在看 man page （ man clone ）时有这样一句：

Since the child and calling process may share memory, it is not possible for the child process to execute in the same stack as the calling process.

中文翻译是：由于子进程和调用进程可能共享内存，因此子进程不可能与调用进程在同一堆栈中执行。

为什么会由于共享内存（ share memory ）会导致父子进程运行在相同的栈上？

说了不能啊。

由于子进程和调用进程可能共享内存，因此子进程不可能与调用进程在同一堆栈中执行。=>共享内存会导致父子进程运行在相同的栈。
想知道这个因果关系是如何得出来的
不应该是 如果父子进程运行在相同的栈上，就无法共享内存吗……（虽然这个命题好像也是不对的

每个 process 有自己的 stack pointer （ SP ）。如果两个 process 共享 memory space 还共享 stack 的话，互相有可能会 overwrite 掉对方的 stack。
e.g.
Process 1 has initial sp1 = 0xff
Process 1 advances sp1 to 0xf0
Process 1 calls clone
Process 2 is created with initial sp = 0xff
Context switch to Process 2
Process 2 advances sp2 to 0xa0
Context switch to Process 1
Process 1 continues executing with sp1 = 0xf0
--> Content from 0xff to 0xf0 are overwritten by Process 2
所以：由于子进程和调用进程可能共享内存 => 因此子进程不可能与调用进程在同一堆栈中执行

这里并没有后半个因果关系: “由于子进程和调用进程可能共享内存，因此子进程不可能与调用进程在同一堆栈中执行。=> 共享内存会导致父子进程运行在相同的栈”

上面是一本正经的瞎猜，说错了请多包涵

例子的第 4 行有个 typo，应该是“ with initial sp2 = 0xff ”

人家说的明明是不能。

补充说明，问题的语境是
```c
_GNU_SOURCE

int clone(int (*fn)(void *), void *child_stack,
int flags, void *arg, ...
/* pid_t *ptid, void *newtls, pid_t *ctid */ );
```
... ...
The child_stack argument specifies the location of the stack used by the child process. Since the child and calling process may share memory, it is not possible for the child process to execute in the same stack as the calling process. The calling process must therefore set up memory space for the child stack and pass a pointer to this space to clone(). Stacks grow downward on all processors that run Linux (except the HP PA processors), so child_stack usually points to the topmost address of the memory space set up for the child stack.
... ...
现在觉的“ Since the child and calling process may share memory, it is not possible for the child process to execute in the same stack as the calling process. ”这句话没有营养，关键在后面，调用 clone 要指定 child_stack。