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

By default, GCC compiles C++-code for gnu++98, which is a fancy way of saying the C++98 standard plus lots of gnu extenstions.

You use -std=??? to say to the compiler what standard it should follow.
Don't omit -pedantic though, or it will squint on standards-conformance.

The options you could choose:

standard          with gnu extensions

c++98             gnu++98
c++03             gnu++03
c++11 (c++0x)     gnu++11 (gnu++0x)
c++14 (c++1y)     gnu++14 (gnu++1y)

Coming up:

c++1z             gnu++1z (Planned for release sometime in 2017, might even make it.)

GCC manual: https://gcc.gnu.org/onlinedocs/gcc-4.9.2/gcc/Standards.html#Standards

Also, ask for full warnings, so add -Wall -Wextra.

There are preprocessor-defines for making the library include additional checks:

• _GLIBCXX_CONCEPT_CHECKS to add additional compile-time-checks for some templates prerequisites. Beware that those checks don't actually always do what they should, and are thus deprecated.
• _GLIBCXX_DEBUG. Enable the libraries debug-mode. This has considerable runtime-overhead.
• _GLIBCXX_DEBUG_PEDANTIC Same as above, but checks against the standards requirements instead of only against the implementations.