@programminginc - страница 1776

Stanislav

constexpr std::string & std::vector

Stanislav

operator <=>

Stanislav

and finally math constants is in standard

joel

you're not stating differences you're just being random

Stanislav

https://en.wikipedia.org/wiki/C%2B%2B20

Stanislav

https://en.wikipedia.org/wiki/C%2B%2B20

🤔 lerp is not described

Robin

and finally math constants is in standard

ok, great

Anonymous

Can anyone plz tell me which is the best ide for c programming??

Mat

Can anyone plz tell me which is the best ide for c programming??

Neovim :P

Osa

How you learn c language

Mat

How you learn c language

Learning it... Studying the theory, practicing and doing some little projects

Osa

Can you teach me c lenguage please

Dima

What the hell

Anonymous

Learning it... Studying the theory, practicing and doing some little projects

Ohk.. thanks for suggesting me..

Mat

Can you teach me c lenguage please

Gosh, are you willing to pay?

Osa

Yaaa

Anonymous

Thanks.

Mat

Yaaa

There are a lot of books and courses on the net. Use them :) usually the prices are kinda low for what they teach, and sometimes they're free, too.

Osa

Can you send that links

Francisco

Can you send that links

Can't you use Google like anyone else?

Osa

Ok

Osa

But how can i learn c language

Dima

But how can i learn c language

#googleit

Osa

If u dont mine can u teach

Dima

Nobody will

jack

How to make this pattern in c?? Plz hlp.. tell me only logic 1 23 456 78910 11 12 13 14 15

for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { printf("x++"); } printf(" "); }

Dima

for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { printf("x++"); } printf(" "); }

Oh No

jack

Oh No

?

joel

If u dont mine can u teach

learn it yourself you have a brain

joel

for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { printf("x++"); } printf(" "); }

no

jack

no

ok

jack

👍

joel

there is something you gotta do within the inner loop

joel

declare a temp var outside the two loops

joel

and assign 0 to it

jack

i just told the logic

joel

no it's wrong

Francisco

i just told the logic

You gave a completely wrong code snippet

jack

forgot to use \n at where i gave space for outer loop

jack

for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { printf("x++"); } printf("\n"); }

Dima

for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { printf("x++"); } printf("\n"); }

jack

for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { printf("x++"); } printf("\n"); }

assign x=1 at the time of declaration

joel

for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { printf("x++"); } printf("\n"); }

now that's correct

Anonymous

#include<iostream> using namespace std; int* fun() { int arr[2]={1,2}; return arr; } int main() { int *arr; arr=fun(); cout<<arr[0]<<" "<<arr[1]; }

Anonymous

why this is working??

Anonymous

how local variable address can be returned

Anonymous

i think it get's destroyed after the call.

Mat

i think it get's destroyed after the call.

Yep

Anonymous

#include<iostream> using namespace std; int* fun() { int arr[2]={1,2}; return arr; } int main() { int *arr; arr=fun(); cout<<arr[0]<<" "<<arr[1]; }

then how it returns the address

Anonymous

if the arr is getting destroyed

Mat

arr inside fun is destroyed, but the value is copied into the arr inside main before that

Anonymous

then how it can return it

Anonymous

arr inside fun is destroyed, but the value is copied into the arr inside main before that

bro i used pointer to return a address

Mat

bro i used pointer to return a address

So what? An address is a value

Anonymous

but it the arr space i created if it is getting destroyed how it can display

Anonymous

if address is copied in main arr

Anonymous

if the values on that array gets destroyed

Anonymous

how can we get the value

Mat

The variable named arr inside fun is destroyed, but the array is still there. Before that, fun's arr is copied inside main's arr. Now main's arr holds the address of that array

Anonymous

The variable named arr inside fun is destroyed, but the array is still there. Before that, fun's arr is copied inside main's arr. Now main's arr holds the address of that array

array will not gets destroyed??

Mat

array will not gets destroyed??

It's not like someone will go and wipe all that memory.

Anonymous

https://www.youtube.com/watch?v=RWNM7CzDNyY watch this video in this the same code gives array.

Anonymous

sry gives * error.

Mat

sry gives * error.

Which minute?

Anonymous

Which minute?

0:53

Ariana

It should be easier to think of it this way Functions return one number, it could represent a integer, float, character, pointer etc. Any local variables can effectively be considered gone after returning So since functions can’t return like multiple numbers, you’ll got to return a pointer

Mat

0:53

I see a warning, not an error. They're different.

Anonymous

I see a warning, not an error. They're different.

the program stopped after that

I_Interface

I see a warning, not an error. They're different.

warning = ub

Anonymous

It should be easier to think of it this way Functions return one number, it could represent a integer, float, character, pointer etc. Any local variables can effectively be considered gone after returning So since functions can’t return like multiple numbers, you’ll got to return a pointer

so arr will not get deleted??

Anonymous

after function call.

Anonymous

the memory space created will not get wiped out??

Ariana

Local variables can be considered gone after returning Local variables are 'pushed' onto the program stacked and 'popped' once you return

Mat

the program stopped after that

Fact is you should hope the right address is passed to main's arr, as that address (the one inside fun's arr) has only fun's scope. Usually you create the pointer in your calling function (main in this case) and you pass it to your called function (fun in this case)

Anonymous

Local variables can be considered gone after returning Local variables are 'pushed' onto the program stacked and 'popped' once you return

so what about arr?

Ariana

#include<iostream> using namespace std; int* fun() { int arr[2]={1,2}; return arr; } int main() { int *arr; arr=fun(); cout<<arr[0]<<" "<<arr[1]; }

so here you declared it locally right, it'll be gone after returning

Anonymous

so here you declared it locally right, it'll be gone after returning

i returned its address to main

Anonymous

so why the elements are still present in same array

Ariana

So basically how local variables work is that it gets pushed to a stack, then it gets popped So what it means to get popped is that the area can be overwritten, so the pointer may still be valid, but if you say call another function you’ll have almost no clue what is the pointer doing

Anonymous

So basically how local variables work is that it gets pushed to a stack, then it gets popped So what it means to get popped is that the area can be overwritten, so the pointer may still be valid, but if you say call another function you’ll have almost no clue what is the pointer doing

thnx man'