Operators confusion

Question

This is where I am getting completely lost...

Int x = 34 (declaring x is equal to 34)
Int y = ++x; (no idea what this is declaring) //y is 35 (this is a note)

Int x = 34; (stating x is 34)
int y = x++ (Again, no idea what this is trying to declare) // y is 34 (another note)

How the heck does this work, what are we doing/accomplishing by putting this in?.. I do not understand. Anyone please help with clarification.

Also looking at the increment and decrements lesson I am also clueless as to what we are stating / trying to achieve with the formula. Or how it even works.. Please anyone lol

Answer 1

Past explanations: (there are lots of answers to this question already here)

<redacted list, original post spans 20+ threads on prefix/postfix operators. Newer threads were removed due to duplication>

https://www.sololearn.com/Discuss/160327/?ref=app
https://www.sololearn.com/Discuss/69892/?ref=app
https://www.sololearn.com/Discuss/252933/?ref=app
https://www.sololearn.com/Discuss/147409/?ref=app

Answer 2

Explanation:

int x = 34;
Does NOT "declare x is equal to 34"

It declares x as an integer variable and assigns 34 to it.

int y = ++x;
This declares y as an integer variable and assigns the value of x+1 to it.

++x is the same as x+1
This is known as a Pre-increment and it increases the value of x before executing the statement.
In this case ++x means that the value of x has increased by 1 and is now 35.
So, y=35;

int x = 34; (34 is assigned to x)

int y = x++;
This declares y as an integer variable and assigns the value of x to it but the value will be increased by one after being used.

x++ is also the same as x+1
This is known as a Post-increment and unlike the Pre-increment, it increases the value of x after executing the statement.
In this case x++ means that the value of x is still unchanged until after it is first used.
So,

//1st execution
System.out.println("y="+y);

//y=34

//2nd execution to test the value of x
System.out.println("x="+x);

//x=35

https://code.sololearn.com/csst17nt85Iz/?ref=app

Answer 3

Haha.. @Restoring faith...😂😂😂
Savage

Answer 4

Haha .. True @Shamima

Answer 5

@Rrestoring Faith. Faith has officially been restored. +1

Someone please bookmark this thread.

Answer 6

@Rrestoring Faith, it'd be easier for you if you explained instead 😂

Answer 7

Restoring Faith lolol I might as well add you XD you always answer quickly and effectively. Thank you!! I appreciate you.

Answer 8

Please refer to this code for reference

https://code.sololearn.com/W9BBJnbCBEv7/?ref=app

Answer 9

😷😷 following

Answer 10

that is an impressively long list lmao

Answer 11

https://www.sololearn.com/discuss/489143/?ref=app

Answer 12

++a and a++ are equals... in there own why
Int a = 3, b = 5 ;
++a;
Cout << a;
B++;
Cout << b;
Output will be
4 6
(A nd b will both increment by 1)

But post increment or decrement will use with assingment operator " = " then ++a(pre increment) & a++(post increment) are have different reasults lets see

Int a = b = c = d = 0;
a = b++;
c = ++d;
Cout << a << endl << b << endl << c
<< endl << d;

Output :
0(here post increment assing its value to a and the increamnt by 1)
1 (now B's value is 1 bcz of increament)
1( Here first D is increment bye 1 then assing its value to C
1

I hope u will understand

Answer 13

so when you use ++ before an integer, 1 is added to that integer on that line.

therefore,
x = 16
y = ++x; ( y equals x plus 1)
then y = 17
and x = 17

but if you use ++ after an integer the plus one wont be added until after the line it was used on.

so,
x = 80
y = x++ (y equals x and plus one is added afterwards so y is just...)
y= 80
but x = 81 because we still added one.

Answer 14

When we use the increment at the beginning ie pre increment it will first increment the value and then store it to the variable while in the other case it will first assign the value to the variable and after that it increments the x

Answer 15

Programming in c++

Answer 16

Programacion lineal

Answer 17

Hmm

Answer 18

Look
It is very simple, just try to understand it with a calm mind.

In your first example :

int x=34;
int y= ++x;

// this is a prefix increment operator.
This operator first increments the value of x by 1 and then uses the new value.
A simple way to remeber this is :
Prefix : increment and then use

In your second example :

int x = 34 ;
int y = x++;

// this is a postfix increment operator.
This uses the initial value first (in this case 34) and then increments the value.

Take another example of postfix :

int x = 34 ;
int y = x++ ;
int z = x ;

In is code, y =34 and z=35
This shows that the value of "x" is used first to declare value of "y" , then the value of "x" is incremented by 1. Then the value of "x" is assigned to "z".

An easy way to remeber postfix increment operator is :
Use and then increment

Hope your doubt is cleared.
If you still have any doubts, you can reach out to me, I'll clear them all.

Answer 19

Вообще понять не могу все это

Answer 20

Find and write the output of the following python code:
def fun(s):
k=len(s)
m=" "
for i in range(0,k):
if(s[i].isupper()):
m=m+s[i].lower()
elif s[i].isalpha():
m=m+s[i].upper()
else:
m=m+'bb'
print(m)
fun('school2@com')

How the output is SCHOOLbbbbCOM