程序退出状态
我有一个关于 Linux 中程序退出状态的问题。在我的程序中，我分叉了一个子进程并调用 waitpid 来获取它。当 waitpid 返回时，我想检查子进程的退出状态。我转向手册寻求帮助，发现 waitpid 的第二个参数将保存退出状态，我可以使用宏 WEXITSTATE 来读取它。然而，这个宏只是提取真实退出状态的最低有效8位，而在函数exit(int ret_val)手册中，它将以ret_val & 退出。 0x377，而不是最低有效 8 位。
我的问题是，其他更多位在哪里？我们就简单地放弃它们吗？为什么Linux采用这种策略？这样的实现不会给我们的程序带来麻烦吗？
谢谢并致以最诚挚的问候。
I have a question about program exit state in Linux. In my program, I fork a child process and invoke waitpid to reap it. When waitpid returns, I wanna check exit state of my child process. I turn to manual for help and find that the second argument of waitpid will hold exit state and I can use macro WEXITSTATE to read it. However, this macro just extract least significant 8 bits of the real exit state, while in manual of function exit(int ret_val), it will exit with ret_val & 0x377, instead of least significant 8 bits.
My question is, where is the other more bits? Do we simply drop them? Why Linux employ this strategy? Doesn't this implementation introduce trouble to our program?
Thanks and Best Regards.
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我想您会发现
0x377确实是，或者应该是0377。它是八进制的，所以 3778 是8位。
I think you will find that
0x377is really, or should have been,0377.It's octal, so 3778 is 8 bits.
根据 POSIX 规范，退出返回值仅假定在 0 到 255 之间。您不应该返回高于该值的值（换句话说，符合 POSIX 标准的操作系统将只关心退出返回值的低八位，这就是传递给父进程的全部内容）。
Exit return values are only suppose to be between 0 and 255 per the POSIX spec. You shouldn't be returning values higher than that (in other words a POSIX-compliant OS will only be concerned with the lower eight-bits of your exit return value, and that's all that will be passed to the parent process).
Unix/POSIX 仅支持 8 位。 10 位将是一个奇数（在数学和逻辑意义上）值，所以我必须同意@DigitalRoss。
Unix/POSIX only supports 8 bits. 10 bits would be an odd (in both mathematical and logical senses) value, so I'd have to agree with @DigitalRoss.