当您需要访问堆栈中“在其下方”的参数时,处理调用函数的返回地址位于已放置在堆栈中的任何参数之上的最佳方法是什么?我正在使用S12 Motorola / Freescale处理器。(S12具有:由A和B寄存器组成的16位D寄存器,每个寄存器具有8位。X和Y索引寄存器分别具有16位,程序计数器和堆栈指针)代码示例为:
MAIN ORG $1500 ;Set the start of the program
LDD #SomeValue ;Load the D register with a value
PSHD ;Push D onto the stack
JSR ROUTINE ;Go to the subroutine - pushes the PC to the SP
END_MAIN END
ROUTINE
PULD ;Pull the top of the stack into the D register
;D now holds the address for returning to the
;main function.
PSHD ;Push the return address back onto the stack
END_ROUTINE RTS ;Return to Main routine
问题在于,堆栈的顶部保留着下一条指令的地址,这使操作变得困难。例如,如果我需要一个位于地址下面的参数,则必须手动调整SP(看起来很hacky),或者必须拉开堆栈的顶部并将其存储在一个寄存器中,该寄存器需要空间。最后一种方法的一个变种是将返回地址存储在变量中,不幸的是,此处声明的变量在范围上是全局的,这并不理想。
ReturnAddress EQU $2000 ;Declare variable at address $2000
STD ReturnAddress ;Store the D register's contents in variable
还有其他我看不到的选择吗?
得益于Jester的一些投入,观察堆栈中发生的事情并使用其中包含的参数非常简单。我编写了一个简单的程序,演示了按值和引用传递参数的方法。
QUOTIENT EQU $1000 ;Variable to hold our QUOTIENT
REMAINDER EQU $1002 ;Variable to hold our REMAINDER
MAIN ORG $2000 ;Set the start of the program
LDD #!50 ;Load the D register with a value (16 bits)
PSHD ;Push D onto the stack (16 bits)
LDD #!10 ;Load the D register with a value (16 bits)
PSHD ;Push D onto the stack (16 bits)
LDD #QUOTIENT ;Load the D register with an address (16 bits)
PSHD ;Push D onto the stack (16 bits)
LDD #REMAINDER ;Load the D register with an address (16 bits)
PSHD ;Push D onto the stack (16 bits)
JSR DIVIDE ;Push the PC onto the stack (16 bits).
LEAS $0A, SP ;Instead of PULD 5x, thanks Jester
END_MAIN END
;******************************************************************************;
;DIVIDE - This routine expects the following parameters to be on the stack:
; STACK
;( 0 ) - (16 bits for return address)
;( 2 ) - (16 bits of address to store the REMAINDER (ANSWER var reference))
;( 4 ) - (16 bits of address to store the QUOTIENT (ANSWER var reference))
;( 6 ) - (16 bits for divisor)
;( 8 ) - (16 bits for dividend)
;******************************************************************************;
DIVIDE LDD 8,SP ;Loads the dividend (50)
LDX 6,SP ;Loads the divisor (10)
IDIV ;Divide
LDY 4,SP ;Get the address of the Quotient
STX 0,Y ;Store at that address
LDY 2,SP ;Get the address of the remainder
STD 0,Y ;Store at that address
END_DIVIDE RTS
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