The output of the program is always '0'? - c++

Question

I want to find the sum up to the 'n'th term for the following series:
(1/2)+((1*3)/(2*4))+((1*3*5)/(2*4*6))....
So, I wrote the following program in c++ :
#include <bits/stdc++.h>
#include <conio.h>
using namespace std;
int main()
{
int p=1, k=1, n=0;
float h=0;
cout<<"Enter the term: ";
cin>>n;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=i; j++)
{
p*=((2*j)-1);
k*=(2*j);
}
h+=(p/k);
p=1;
k=1;
}
cout<<"The sum is : "<<h;
return 0;
getch();
}
However, the output of the program always gives me '0'. I can't figure out the problem with the program.
N.B. I'm new to programming.

Answer 1

The problem here is that you haven't declared p and k as float or doubleor explicitly cast them as such before the calculation and assignment to h.
What's happening is for every iteration of the loop p < k (by nature of the problem) since p and k are both declared as int, p / k = 0. So you're just summing 0 for every iteration.
Either declare p and k as float or double or do this:
h += ((float) p) / ((float) k)
Also, for this specific problem I assume you're looking for precision, so be wary and look into that as well Should I use double or float?

Answer 2

implicit conversion and type casting are a trap where all newbies fall.
in the instruction:
h += p/k;
the compiler performs an integer division first, then a promotion of the result to floating point type.
and since:
p < k ; for all i,j < n
then:
res = (p / k) < 1 => truncates to 0; // by integer division
thus:
sum(1->n) of p/k = sum (1->n) 0 = 0;
finally:
h = conversion to float of (0) = 0.0f;
that's why you have the result of 0.0f at the end.
the solution:
1- first of all you need to use the natural type for floating point of c++ which is "double" (under the hood c++ promotes float to double, so use it directly).
2- declare all your variable as double, except the number of terms n:
3- the number of terms is never negative, you need to express that in your code by declaring it as an unsigned int.
4- if you do step 3, make sure to catch overflow errors, that is if the user enters a negative number your risk to have a very big number in "n", expel : n =-1 converts to 0xffffffff positive number.
5- engineer your code sometimes is better.
6- include only the headers that you need, and avoid a importing any namespace in your global namespace.
here is how i think you should write your program.
#include <iostream>
double sum_serie(unsigned int n)
{
double prod = 1.0, sum = 0.0;
for (double c=1; c<=n ; c++)
{
prod *= ( ( 2*c ) - 1 ) / ( 2*c ); // remark the parenthesis
sum += prod;
}
return sum;
}
int main()
{
unsigned int n = 0;
int temp = 0;
std::cout << " enter the number of terms n: ";
std::cin >> temp;
if (temp > 0)
n = temp; // this is how you catch overflow
else
{
std::cout << " n < 0, no result calculated " << std::endl;
return 0;
}
std::cout << " the result is sum = " << sum_serie(n) << std::endl;
return 0;
}
I know that the question was about the implicit conversion and casting in C++, but even the way of writing a code can show you what bugs you have in it, so try to learn a proper way of expressing your ideas into code, debugging comes natural afterward.
Good Luck