What's the point of including -std=c++0x in a G++ compile command?

Question

I have recently started learning C++ and, since I\'m on Linux, I\'m compiling using G++.

Now, the tutorial I\'m following says

If you happen t

Answer 1

Source for your reference:

main.cpp

#include 
using namespace std;

int main()
{
    cout << "Test main CPP" << endl;
    return 0;
}

build.sh

rm demoASI*
echo "**cleaned !!**"

##### C++ 11 Compliance #####
# type ONE
g++ -o demoASI_1 -std=c++0x main.cpp
echo "**rebuild-main-done (C++ 11 Compilation) !**"

# type TWO
g++ -o demoASI_2 -std=c++11 main.cpp
echo "**rebuild-main-done (C++ 11 Compilation) !**"

##### C++ 11+ Compliance #####
# type THREE
g++ -o demoASI_3 -std=c++1y main.cpp
echo "**rebuild-main-done (C++ 11+ (i.e. 1y, but not C++14) Compilation) !**"

###### C++ 14 Compliance  ######
# type FOUR
g++ -o demoASI_4 -std=c++14 main.cpp
if [ $? -eq 0 ]
then
    echo "**rebuild-main-done (C++ 14 Compilation) !** :: SUCCESS"
else
    echo "**rebuild-main-done (C++ 14 Compilation) !** :: FAILED"
fi

Now, execute the script as; ./build.sh (assuming build.sh has execution permission)

You can first check the version of your g++ compiler, as;

g++ --version

The version of g++, after 4.3, has support for the c++11.

Please see, for c++14 support info in compiler.