Size: a a a

2020 January 24

M

Mr.Temirlan in Python KZ
😂
источник

AS

AlI Sergazy in Python KZ
костыли пофиг на чем писать
источник

AS

AlI Sergazy in Python KZ
питон нормально для этого /s
источник

АА

Алихан Амандык in Python KZ
redhemd
transfer your question here @python
собстна он оттуда и пришёл сюда)
источник

r

redhemd in Python KZ
Алихан Амандык
собстна он оттуда и пришёл сюда)
в сообщ с 51к чел куда вероятней получить ответ, да еще на англ, чем тут ))
источник

r

redhemd in Python KZ
не понял этого вообще
источник

АА

Алихан Амандык in Python KZ
я сначала подумал может это тролль такой, который спамит со своим вопросом)
источник

.

. in Python KZ
Аlisher Temirov
I needn't break
источник

SA

Sultan Abilda in Python KZ
.
Hi Harry. Sorry for the question. But what are you trying to do?
источник

SA

Sultan Abilda in Python KZ
In your example you just printing numbers from 0 to 3. Btw, you can just use break instead of changing i.
источник

.

. in Python KZ
Sultan Abilda
Hi Harry. Sorry for the question. But what are you trying to do?
// C++, there's two times loops
for(int i = 0; i < 4; i ++)
   if(i == 1)
       i = 8;
   

// Python, there's four times loops
for i in range(4):
   if i == 1:
       i = 8
источник

.

. in Python KZ
.
// C++, there's two times loops
for(int i = 0; i < 4; i ++)
   if(i == 1)
       i = 8;
   

// Python, there's four times loops
for i in range(4):
   if i == 1:
       i = 8
I'm gonna convert c++ code to python, but it doesn't work
источник

SA

Sultan Abilda in Python KZ
.
// C++, there's two times loops
for(int i = 0; i < 4; i ++)
   if(i == 1)
       i = 8;
   

// Python, there's four times loops
for i in range(4):
   if i == 1:
       i = 8
Because in c++ implementation condition checks i variable, but in python implementation you are using range(4) iterator which will generate numbers from 0 to 3 included. So in python assigning i = 8 will not affect loop.
источник

SA

Sultan Abilda in Python KZ
They have little bit different mechanic.
источник

.

. in Python KZ
Sultan Abilda
Because in c++ implementation condition checks i variable, but in python implementation you are using range(4) iterator which will generate numbers from 0 to 3 included. So in python assigning i = 8 will not affect loop.
so, how can I do it?
источник

SA

Sultan Abilda in Python KZ
As I already said, you are using assignment to exit the loop. But better approach will be using the break statement
источник

SA

Sultan Abilda in Python KZ
So just replace "i = 8" with "break"
источник

SA

Sultan Abilda in Python KZ
Make sense?
источник

.

. in Python KZ
Sultan Abilda
As I already said, you are using assignment to exit the loop. But better approach will be using the break statement
is there another way? because I really needn't break
источник

SA

Sultan Abilda in Python KZ
.
is there another way? because I really needn't break
To offer you the better approach. Can you explain me the logic of your code? Because right now it's just changing the loop variable which leads the exit from loop.
источник