Question

provide na example of  x86 assembly instruction simulation that involves %esp, %ebp, and %eip machine registers

Answer 1

The subroutine attached below performs the following actions:

• saving a copy of the stack pointer in the base pointer (EBP)
• allocating local variables by changing the stack pointer
• saving register values on the stack

/*Start the code section/ .text /Define myFunc as a global (exported) function. */ globl myFunc type myFunc, @function myFunc: /*Subroutine Prologue */ push %ebp mov %esp, %ebp /* Set the new base pointer value. */ sub $4, %esp /* Make room for one 4-byte local variable. */ push %edi push %esi /* (no need to save EBX, EBP, or ESP) */ * Save the old base pointer value. '* /Save the values of registers that the function*/ /* will modify. This function uses EDI and ESI */ *Subroutine Body mov 8(%ebp), %eax /* Move value of parameter 1 into EAX. */ mov 12(%ebp), %es! /* Move value of parameter 2 into ESI. */ mov 16(%ebp), %edi /* Move value of parameter 3 into EDI. */ mov %edi, -4(%ebp) /* Move EDI into the local variable. */ add %esi, -4(Xebp) /* Add ESI into the local variable. */ add -4(%ebp), %eax /* Add the contents of the local variable */ /*into EAX (final result).*/ /*Subroutine Epilogue */ pop %esi pop %edi mov %ebp, %esp /* Deallocate the local variable. */ /*Recover register values. */ op %eb p p ret * Restore the caller's base pointer value. *

myFunc is a label and so elsewhere in the code, we can refer to its memory location using the more convenient symbolic name instead of expressing the location in its 32-bit value. This causes the EIP register to update implicitly by the provided control flow instructions. It cannot be manipulated directly therefore it is not mentioned in the code explicitly.