Re: Newbie needs help with an exercise from the Book Programming in Objective-C 2.0
Re: Newbie needs help with an exercise from the Book Programming in Objective-C 2.0
- Subject: Re: Newbie needs help with an exercise from the Book Programming in Objective-C 2.0
- From: Tommy Nordgren <email@hidden>
- Date: Tue, 3 Mar 2009 03:00:13 +0100
On Mar 2, 2009, at 6:56 PM, James Carroll wrote:
Hi there,
I hope this is the right way to go about getting
help! I am methodically going through the book Programming in
Objective-C 2.0 and have hit an exercise (for anyone who has it,
page 96 Exercise 2) I'm stuck on. Now I am going to go back over
this chapter anyway as there are clearly a few things I need to go
back through, but was wondering if anyone can help tell me what I am
doing wrong with this code so at least I can see what I did wrong
and don't compound the mistake as I have written it a few ways and
can't make it work!
I am looking for every fifth triangular numbers (5,10,15 etc to 50)
using this formula;
triangleNumber = n (n + 1) / 2
For example to find the 10th triangle number you swap n for 10.
The bit that seems to be getting me stuck is the actual formula, no
matter what I do the error comes up "n is not a function", could
someone show me what I'm doing wrong?Here is my last attempt at the
bit of code!
#import <Foundation/Foundation.h>
int main (int argc, const char * argv[]) {
NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init];
int n, number, triangleNumber;
// insert code here...
NSLog(@"triangle numbers 5 to 55");
triangleNumber = 0;
do {
triangleNumber = n (n + 1) / 2;
Only Symbolic Math Programs like Mathematica allows multiplication
without an explicit multiplication operator
triangleNumber = n * (n +1) / 2;
NSLog(@"i% i%", number, triangleNumber);
number = n+5;
}
while ( number <=50);
[pool drain];
return 0;
}
Thanks in advcnace.
James Carroll Director
evil.twin.artworks
email@hidden
http://www.eviltwinartworks.com
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"Home is not where you are born, but where your heart finds peace" -
Tommy Nordgren, "The dying old crone"
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