std::showbase, std::noshowbase
From cppreference.com
| Defined in header <ios>
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std::ios_base& showbase( std::ios_base& str );
|
(1) | |
std::ios_base& noshowbase( std::ios_base& str );
|
(2) | |
1) Enables the
showbase flag in the stream str as if by calling str.setf(std::ios_base::showbase).2) Disables the
showbase flag in the stream str as if by calling str.unsetf(std::ios_base::showbase).This is an I/O manipulator, it may be called with an expression such as out << std::showbase for any out of type std::basic_ostream or with an expression such as in >> std::showbase for any in of type std::basic_istream.
The showbase flag affects the behavior of integer output (see std::num_put::put), monetary input (see std::money_get::get) and monetary output (see std::money_put::put).
Parameters
| str | - | reference to I/O stream |
Return value
str (reference to the stream after manipulation).
Notes
As specifed in std::num_put::put, the showbase flag in integer output acts like the # format specifier in std::printf, which means the numeric base prefix is not added when outputting the value zero.
Example
Run this code
#include <iomanip>
#include <iostream>
#include <locale>
#include <sstream>
int main()
{
// showbase affects the output of octals and hexadecimals
std::cout << std::hex
<< "showbase: " << std::showbase << 42 << '\n'
<< "noshowbase: " << std::noshowbase << 42 << '\n';
// and both input and output of monetary values
std::locale::global(std::locale("en_US.UTF8"));
long double val = 0;
std::istringstream("3.14") >> std::showbase >> std::get_money(val);
std::cout << "With showbase, parsing 3.14 as money gives " << val << '\n';
std::istringstream("3.14") >> std::noshowbase >> std::get_money(val);
std::cout << "Without showbase, parsing 3.14 as money gives " << val << '\n';
}
Output:
showbase: 0x2a
noshowbase: 2a
With showbase, parsing 3.14 as money gives 0
Without showbase, parsing 3.14 as money gives 314
See also
| clears the specified ios_base flags (function) | |
sets the specified ios_base flags (function) |