Cs50 PSET 1 马里奥不太舒服

Question

hey thanks for the help: screenshot

New to programming and trying to understand the logic why the code in line 17 produces the desire result of a left-aligned pyramid. Just trying to fully understand why x= height - y - 1 gives the desired outcome.

Answer 1

This snippet

for (int x = height - y - 1; x < height; x++)
{
    print("#");
}

does not have anything to do with alignment. The pyramid will always be left aligned without a space( " " ) printing loop, which you don't have in your code and you probably don't need either.

Now let's walk through the iteration to understand what's going on

Consider the height to be 5

In the first iteration of the outer loop (ie for (int y = 0; y < height; y++) )

• y = 0
• x = height - y - 1 = 5 - 0 - 1 = 4

So x starts at 4 and stops before reaching height , ie 5. So this loop will be executed exactly 1 time . Which means it'll print a singular # .

In the second iteration of the outer loop

• y = 1
• x = height - y - 1 = 5 - 1 - 1 = 3

So x starts at 3 and stops before reaching height , ie 5. So this loop will be executed exactly 2 times . Which means it'll print # twice .

In the third iteration of the outer loop

• y = 2
• x = height - y - 1 = 5 - 2 - 1 = 2

So x starts at 2 and stops before reaching height , ie 5. So this loop will be executed exactly 3 times . Which means it'll print # thrice .

In the fourth iteration of the outer loop

• y = 3
• x = height - y - 1 = 5 - 3 - 1 = 1

So x starts at 1 and stops before reaching height , ie 5. So this loop will be executed exactly 4 times . Which means it'll print # 4 times.

In the fifth and final iteration of the outer loop

• y = 4
• x = height - y - 1 = 5 - 4 - 1 = 0

So x starts at 0 and stops before reaching height , ie 5. So this loop will be executed exactly 5 times . Which means it'll print # 5 times .

So to achieve that logic, is the reasoning behind using x = height - y - 1 . However there are other ways to do this too-

for (int x = 0; x < y + 1; x++)
{
    print("#");
}

This will also work in the same logic, but hopefully with less confusion.

Notice, how the number of characters printed in each line matches with the line number . So the first line has 1 hash, 2nd has 2 and so on. We can deduce the line number from y . For the first line y = 0 , for the second line y = 1 and so on. So we can simply add 1 to y and set that as our upper bound to print the hashes.