I do not get it...

Question

If

int a=3;
int b=2;
b=a++;
cout<<++b;

then WHY THE HELL the result IS EQUAL WITH 4 ??????

Answer 1

int a = 3;
int b = 2;

// First, b = a, then a increment by 1
// b = 3, a = 3
b = a++;

// First, b increment by 1, then print out the b to the console
// b = 4, Output : 4
cout << ++b;

Important note:

120% avoid doing the incremental/decremental operation "in a row" inside the cout. That causes undefined behavior.

Example: Compare two groups together. They both do the same operation but as you clearly see, former expresion's behavior is completely undefiend.

#include <iostream>
using namespace std;

int main()
{
int x = 1;
int y = 2;

cout << x++ << " " << ++x << " " << --y << " " << y++ << "\n\n";

x = 1;
y = 2;

cout << x++ << endl;
cout << ++x << endl;
cout << --y << endl;
cout << y++ << endl;
}

Output:

2 3 2 2

1
3
1
1

Discussion : [https://stackoverflow.com/questions/33445796/increment-and-decrement-with-cout-in-c]
Live link : [http://cpp.sh/8xuaw]

Answer 2

@Enurien it's work like this
int a=3 ; //a has assigned value 3 initially
int b=2; // b has assigned value 2 initially
b=a++; // a++ is post increment so first value is used then incremented so value 3 is assigned to b
b=3 now
cout<<++b; //this is pretty increment first value is incremented by 1 then store so b=4
and it will be print as output b=4

Answer 3

https://www.sololearn.com/Discuss/407846/?ref=app

Answer 4

b=a++ \\first it will assign the value of 'a' to 'b' and then add up 'a+1'.
so b=3 and a=4

in 2nd case '++b' it will add up to 1 and then assign.
so 'cout<<++b' means 'b+1' and print it.