Why output showing 0 instead of 0.0? - c++

Question

I declared an array of 10 elements and initialized it with 0.0 .. 9.9 , its output is perfect except 0.0 has been changed into 0, why is it so?
#include <iostream>
using namespace std;
int main(void)
{
int const SIZE = 10;
double number[SIZE] = {0.0,1.1,2.2,3.3,4.4,5.5,6.6,7.7,8.8,9.9};
for(int i(0) ; i < SIZE ; i++)
{
cout << number[i] << endl ;
}
system("PAUSE");
}
Thanks,
output:
0 //it should be 0.0 not 0//
1.1
2.2
.
.
.
9.9

Answer 1

std::cout by default will output up to 6 digits, which in this case is one (the zero) since no decimals are needed. std::cout << (double)1.0; would display 1, for instance. You can use std::setprecision from the <iomanip> header and std::fixed to keep the decimal.
#include <iostream>
#include <iomanip>
int main(void)
{
int const SIZE = 10;
double number[SIZE] = {0.0,1.1,2.2,3.3,4.4,5.5,6.6,7.7,8.8,9.9};
std::cout << std::setprecision(1) << std::fixed;
for(int i(0) ; i < SIZE ; i++)
{
std::cout << number[i] << std::endl ;
}
}
There are lots of helpful i/o manipulators. Here's a reference to check it out.

Answer 2

“0.0” and “0” are two numerals for the same number. A double stores only the number1; it does not store the original numeral.
When the source text of a C++ program contains a numeral such as 0.0 or 1.1, the compiler converts it from that numeral (which is a string of characters) to a double. The double format only represents numbers, not the strings they came from. So, a zero in double is just zero; it is not “0” or “0.0” or “0.000”. When you print it, there is no way for software that prints a double to know whether the original numeral was “0” or “0.0”. It just prints it according to the rules for printing a double.
By default, a double value of zero is printed as “0”. If you wish it to be printed differently, you can use I/O manipulators to ask that it be formatted differently. For example, after you #include <iomanip>, you can use std::cout << std::setprecision(1) << std::fixed; to set the floating-point output format to one digit in a fixed (versus scientific) format. Then printing a double value of zero will produce “0.0”.
1 Except, for zero, double can distinguish +0 and –0 (when IEEE 754 is used for floating-point).

Answer 3

That is simply how cout works by default. To change that, you can either:
use the precision() method:
cout.precision(1);
for(int i(0) ; i < SIZE ; i++)
{
cout << number[i] << endl ;
}
Use setprecision():
#include <iomanip>
cout << setprecision(1) << fixed;
for(int i(0) ; i < SIZE ; i++)
{
cout << number[i] << endl ;
}

Answer 4

If you really need to output 0.0 rather than 0, look into #include <iomanip>

Answer 5

The normal format will use a representation a compact as possible in most cases. Since binary floating point numbers are normalized, they don't represent any trailing digits. If you want to have an extra fractional position, you might want to use these flags:
std::cout << std::fixed << std::setprecision(1);
The entire program would look like this:
#include <iomanip>
#include <iostream>
int main()
{
int const SIZE = 10;
double number[SIZE] = {0.0,1.1,2.2,3.3,4.4,5.5,6.6,7.7,8.8,9.9};
std::cout << std::fixed << std::setprecision(1);
for(int i(0) ; i != SIZE ; ++i)
{
std::cout << number[i] << '\n';
}
std::cin.ignore();
}