Because I like it and this particular machine doesn't support newer OS's. I have a PC with win10 too, and another one with Elementary OS. But I am fond of this computer in particular.

Well kinda reasonable, sorry :)

Okay. And to make a window in a Linux system? QT works too?

for(int i=0;i<=300;i+=20){ printf("%d\t%d\n",i, (i-32)*(5/9)); } Why this outputs "0 0" instead of "0 -17"?

You're incrementing i by 20 in each iteration.

First iteration doesn't increment

You're incrementing i by 20 in each iteration.

for(int i=0;i<=300;i+=20){ printf("%d\t%d\n",i, (i-32)*(5/9)); } Why this outputs "0 0" instead of "0 -17"?

Yeah but why (i-32)*(5/9) gives me 0 insted of -17?

for(int i=0;i<=300;i+=20){ printf("%d\t%d\n",i, (i-32)*(5/9)); } Why this outputs "0 0" instead of "0 -17"?

maybe let's not confuse anyone with what may look like ./ :D

http://ideone.com/4N6vYE

because 5/9 == 0

I didn't know that. I come from Java

I didn't know that. I come from Java

@Mouhieddine

In Java, int / int == int too, as far as I know

Works

It's bug prone actually. Depends on the compiler IIRC

Are you talking about C++?

Java

What compilers implicitly cast something that should obviously not be implicitly cast? O_O

I try to follow Java style conventions (except for starting function names from a lowercase letter) in C O_o

It’s hard to read your output

You get used to it, this is it

i dont like to spend my time in only job sorry!

int* fun(int *arr, int len){ int *b = new int[5]; //inserting values into array b return b; } int main(){ //made an array arr and the int *b = fun(arr,len); //How to iterate through array b } If I want to iterate through array b, then how do i know the length of b in main?

do what you want

you don't. But have len passed to the function. Otherwise you would need to pass it along.

ok. had this been the char* then, wouldn't it be easy to know that. loop till you find '\0'

Shut up man

and you cannot change that! of course... i like coding but i also like more eat delicios food as example..

you don't. But have len passed to the function. Otherwise you would need to pass it along.

didn't you want to say return it along than pass it along?

yes, return it along. Or in general pass it along with the pointer by e.g. creating a structure containing the size and the pointer (kinda similar to std::array)

So, I was reading the difference between arr and &arr. And it showed that arr mean the address of the first element and &arr mean the address of the entire array.

So, I was reading the difference between arr and &arr. And it showed that arr mean the address of the first element and &arr mean the address of the entire array.

int* fun(int *arr, int len){ int *b = new int[5]; //inserting values into array b return b; } int main(){ //made an array arr and the int *b = fun(arr,len); //How to iterate through array b } If I want to iterate through array b, then how do i know the length of b in main?

arr is the pointer to the first element and where the array starts. &arr is the address of the pointer to the array / memory region

sizeof(arr)/sizeof(arr[0])

I was going to say if this is correct of not.

it's not. since arr is a pointer (hence size 8 or 4...)

No