std::showbase, std::noshowbase
From cppreference.com
Defined in header <ios>
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std::ios_base& showbase( std::ios_base& str ); |
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std::ios_base& noshowbase( std::ios_base& str ); |
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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).
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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 is("3.14"); is >> std::showbase >> std::get_money(val); std::cout << "With showbase, parsing 3.14 as money gives " << val << '\n'; is.seekg(0); is >> 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) |