Please Explain what is difference between i=i++; and i++. Please look at code below and make me understand how it works in case of first condition?

Question

look at these two conditions in the code and output respectively:

#include <iostream>
using namespace std;

int main()
{
    int i=1;

   i=i++;  // first condition 
    i++; //2nd condition 

   cout << i << endl; //it will print 2 in second condition **but why print 1 in first condition?**
 
    return 0;
}

Answer 1

i++ will increment i, but will return the original value. So,

int i=0;
int x = i++; // x = 0, i = 1

if you wanted to return the new value, use ++i instead.

int i=0;
int x = ++i; // x=1, i=1

Answer 2

the statement i++ increases i after the statement is executed. in contrast to ++i which increases i before the statement is executed. so the code

int i=1;
i=i++;

does:

1. (line 1) initialize variable i with value 1.
2. (line 2) insert the current value of i (1) to the variable i (it is not yet changed).
3. (line 2) increases i from 1 to 2.

the code

int i=1;
i++;

does:

1. (line 1) initialize variable i with value 1.
2. (line 2) increases i from 1 to 2.

The difference is more clear in those examples:

example a

int i=1;
i=10+(i++);
std::cout << i << std::endl;

does:

1. (line 1) initialize variable i with value 1.
2. (line 2) compute the result of 10 plus the current value of i(1) (result 11).
3. (line 2) insert result 11 of last computation to i.
4. (line 2) increases i from 11 to 12.
5. (line 3) print 12.

example b

int i=1;
int j = i++
i=10+j;
std::cout << i << std::endl;

does:

1. (line 1) initialize variable i with value 1.
2. (line 2) insert the current value of i(=1) to j.
3. (line 2) increases i from 1 to 2
4. (line 3) compute the result of 10 plus the current value of j(=1) (result 11).
5. (line 3) insert result 11 of last computation to i.
6. (line 3) print 11

example c

int i=1;
int j = i++
i=10+i;
std::cout << i << std::endl;

does:

1. (line 1) initialize variable i with value 1.
2. (line 2) insert the current value of i(=1) to j.
3. (line 2) increases i from 1 to 2
4. (line 3) compute the result of 10 plus the current value of i(=2) (result 12).
5. (line 3) insert result 112 of last computation to i.
6. (line 3) print 12

summery

All of the following codes leave i with the same value at the end:

a)

int i = 1;
i++;

b)

int i = 1;
i = i++;

c)

int i = 1;
int k = i++;