问题
When exit(0) is used to exit from program, destructors for locally scoped non-static objects are not called. But destructors are called if return 0 is used.Note that static objects will be cleaned up even if we call exit().
There should be some reason behind this logic. i just want to know what it is? Thank you.
回答1:
In the case of exit( 0 ), you're calling a function. You
don't expect the destructors of local variables to be called if
you're calling a function. And the compiler doesn't know,
a priori, that there is anything special about exit( 0 ).
In fact, this rationale really only applies to C++ before
exceptions. The standard could redefine exit() to throw an
implementation defined exception with the argument, and specify
that the call to main is wrapped in a try block which catches
this exception, and passes the return code back to the system.
This would mean that exit have a completely different
semantics in C and in C++, however; at any rate, there's been no
proposal before the committee to make this change.
来源：https://stackoverflow.com/questions/15945607/exit0-vs-return-0