execve（）是否设置寄存器以调用动态链接器或要执行的可执行文件？

Question

Understanding the Linux Kernel says execve() calls load_binary() of each linux_binfmt object, and load_binary()

Invokes the start_thread( ) macro to modify the values of the User Mode registers eip and esp saved on the Kernel Mode stack, so that they point to the entry point of the dynamic linker and to the top of the new User Mode stack, respectively.

mosvy wrote:

What happens is that the kernel arranges for the execve system call, upon returning to user mode, to have the IP (instruction pointer) register set to point to the beginning of the _start function , and the SP (stack pointer) register set to point to the beginning of the argv + env string list, so the effect from the point of view of user mode is as if someone had called the _start function as:

 _start(argc, argv0, argv1, ... , NULL, env0, env1, ... NULL) 

in a calling convention where all arguments are passed on the stack.

By "the beginning of the _start function", did he mean the entry point of the executable to be executed ?

Which one is correct? Ie does execve() set up registers to invoke dynamic linker or the executable to be executed?

Answer 1

_start is the entry point of an executable. In C you may think that main is the entry point, but this is just the point of view of the programmer. The compiler may insert other code in _start, to run before calling main.

My interpretation is that everything described in you question, happens before calling _start (or seeming to call _start).