for loop help. Terminates when it isnt supposed to. c++

Question

I'm new to stackoverflow, but i did try to look for an answer and could not find it. I also can't seem to figure it out myself. So for a school C++ project, we need to find the area under a curve. I have all the formulas hardcoded in, so don't worry about that. And so the program is supposed to give a higher precision answer with a higher value for (n). But it seems that when I put a value for (n) thats higher than (b), the program just loops a 0 and does not terminate. Could you guys help me please. Thank you. Heres the code:

/* David */

#include <iostream>
using namespace std;

int main()
{

cout << "Please Enter Lower Limit: " << endl;
int a;
cin >> a;

cout << "Please Enter Upper Limit: " << endl;
int b;
cin >> b;

cout << "Please Enter Sub Intervals: " << endl;
int n;
cin >> n;

double Dx = (b - a) / n;
double A = 0;
double X = a;

for (X = a; X <= (b - Dx); X += Dx)
{
    A = A + (X*X*Dx);
    X = X * Dx;
    cout << A << endl;
}

cout << "The area under the curve is: " << A << endl;

return 0;
}

`X = X * Dx;` What is this one for? It doesn't make much sense. — Igor Tandetnik, Feb 08 '15 at 22:57
That was one mistake that I ended up finding. It's supposed to be x=x+DX; — David Kluszczynski, Feb 10 '15 at 21:20
But you already have `X += Dx` in the `for` loop. Do you want to increment `X` twice? — Igor Tandetnik, Feb 10 '15 at 21:43

JuniorCompressor · Accepted Answer · 2015-02-08T23:12:29.943

1

a, b, n are integers. So the following:

(b - a) / n

is probably 0. You can replace it with:

double(b - a) / n

edited Feb 08 '15 at 23:12

answered Feb 08 '15 at 22:56

JuniorCompressor

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score 1 · Answer 2 · answered Feb 08 '15 at 22:57

Since all the variables in (b - a) / n are int, you're doing integer division, which discards fractions in the result. Assigning to a double doesn't change this.

You should convert at least one of the variables to double so that you'll get a floating point result with the fractions retained:

double Dx = (b - a) / (double)n;

score 0 · Answer 3 · edited May 23 '17 at 11:57

0

The other answers are correct. Your problem is probably integer division. You have to cast on of the operands to double.

But you should use static_cast<> instead of C-style casts. Namely use

static_cast<double>(b - a) / n

instead of double(b - a) / n or ((double) (b - a)) / n.

edited May 23 '17 at 11:57

Community

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answered Feb 08 '15 at 23:01

Lukas Schmelzeisen

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So you rather add 10 more characters to do an int to double conversion? – JuniorCompressor Feb 08 '15 at 23:10
Yes, I definitely would. C style casts are discouraged in C++ code. And not using them at all allows you do enable compile flags like `-Wold-style-cast` that could help you spot accidental C style casts that might be more error prone than int to double. – Lukas Schmelzeisen Feb 08 '15 at 23:15
I think it's wise to wear a seat belt but not to drive at 5 mph. – JuniorCompressor Feb 08 '15 at 23:28

score 0 · Answer 4 · answered Feb 08 '15 at 23:02

You are performing integer division. Integer division will only return whole numbers by cutting off the decimal:

3/2 == 1 //Because 1.5 will get cut to 1
3/3 == 1
3/4 == 0 //Because 0.5 will get cut to 0

You need to have at least one of the two values on the left or right of the "/" be a decimal type.

3 / 2.0f == 1.5f
3.0f / 2 == 1.5f
3.0f / 2.0f == 1.5f

for loop help. Terminates when it isnt supposed to. c++

4 Answers4