Can any one explain how this output comes?

Question

What is the output of this code?
int i=1, j=2, k=3;
int w;
w=(i,j,k);
cout<< w;
int r;
{r=i,j,k};
cout << r;
int q;
q= i, j, k;
cout << q;

//Output 311

Answer 1

🇮🇳Omkar🕉 My understanding regarding variable declaration and definition in C++ are as follows:

A variable declaration involves declaring the type and the symbol name for a variable.

A variable definition involves defining the storage location for that variable which might be defined in another file or when the variable is declared.

Therefore, the first line of the code below could be considered both declared and defined. It declares the type and symbol name and defines the variable is global.

int i;
int main() {
i = 42;
}

On the other hand, as pointed out by ~ swim ~, the line below only declares the variable which is defined elsewhere:

extern int i;

The article linked below explains these differences quite well:

https://www.cprogramming.com/declare_vs_define.html

Answer 2

According to Wikipedia
https://en.m.wikipedia.org/wiki/Comma_operator
The comma operator evaluates it's first operand and discards its result and returns result of second operand.

So the expression (i, j, k) will return only k other operands like i and j will be discarded.

therefore w is initialized with k (3)
So cout<<w; gives you 3.

🤔Most important thing to remember here is that comma operator has least precedence in all operators! So in order to make sure that right hand side gets evaluated first they are using perantheses ().

https://en.cppreference.com/w/c/language/operator_precedence

next r=i,j,k;

As stated earlier comma has least precedence. So = operates first and r gets initialized with i (1)

printing it you get 1.

The same logic as above for next
q =i, j, k;
q = i; is evaluated first due to operator precedence.
Printing q you get 1

Hence 311
Happy learning.
Wish you success!

Answer 3

~ swim ~ sometimes does words confuse me too 😂😂😂

Answer 4

In w = (i,j,k) we assigned each value to w so in this case last value will be assigned to w so value is 3

In

{
r = i, j, k
}

We assigned the value of i to r within the block so r will be 1

In q = i, j, k we assigned the value of i to q without blocks so q will be 1

curly braces doesn't matter to assign value. It just confuse you.

Answer 5

A J
In simplest words.
int x; //declaration

int a =5; // initialization

int foo;
foo = 5; //assignment.

Declaration will introduce an identifier,
int x; will make x be variable of type x
void bar(); will make bar a function identifier //will introduce prototype.
class C; will make C a class.
Etc. It's basically introduction of identifier.

initialization - assigning object, variable it's initial value.
int x =5;

Assignment - assigning a value after declaration.
Assignment can happen any timeas long as identifier is in scope. But initialization is done when identifier is introduced.
Initialization can be called combination of assignments and declaration.

*Anyone correct if wrong.

Answer 6

~ swim ~ now definition is fourth. Now one more confusion 😁

Answer 7

A J
"In w, (i, j, k) we assigned each value to w so in case last valye will be assigned to w so value is 3 " - AJ

r is not assigned all values, it gets initialized with value returned by expression within parantheses on right hand side.

Next
*Block don't make any change here. It's precedence that causes r to get initialized by i, before operation of comma.

Same for last statement.

Thank you.
Keep helping 🤗

Answer 8

Thanks 🇮🇳Omkar🕉 and ~ swim ~ for explanation. I need to learn more on c and c++.

But assigned by i and initialised by i both are same. in both case others value will change on change of i

Answer 9

~ swim ~ 😢

😁thank you friend.
Still I don't think it's definition. 😅

Because it declares x as an integer.

Let me give an example,

class MyClass; //declare

class SwimsClass{
int answers;
};
// I think of SwimsClass as definition 🤔

So in int x; we are not explaining(defining)what a int be like. just telling that it'll be int

I have mostly seen declarations , definition words in case of functions.

declaration - just prototype.
definition - explain working , body
edit:
int getAge(); //decalre
int getAge() //definition, includes body.
{
return age;
}
*Again free for correcting. 😇

Answer 10

~ swim ~ initialization in block scope don't make any change 🤔(only declaration will do)
Am I right?

Answer 11

Initialize, declaration and assignment

🇮🇳Omkar🕉 ~ swim ~ Can you tell the difference between them. It's little bit confusing.

Answer 12

Malem Yengkhom It's related to this question so we can.

Answer 13

~ swim ~
Reference site 📝call declaring/defining primitive type variables as declaration

http://www.cplusplus.com/doc/tutorial/variables/
(Section :Declaration of variables )

Quoting a sentence
" C++ is a strongly-typed language, and requires every variable to be declared with its type before its first use. "

Anyways ,
That's lot of confusion.

We can call it whatever we like.

Only thing that matters is code should work😁👌

Thanks for all corrections and your time 🤝

Answer 14

But this kind of topic is not discuss in sololearn here

Answer 15

Тоджинисо Бердиева What mmm?