Well you forgot to initialise variables

int a; a = add(&linkedlist, sayi); i did that and it that work

int a; a = add(&linkedlist, sayi); i did that and it that work

why should i do that like this? what is the reason

int a; a = add(&linkedlist, sayi); i did that and it that work

No, that's not what I meant

You have to initialise your variable "linkedlist"

I don't really know how to do that in your case a I did not dig deep into your code but that's what the error says

LL linkedlist; init(&linkedlist); i initialise that way, call by reference

simply do: int secim = 0;

Also you should do this

LL linkedlist; init(&linkedlist); i initialise that way, call by reference

Okay, give me some time taking a look...

Btw I did not read your error correctly first, so sorry for the confusion

@alkalemir okay, try that first: int secim = 0; I have the feeling that'll solve it

Even though it looks like it's unrelated, but error messages are weird sometimes.

int a; a = add(&linkedlist, sayi); i did that and it that work

i solve this like that

but i dont know reason

Okay, interesting 🤔

but i dont know reason

I wonder about it too

I wonder about it too

maybe someone can help us about this error 😄

Anyone know that

help me

Got it

#include<studio.h> int main() { printf("%c", 5["CodingLogic"]); return 0; } Can someone explain the printf statement?

#include<studio.h> int main() { printf("%c", 5["CodingLogic"]); return 0; } Can someone explain the printf statement?

Yes

Arrays are basically syntactic sugar in a way

Basically assume an array a[] a[5] actually is *(a+5)

But here's the interesting thing

5[a] becomes *(5+a)

Which is basically the same thing

So u mean to say *(a+5) and *(5+a) both are same?

Tl;dr a[5] = 5[a] = *(a+5) = *(5+a)

They're the same

"Codinglogic" is a string literal but it can be used as an array too

So 5["Codinglogic"] means the 6th element from the string literal and the %c printf format specifier prints the 6th element

Remember that arrays start at 0

The output is g

Ok thanks bro

Also other mistakes you made: Did you mean #include <stdio.h> You don't need to return 0 for main (not a mistake) Keep the { on the same line unless the function name is longer than 80 characters

Also other mistakes you made: Did you mean #include <stdio.h> You don't need to return 0 for main (not a mistake) Keep the { on the same line unless the function name is longer than 80 characters

returning 0 is the best practice it says the operating system that the program is executed and when you close the terminal in background it stops the execution of the program.

returning 0 is the best practice it says the operating system that the program is executed and when you close the terminal in background it stops the execution of the program.

No

Main is a special case If you reach the end of main it is automatically returned

There are cases such as looping code inside main where return 0 at the end would become unreachable code

For any other function with a non void return type not returning a value is undefined behavior

As for return types

Values are not sent to the operating system

The return value goes to the caller

When you return from main you're returning a value to whoever called the program

And you don't use the operating system to call your program You'd use a terminal, a file explorer or something else to call the linker entry point

And you don't use the operating system to call your program You'd use a terminal, a file explorer or something else to call the linker entry point

Ohh I didn't know these things.

Ohh I didn't know these things.

There was a time when they did but these days you don't even see the operating system

There was a time when they did but these days you don't even see the operating system

Yes I read this in a book that's why i am teeling this.

Main is a special case If you reach the end of main it is automatically returned

But main will return any value other (garbage value)

But main will return any value other (garbage value)

No

It will be 0

No

So waht will be the returning value and on what factors it depends

Your book is probably teaching C89

You should read up on C99

It will be 0

but if the program is exectued successfully.

Your book is probably teaching C89

Yes

You should read up on C99

The book second part contain this.

So waht will be the returning value and on what factors it depends

If the program is not executed successfully what will be tge returning value.

If the program is not executed successfully what will be tge returning value.

Always 0

#include <error.h> int main() { errno=5; }

Will return 0

Always 0

So, Why we use returning values in C ?

So, Why we use returning values in C ?

For that you must learn about the processor

When you call something the caller transfers control to what was called When a return is used the control is sent back to the caller

When you call something the caller transfers control to what was called When a return is used the control is sent back to the caller

yes i read it in functions but what is the use of return of main

yes i read it in functions but what is the use of return of main

Well something has to call main to start the program

When you return you are returning control back to the caller

This is much more visible in a terminal

When you return you are returning control back to the caller

So who is the caller in case of main

When the program runs you don't see the shell

So who is the caller in case of main

The shell

If you're in bash and run this program the return value goes in ?

The shell

it is for windows/linux

You can access ? by $? echo $?

Or printf $?

Can you suggest me something book/blog for what happens in behind when a program is executed ?

#include <stdio.h> int main() { int n1,n2; char ope; printf("Enter first number : "); scanf("%d",&n1); printf("Enter second number : "); scanf("%d",&n2); printf("Enter opertaion + , - , * \n"); scanf("%s",&ope); switch(ope) { case '+' : printf("\n\nAddition is : %d ",n1); break; case '-' : printf("\n\nSubtraction is %d ",n1-n2); break; case '*' : printf("\n\nMultiplication is %d ",n1*n2); break; default : printf("\n\nInvalid operation entered"); break; } return 0; }

Can you please explain why my output is coming wrong ?