std::showbase, std::noshowbase
From cppreference.com
Defined in header <ios> | ||
std::ios_base& showbase(std::ios_base& str ); | (1) | |
std::ios_base& noshowbase(std::ios_base& str ); | (2) | |
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).
Contents |
[edit]Parameters
str | - | reference to I/O stream |
[edit]Return value
str (reference to the stream after manipulation).
[edit]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.
[edit]Example
Run this code
#include <iomanip>#include <iostream>#include <locale>#include <sstream> int main(){// showbase affects the output of octals and hexadecimalsstd::cout<<std::hex<<"showbase: "<< std::showbase<<42<<'\n'<<"noshowbase: "<< std::noshowbase<<42<<'\n'; // and both input and output of monetary valuesstd::locale::global(std::locale("en_US.UTF8"));longdouble 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
[edit]See also
clears the specified ios_base flags (function) | |
sets the specified ios_base flags (function) |