I can't understand how and what is added here :( Не могу понять как и что тут прибавляется :( how to understand sum+=i++

Question

I welcome everyone. dear java gurus, tell a beginner on the fingers how to calculate this operator expression.I was reading Herbert Schildt's Java 8 beginner's guide and then I saw an interesting line. I don't understand how the final amount is calculated. Help me understand, thanks in advance
int i;
int sum = 0;
for(i = 1; i <= 5; sum += i++) ; <---------
System.out.println("Sum: " + sum);
15 🤔
/************************************************/
Приветствую всех. Уважаемые java-гуру, расскажите новичку на пальцах, как вычислить этот оператор выражение. Я читаю книжку Герберта Шильдта руководстводля начинающих Java 8, и тут я увидел интересную строку. Я не понимаю, как рассчитывается окончательная сумма. Помогите разобраться, заранее спасибо
int i;
int sum = 0;
для (i = 1; i <= 5; sum + = i ++); <---------
System.out.println ("Sum:" + sum);
15 🤔 как получилось 15?? 1+1+1=3 *5? Так? Но когда меняю на другие цифры я считаю не правильно. Буду благодарен за помощь

Answer 1

and last b = 3 + 3

Answer 2

oh, if you don't know, a += b is a shortcut for a = a + b...

Answer 3

for (i=1; i<=5; sum+=i++);

is a loop, wich do nothing at each body iteration (notice the semi-colon next to closing parenthesis)...

however, that doesn't mean the loop run without doing anything: the parenthesis contains 3 parts separated by semi-colons:

1) initialization (executed only once just before loop start), usually initializing a counter variable

2) condition (checked at each iteration start)

3) part executed at each iteration end (usually dec/incrementing counter)

in this case, counter i is initialized to 1, and loop run while i<=5...
at each iteration, i is added to sum, then incremented (post-increment)

so the loop compute sum of all whole number from 1 to 5 (included) in variable sum (wich is initialized to zero just before the 'for' statement ;)

Answer 4

if you initialize b to zero, each steps are:

b = 0, a = 1 (start)
b = 0 + 1, a = 2 (loop 1)
b = 1 + 2, a = 3 (loop 2)
(end, as a >= 3)

so there's only two iterations, b ends holding 3 and a hold 3 also ;)

Answer 5

yes if you put ++ after b...

Answer 6

after i have added a + b then i do increment b?

Answer 7

Int a,b;
For(a=1; a<3; b+=a++);
I think at first: a = 1 + b = 1 a ++ == 3?
and so every 3 iteration?

Answer 8

I just started to see the light at the end of the tunnel🥴 I take more information on examples, as you did the last

Answer 9

a < 4
b = 0, a = 1 (start)
b = 0 + 1, a = 2 (loop 1)
b = 1 + 2, a = 3 (loop 2)
b = 2 + 3, a = 6 (loop 3)
(end, as a >= 4)
did I make the right decision?😍

Answer 10

yes ;)

Answer 11

oh, not exactly: last a == 4 not 6 (a get only 1 added each turn ^^)

Answer 12

Yeaaaaaaaahhhhhhh🥰🕺🕺🕺thank you friend

Answer 13

a < 4
b = 0, a = 1 (start)
b = 0 + 1, a = 2 (loop 1)
b = 1 + 2, a = 3 (loop 2)
b = 2 + 3, a = 4 (loop 3)
b = 3 + 4, a = 5
B = 4 + 5, a = 6, loop5
B = 5+6, a = 7, loop6
B = 6 +7,a = 8 loop 7
(end, as a >= 7)
B = 6+7=13, a = 8 (13+8=21)🤩🤩🤩🤩🤩🤩

Answer 14

if your condition is a<7 not a<4, then there's no 7th loop...

and your b is false since 4th loop:

a = 1, b =0
for (; a<7; b += a++);

1) b = 0 (b) + 1 (a) = 1, a = a + 1 = 2
2) b = 1 (b) + 2 (a) = 3, a = a + 1 = 3
3) b = 3 (b) + 3 (a) = 6, a = a + 1 = 4
4) b = 6 (b) + 4 (a) = 10, a = a + 1 = 5
5) b = 10 (b) + 5 (a) = 15, a = a + 1 = 6
6) b = 15 (b) + 6 (a) = 21, a = a + 1 = 7

end as a>=7