Hi, is it expected that the attached program only works when no optimization is used? I do not understand why the if-test is ever entered in this program. The variables that are compared should be exactly the same, since one is used to set the value of the other another place in the program. I will attach the whole program below. In short, how can it be that an if-test of the following form can print two exactly equal numbers?: if( a != b) { cout << setprecision(70); cout << a << " " << b << endl; } Btw, I know that normally one should do a comparison within a certain error when comparing doubles, but I am only interested in how the if-test above can print two equal numbers? Here is the program (the if-test is entered when N=4, but not when N=2,3). In short, first pot[m][n] is set using calc(m,n), but afterwards the if-test says that they are different, if I compile with -O1: #include <iostream> #include <iomanip> using namespace std; const int M = 2, N = 4; double calc(const int m, const int n) { double pot = 0.0; for(int n2 = 0; n2 < n; ++n2) { pot += 0.1; } return(pot); } void check(double pot[][N]) { for(int m = 0; m != M; ++m) { for(int n = 0; n != N; ++n) { cout << "n = " << n << endl; if( calc(m,n) != pot[m][n] ) { cout << "Error!" << endl; cout << setprecision(70); cout << pot[m][n] << " and " << calc(m,n) << endl; } } } } int main() { double pot[M][N]; for(int m = 0; m != M; ++m) { for(int n = 0; n != N; ++n) { pot[m][n] = calc(m,n); } } check(pot); return(0); }