在没有文件的情况下编译 C++ 代码 [英] Compiling C++-code without a file

问题描述

我正在尝试使用您的标准 g++ 编译器编译一些 C++ 代码.但是，不是从文件编译:

I'm trying to compile some C++ code using your standard g++ compiler. However, rather than compiling from a file:

main.cpp:

#include <iostream>

int main(){
    std::cout << "Hello World!
";
    return 0;
}

我更喜欢做类似的事情

g++ ... "#include <iostream>
 int main(){ std::cout << "Hello World!
"; return 0;}"

之前来自 stackoverflow 的帖子表明

A previous post from stackoverflow showed that

echo "int main(){}" | gcc -Wall -o testbinary -xc++ -

有效，但我想知道它是如何工作的，并且更好，如果有办法做到这一点而无需管道内容.

works but I would like to know how it works and better yet, if there is a way to do this without the need to pipe the contents.

我正在生成运行时代码，我需要生成共享库并加载创建的函数.

I'm doing run-time code generation where I need to generate a shared library and load the functions created.

我以为会有一个标志告诉编译器嘿，我给你的是源代码而不是文件".

I thought there would be a flag to tell the compiler "hey, I'm giving you the source code and not the file".

再次感谢您的帮助！

推荐答案

echo "int main(){}";|gcc -Wall -o testbinary -xc++ -

echo "int main(){}" | gcc -Wall -o testbinary -xc++ -

有效，但我想知道它是如何工作的，并且更好，如果有办法做到这一点而无需管道内容.

works but I would like to know how it works and better yet, if there is a way to do this without the need to pipe the contents.

或者你可以说(例如在 shell 脚本中):

Alternatively you can say (e.g. in a shell-script):

gcc -Wall -o testbinary -xc++ - << EOF
int main(){}
EOF

这篇关于在没有文件的情况下编译 C++ 代码的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！