sample questions

**Ron** · 06-21-2008

Hey,,,
I was answering some sample questions and came across this problem. Its based on simple 'for' loop.

Code:

int i, n = 20;
for(i = 0;i < n;i--)
printf(“-“);

I have to change only one character to make this program work. The solution says there are 3 ways to solve this problem.

I have tried to work around this. But I need to atleast to change 2 characters.

Please help

**Elysia** · 06-21-2008

You failed to mention what the problem was or what you were expecting it to do?
Is the problem something along the lines of that illegal double quote character in the printf (“)?

**Ron** · 06-21-2008

as seen in the code....
the code will run an infinite loop because with i--, i will always be less than n

**Elysia** · 06-21-2008

Oh no, it probably won't be an infinite loop. It will stop after a while, I'm guessing, but probably not as you want.

**dwks** · 06-21-2008

One way: add a semicolon to the end of the for loop. The compiler should be able to optimise that loop out, assuming you enable optimisations.

Another way: initialize i to 90 instead of 0, by adding a "9". The loop will never execute. (I know this is adding instead of changing, but still . . . .)

You could always change the "<" to ">", to get the loop to never execute.

But I think the answer they're looking for is to change "i--" to "n--".

That's your way, Elysia.

**Elysia** · 06-21-2008

It's amazing how such apparent solutions can be transparent when you aren't the one writing the code. And even if you, they can still be transparent.

**dwks** · 06-21-2008

I only thought of it by chance. I was trying to devise a way to get a truly infinite loop, and I thought that the only way to get that would be to prevent i from underflowing, and the only way to do that would be to change another variable instead, i.e. n. By coincidence, that makes the loop execute 20 instead of infinite times.

[edit] I think neither of us noticed it immediately because that's not the way for loops usually work. I guess it's a good question in that it makes you think about that. [/edit]

**Ron** · 06-21-2008

that was smart changing i-- to n--

But initializing i to 90 wont work as i<n (n is 20)

Code:

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

OK the purpose of this program is to print 20 dashes.

**dwks** · 06-21-2008

But initializing i to 90 wont work as i<n (n is 20)

It depends on what you mean by "work". I stated that it would make the loop never execute, which it does.

Code:

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

OK the purpose of this program is to print 20 dashes.

Right, that's what we guessed . . . but stating that at the beginning would have made responses to your post much more useful.

Anyway . . . I can't see what the other two solutions might be at the moment.

**mkruk** · 06-21-2008

The second solution is

Code:

int i, n = 20;
for(i = 0; i + n; i--)

this for loop will be executed while (i + n) != 0 (integral promotion!)

I just figured a third solution out: Though it prints just 19 hypens:

Code:

int i, n = 20;
for(i = 0; ~i < n; i--)

that'd be bitwise complement. But I dont know if this is okay, because it laks one hypen.

**C_ntua** · 06-21-2008

3rd solution

Code:

for (i=0; i % n; i--)
printf("-");

I suppose it works, if there is no problem with negatives and %

**mkruk** · 06-21-2008

Originally Posted by C_ntua

3rd solution

Code:

for (i=0; i &#37; n; i--)
printf("-");

I suppose it works, if there is no problem with negatives and %

nope. on the first execution this will be 0 / 20, which is 0 and a remain of 0.
because of integral promotion (0 is FALSE, 1 is TRUE), this will evaluate FALSE and stop the loop.

**King Mir** · 06-21-2008

Code:

int i, n = 20;
for(i = 0;-i < n;i--)
printf(“-“);

**Ron** · 06-22-2008

I dont think 2nd anbd 3rd soltion are rght. THese are repetitive loops that will never end.

**dwks** · 06-22-2008

mkruk's first solution and King Mir's solutions both work, I think.

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