如何摆脱 g++ 命令行中的 -std=c++17 ？

Question

I have encountered an error, saying that std::size() is not a member of std. Then I found this link: Why std::size() is not a member of std in gcc 8.2.0 Eventually, my code works well after I compile the file with g++ -std=c++17 myFile.cpp . Now, I tried g++ -std=c++17 6.cpp -o - 6.out as the previous link suggested, and it did not work. Any comments?

Answer 1

Compile with the following command:

g++ -std=c++17 yourFile.cpp -o output_executable_name

Replace yourFile.cpp by the name of your .cpp file, and output_executable_name by the name you want, it will generate an executable with this name.

Find here more information on g++ compiler.

Answer 2

This can be done with a shell alias.

If you really want to get rid of having to type -std=c++17 you can set an alias in your .bashrc or .zshrc file (or whatever rc file corresponding to the shell you use).

I mainly use zsh but it's the same process for almost all shells...

Find out which shell your using with echo $SHELL this will likely either be /bin/zsh or /bin/bash . then proceed with the following depending on your shell

vim ~/.zshrc or vim ~/.bashrc (depending on your shell)

add alias g++="g++ -std=c++17" then save the file

restart your current shell/terminal window (close and reopen)

next time you go to compile with:

g++... the shell will replace it with g++ -std=c++17 and saved yourself 10 keystrokes.

Note: This isn't advised if you're going to be working with various compliler versions for different projects down the road but if you're asking this question it likely will never matter. glhf!