my question is how do I replace the value of b and run the loop over again untill b = 100Code:#include <stdio.h> int main() { double a=1; double b=0.01; double c; c = a * b; while ( a <= 10000) { printf("%f\t%f\n", a, c); a = a++; c = a*b; } } return 0; }
thanks
Put into words what it is you actually are trying to do. You can't code for what you want if you don't know what it is you want to do. How are a and b related? Why are you using a to control the loop if you really care about what b is doing?
Quzah.
Hope is the first step on the road to disappointment.
By having a second, outer loop.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
b is 1% and a is 1 dollar
i used this loop to get a to go up to 10,000 dollars and return the % amount for each dollar
i don't know how to change the value of b and run the loop again
if I change the value of b to .02 the loop will return the % of the given dollar value
sorry I should of put comments in code
Here's a hint... If the loop were in a function, you could feed it seed values and run it as often as you like.
> a = a++;
This is wrong -> see Question 3.3
> b is 1% and a is 1 dollar
Then you'll want to use integers, not floats. floats are approximate numbers with all sorts of rounding errors. If you want to keep things nice and accurate, do something like
int dollars = 100;
int cents = 1;
...
dollars += 100; // a new dollar
cents++; // a new cent
Notice how I've uses meaningful variable names, not meaningless a,b,c
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.
