计算机程序运行时会发生什么?

Other systems can be very different. DOS, for instance, ran in real mode, and its memory allocation when running programs looked much differently:

+-----------+ top of memory
| extended  | above the high memory area, and up to your total memory; needed drivers to
|           | be able to access it.
+-----------+ 0x110000
|  high     | just over 1MB->1MB+64KB, used by 286s and above.
+-----------+ 0x100000
|  upper    | upper memory area, from 640kb->1MB, had mapped memory for video devices, the
|           | DOS "transient" area, etc. some was often free, and could be used for drivers
+-----------+ 0xA0000
| USER PROC | user process address space, from the end of DOS up to 640KB
+-----------+
|command.com| DOS command interpreter
+-----------+ 
|    DOS    | DOS permanent area, kept as small as possible, provided routines for display,
|  kernel   | *basic* hardware access, etc.
+-----------+ 0x600
| BIOS data | BIOS data area, contained simple hardware descriptions, etc.
+-----------+ 0x400
| interrupt | the interrupt vector table, starting from 0 and going to 1k, contained 
|  vector   | the addresses of routines called when interrupts occurred.  e.g.
|  table    | interrupt 0x21 checked the address at 0x21*4 and far-jumped to that 
|           | location to service the interrupt.
+-----------+ 0x0

你可以看到 DOS 允许直接访问操作系统内存，没有保护，这意味着用户空间程序通常可以直接访问或覆盖他们喜欢的任何东西.

int mul( int x, int y ) {
    return x * y;       // this stores the result of MULtiplying the two variables 
                        // from the stack into the return value address previously 
                        // allocated, then issues a RET, which resets the stack frame
                        // based on the arg list, and returns to the address set by
                        // the CALLer.
}

int main() {
    int x = 2, y = 3;   // these variables are stored on the stack
    mul( x, y );        // this pushes y onto the stack, then x, then a return address,
                        // allocates space on the stack for a return value, 
                        // then issues an assembly CALL instruction.
}

f( g( h( i ) ) );