Lets say int health=6;
I would like to display ***** symbles for the value of health
like int health=6;
display ******
if health were to equal 4
display ****
I would like to do this without using IF statments becouse that would take up a lot of space and take forever if I had like health=100;
what would the best way of doing this be, thanks!
You might want to research basic looping structures.
If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein
yeah, that was the first thing I did, I went to
http://www.cprogramming.com/tutorial/lesson3.html
but I still dont know how to print a number of asterisks for an int value
all those seem to do is keep repeating somthing, but I need to print astricks like this ******
and not like
*
*
*
*
*
> and not like
So don't print newlines then
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
f-f-f-f-f-f-f-f-for loop
M-m-m-m-m-m-magic.Code:for (int i = 0; i < hp; i++){ cout << "*"; }
You need a for loop.
An example is:
Edit: Krak posted before me!Code:for(int i=0; i<PlayerHealth; i++){ cout<<"*"}
Stupid things pop singers say
"I get to go to lots of overseas places, like Canada."
- Britney Spears
"I love what you've done with the place!"
-Jessica Simpson upon meeting the Secretary of Interior during tour of the White House
Thanks guys
Either Jlou or Grib posted this solution a while ago.
Code:cout << string(health,'*') << endl;
Since we're providing fun solutions:
[edit]Code:char buf[BUFSIZ] = {0}; memset( buf, '*', health ); cout << buf ;
You could even combine the memset line with the output line...
That looks about right.Code:cout << static_cast<char*>(memset( buf, '*', health ));
[/edit]
Quzah.
Last edited by quzah; 02-03-2005 at 03:59 PM.
Hope is the first step on the road to disappointment.
Oooh! Oooh! Pick me!
Okay, maybe I went overboardCode:char DispHealth[hp]; for(int i=0; i<hp; i++){DispHealth[i]='*'} int toggle=0; for(int i=0; i<hp; i++){ if(toggle==0){ cout<<DispHealth[i]; toggle=1;} else if(toggle==1){ cout<<"*"; toggle=0; } else{ cout<<"\n\nSomebody has problems!\n*Points at Bill Gates* Fix it right now!"} }
This has different methods of displaying the asterisk, one just displaying it and the other reading from a char. Of course, you should know that from looking at my code, but some people...
Anyways, these alternate with the methods they use. I could do murmer murmer murmer...
Hope that confuses you,
fuh
Last edited by fuh; 02-04-2005 at 07:00 AM.
Stupid things pop singers say
"I get to go to lots of overseas places, like Canada."
- Britney Spears
"I love what you've done with the place!"
-Jessica Simpson upon meeting the Secretary of Interior during tour of the White House
>Okay, maybe I went overboard
You also got a little bug:
>else if(toggle=1){
Oops! Problem solved. I just pulled the code up right on the spot.
Stupid things pop singers say
"I get to go to lots of overseas places, like Canada."
- Britney Spears
"I love what you've done with the place!"
-Jessica Simpson upon meeting the Secretary of Interior during tour of the White House
No, no, you got it all wrong. You need to do:
Code:std::fill_n(std::ostream_iterator<char>(std::cout), health, '*');
All the buzzt!
CornedBee
"There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
- Flon's Law
