Comments on your flow chart, which I notice spontaneously at first glance:
You should be. the loops [eg “j=2; j++; as long as j < n”] divide a little further. Otherwise you'll have as much trouble as you… Where does a passage of the loop end, so that the loop begins again with the next passage of the loop hull? Apparently, in your flow chart, the loop hull is always executed, whether the loop condition is fulfilled. [Finally, there is only one arrow left with you, without distinction, whether the loop condition is fulfilled or not.]
Directly after you first marked an i *j, the output will be done with you and the program will be terminated. But… instead it should be checked with [j++; loop condition of the j loop; if necessary, j-slip bodies should be reworked], right?
If an i *j is already recognized as marked in a decision step, you will end up with an “x”. How is it going to continue, or is it supposed to be the end of the program?
You can't write the loop, so you also need to make boxes for i=2 and i++ out and then put the "so long i
Addendum: I assume that your structogram is correct, but that's funny.
For the initialization of the list, I would have expected a loop instead of just one step
For i it is sufficient to count n,
If i is already marked, you can skip the inner j loop
j does not have to start at 2, i*i is sufficient as start value
I would expect the output to be done in a loop of 2…n, and then it is checked whether the number is marked, you have summarized quite a lot in one step
Yeah, I'd like to draw it like that…
===============
Comments on your flow chart, which I notice spontaneously at first glance:
You can't write the loop, so you also need to make boxes for i=2 and i++ out and then put the "so long i- For the initialization of the list, I would have expected a loop instead of just one step
- For i it is sufficient to count n,
- If i is already marked, you can skip the inner j loop
- j does not have to start at 2, i*i is sufficient as start value
- I would expect the output to be done in a loop of 2…n, and then it is checked whether the number is marked, you have summarized quite a lot in one step
Addendum: I assume that your structogram is correct, but that's funny.
Yeah, I can. But where is your own attempt to correct and discuss?
It makes you absolutely NOT when others do your job. You won't learn.
Go through the structogram from above to below and create your nodes with corresponding branches. You can't do that much wrong.
Here mine:
Yes I can
xD