Why is this double value printed as "-0"?

Question

double a = 0; double b = -42; double result = a * b; cout << result; 

The result of a * b is -0, but I expected 0. Where did I go wrong?

Answer 1

The bit representation of -0.0 and 0.0 are different, but they are same value, so -0.0==0.0 would return true. In your case, result is -0.0, because one of the operand is negative.

See this demo:

#include <iostream> #include <iomanip>  void print_bytes(char const *name, double d) {     unsigned char *pd = reinterpret_cast<unsigned char*>(&d);     std::cout << name << " = " << std::setw(2) << d << " => ";     for(int i = 0 ; i < sizeof(d) ; ++i)        std::cout << std::setw(-3) << (unsigned)pd[i] << " ";     std::cout << std::endl; }  #define print_bytes_of(a) print_bytes(#a, a)  int main() {     double a = 0.0;     double b = -0.0;      std::cout << "Value comparison" << std::endl;     std::cout << "(a==b) => " << (a==b)  <<std::endl;     std::cout << "(a!=b) => " << (a!=b)  <<std::endl;       std::cout << "\nValue representation" << std::endl;     print_bytes_of(a);     print_bytes_of(b); } 

Output (demo@ideone):

Value comparison (a==b) => 1 (a!=b) => 0  Value representation a =  0 => 0 0 0 0 0 0 0 0  b = -0 => 0 0 0 0 0 0 0 128  

As you can see yourself, the last byte of -0.0 is different from the last byte of 0.0.

Hope that helps.