <...>; The compiler does something that is the equivalent of *((int*)(((char*)ptr)+2*sizeof(int))) = <...>

Isn't it just an address how the compiler can tell how I'm gonna be using my memory? How about if wanted to implement let's say a strange hash function that needs to access each byte in an array of integers Do I really need to cast the address to an char* then cast it back to int * each time?

It is because when you do a+2, the compiler knows that a is a pointer to int and that you are trying to access an integer two places away from the current int. So it automagically increments the pointer to point to the right integer. This saves us from using sizeof calculations. So basically if you have something like this: int* ptr = <...>; *(ptr+2) = <...>; The compiler does something that is the equivalent of *((int*)(((char*)ptr)+2*sizeof(int))) = <...>

This explains why the compiler would need to know the type of the data type accessed by any pointer

You do it with: array[ index ] or you can use a pointer: int array[3] = { 1 , 2 , 3 }; int * array_ptr = array; Then access it with: *( array_ptr ) //will return 1 *( array_ptr + 1 * sizeof(int) ) //will return 2 *( array_ptr + 2 * sizeof(int) ) //will return 3