• C

loops - is this possible????

If anyone can help I'd be most grateful

I've tried the following statement

for(p=0.05;p<1.05;p + 0.05){

but the compiler tells me that this statement will not be used.  (Probably no surprise to those in the know)

The reason I want to increment p by 0.05 is that I'm trying to write a program to simulate changes in gene frequencies, and I'd like to run the simulation with different starting frequencies, i.e. first time 0.05, second 0.10 third 0.15 and so on.  Gene frequencies always lie between 0 and 1, and rather than scan a whole load of vaules into an array I would like to increment the value of p each time the loop is run.

Thanks Neil
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

What type is p?

double p;
for(p=0.05;p<1.05;p += 0.05){

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Also notice the p += .05
Nile_6670Author Commented:
float        p;

tried suggestion, but is caught in a loop somehwere.  Will have to check it now.

How does += function/what does it do?
Nile_6670Author Commented:
Thanks for the help, I've realised another way of doing this.

float p=0;
int   loopy;


p = p +0.05

p += .05 adds .05 to p and then assigns p .05.  In other words

(p += .05) == (p = p + .05)
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today

From novice to tech pro — start learning today.