Puzzle (coding)

from XKCD wiki thingy

Find three ways to make the C program below print 20 copies of the dash character '-' by changing/adding one character:

Code:

int i, n = 20;
for (i = 0; i < n; i--)
{
    printf("-");
}

I've only found one, but I'm stupid

Spoilerplz.

Also:
* Find a way to get it to print exactly 21 dashes (changing 1 character).
* Find a way to get it to print exactly 1 dash (changing one character).
* Print 20 dashes with the given code, without changing it at all, but adding some preceding code.
* Print 1 dash with the given code, without changing it at all, but adding some preceding code.

i only see one too

hm.
well...

 Solution 1

for (i = 0; -i < n; i--)

Doesn't it already print one dash without changing the code at all?

Doesn't it already print one dash without changing the code at all?

 Spoiler

It prints an infinite number of dashes as-is.

EDIT:

 The possible solutions?

1. -i < n (as you said)
2. i < -n
3. n = -20

This is assuming that C can handle negative indices in for loops. I tried compiling it in Java and it won't work. But I barely know anything about C, so if I'm retarded, just let me know.

EDIT2: To do 21 dashes... you have to change 1 character from the original, not 1 from one of the 20-dash solutions, right? Since the latter is a lot easier...

EDIT3:

 OHWAIT

n--

(??!!)

EDIT4: Nvm, my EDIT1 solutions don't work, because then there'd be more than 3 solutions...

I have the second solution for part 1

 Spoiler

int i, n = 20;
for (i = 0; i < n; n--)
{
printf("-");
}

EDIT: nvm, I was beaten to it

 One dash solution

Code:

int i, n = 20;
for (i = 0; i < [B]0[/B]; i--)
{
    printf("-");
}

Eek, sorry, Heidi. =/

EDIT: Crap, scratch that one-dash solution. Seems like it should work, but it doesn't when I compile it. =/

EDIT: Crap, scratch that one-dash solution. Seems like it should work, but it doesn't when I compile it. =/

Remember that 0 < 0, is not true, so the loop goes false straight away
Even if it was <=, decreasing 0 starting at 0 is always going to be less than or equal to 0, resulting in an infinite loop

Edit: Can't continue doing this, have to go to mall with my father for a bit (Who is in a bad mood )

Yeah, i < -n and n = -20 don't work because you'd have 0 < -20 which is false..

This would be so much easier if it were a do-while loop. Then like everything I thought of would work. =P

Yeah, I was having fun working on it, despite my for loop failures, then I realized I need to study for my Data Structures and Algorithms final (in Java) at 7:30 tomorrow morning, and it's almost 11:30. =/

 Spoiler

~i < n is the 21 output solution.

Find three ways to make the C program below print 20 copies of the dash character '-' by changing/adding one character:

Code:

int i, n = 20;
for (i = 0; i < n; i--)
{
    printf("-");
}

Also:
* Find a way to get it to print exactly 21 dashes (changing 1 character).
* Find a way to get it to print exactly 1 dash (changing one character).
* Print 20 dashes with the given code, without changing it at all, but adding some preceding code.
* Print 1 dash with the given code, without changing it at all, but adding some preceding code.

I've never worked with C before xD

 Solution 1

Code:

int i, n = 20;
for (i = 0; i < n; [color=Red]n[/color]--)
{
    printf("-");
}

 Solution 2

Code:

int i, n = 20;
for (i = 0; [color=Red]-[/color]i < n; i--)
{
    printf("-");
}

 Solution 3

afk

 1 dash, preceding code (solution by GummyBear)

Code:

unsigned int i, n = -1; // \

 20 dashes, preceding code?

Code:

unsigned int i, ii, n = -1;
for (ii = 1; ii < 20; ii++) // \

Oh, and the lulz solution:

 Spoiler

Code:

printf("--------------------"); return;

dont forget the
* Find a way to get it to print exactly 1 dash (changing one character).

int i, n = 20;
for (i = 0; i < n; i--);
{
printf("-");
}