Infinite loop of zero in table - c++

Question

this code is dividing a circle into X part and display a table of radian value of each part.
but
I get an infinite loop of 0 display when I'm using value superior to 6
with value under 6 I get '0 1 2 3 4 5 6'
It seems that displayed value are not float either.
I have the code using degrees and work fine.
#include <iostream>
using namespace std;
#define pi 3.14159265359
#define pi2 6.28318530718
int nbObjets = 0;
void objetsPositionRadian(int tab[], int nbObjets);
int main(){
int tabRadian[] = {0};
std::cout << "Nombre d'objets ? ";
std::cin >> nbObjets;
objetsPositionRadian(tabRadian, nbObjets);
return 0;
}
void objetsPositionRadian(int tab[], int nbObjets){
float radian = (360/nbObjets) * (pi/180);
for (int i = 0; i < pi2; i+=radian){
std::cout << i << " ";
}
std::cout << endl;
}

Answer 1

The obvious move here is to change i in your loop to be a double (or float) instead of an int.
Along with that, when computing radian, 360/nbObjets does integer division, so if nbObjets is > 360, it'll give a result of 0. Changing 360 to 360.0 fixes that problem.
But that leaves another problem: depending on the vagaries of floating point math, if you ask for a lot of objects, there's a good change you'll end up computing positions for one more object than you asked for (and conceivably even more than that if you asked for a really large number of objects).
This problem arises from cumulative errors as you add radian to i in the loop. Rather than doing that, you almost certain want something on this order:
for (int i=0; i<nbObjets; i++)
cout << i * radian;
This way, you always get exactly the number of objects you asked for, and any possible errors in the value don't accumulate from one iteration to the next.

Answer 2

You are doing two dangerous things in the following two lines:
float radian = (360/nbObjets) * (pi/180);
for (int i = 0; i < pi2; i+=radian)
First: nbObject is defined as an integer. As a result 360/nbObjects will be calculated as an integer (e.g. for 7, the result will be 51, not a floating point number).
Next, you define i as an integer. When you add a number to it which is smaller than 1, it will always remain the same.
Therefore I advise you to use more floating point numbers when needed, as in this proposal:
double radian = ((double)360/nbObjets) * (pi/180); // first typecast 360 as a
// floating point, in order
// to enforce floating
// point arithmetic.
for (double i = 0; i < pi2; i+=radian)
This should work better.