《信息安全系统设计基础》家庭作业

《信息安全系统设计基础》家庭作业

4.49

在seq-full.hcl中，修改实现leave指令的控制逻辑块的HCL描述。

首先，leave等价的Y86代码序列为：

	rrmovl %ebp,%esp

	popl %ebp

相应栈帧如图所示：

则实现这个指令所需要执行的计算为：

所以，在对seq-full.hcl的修改中，要将leave指令写入instr_valid，表明该指令为合法指令；在译码与写回阶段，对srcA添加leave指令对应于%ebp，对srcB添加leave指令对应于%ebp，dstE添加leave指令写回于%esp，dstM添加leave指令写回于%ebp；在执行阶段，需要对操作数A赋值为4，即对aluA添加leave指令对应+4,aluB添加leave指令对应valB(即%ebp)；在访存阶段，只读控制信号men_read添加leave指令，地址men_addr添加leave指令对应于valA(即%ebp）；在更新PC阶段无变化

seq-full.hcl的修改后：

代码：

################ Fetch Stage     ###################################

# Determine instruction code
int icode = [
	imem_error: INOP;
	1: imem_icode;		# Default: get from instruction memory
];

# Determine instruction function
int ifun = [
	imem_error: FNONE;
	1: imem_ifun;		# Default: get from instruction memory
];

bool instr_valid = icode in 
	{ INOP, IHALT, IRRMOVL, IIRMOVL, IRMMOVL, IMRMOVL,
	       IOPL, IJXX, ICALL, IRET, IPUSHL, IPOPL ,ILEAVE };

# Does fetched instruction require a regid byte?
bool need_regids =
	icode in { IRRMOVL, IOPL, IPUSHL, IPOPL, 
		     IIRMOVL, IRMMOVL, IMRMOVL  };

# Does fetched instruction require a constant word?
bool need_valC =
	icode in { IIRMOVL, IRMMOVL, IMRMOVL, IJXX, ICALL  };

################ Decode Stage    ###################################

## What register should be used as the A source?
int srcA = [
	icode in { IRRMOVL, IRMMOVL, IOPL, IPUSHL  } : rA;
	icode in { IPOPL, IRET } : RESP;
	icode in { ILEAVE } : REBP;
	1 : RNONE; # Don't need register
];

## What register should be used as the B source?
int srcB = [
	icode in { IOPL, IRMMOVL, IMRMOVL  } : rB;
	icode in { IPUSHL, IPOPL, ICALL, IRET } : RESP;
	icode in { ILEAVE } : REBP;
	1 : RNONE;  # Don't need register
];

## What register should be used as the E destination?
int dstE = [
	icode in { IRRMOVL } && Cnd : rB;
	icode in { IIRMOVL, IOPL} : rB;
	icode in { IPUSHL, IPOPL, ICALL, IRET ,ILEAVE} : RESP;
	1 : RNONE;  # Don't write any register
];

## What register should be used as the M destination?
int dstM = [
	icode in { IMRMOVL, IPOPL } : rA;
	icode in { ILEAVE } : REBP;
	1 : RNONE;  # Don't write any register
];

################ Execute Stage   ###################################

## Select input A to ALU
int aluA = [
	icode in { IRRMOVL, IOPL } : valA;
	icode in { IIRMOVL, IRMMOVL, IMRMOVL } : valC;
	icode in { ICALL, IPUSHL } : -4;
	icode in { IRET, IPOPL ,ILEAVE} : 4;
	# Other instructions don't need ALU
];

## Select input B to ALU
int aluB = [
	icode in { IRMMOVL, IMRMOVL, IOPL, ICALL, 
		      IPUSHL, IRET, IPOPL ，ILEAVE} : valB;	
	icode in { IRRMOVL, IIRMOVL } : 0;
	# Other instructions don't need ALU
];

## Set the ALU function
int alufun = [
	icode == IOPL : ifun;
	1 : ALUADD;
];

## Should the condition codes be updated?
bool set_cc = icode in { IOPL };

################ Memory Stage    ###################################

## Set read control signal
bool mem_read = icode in { IMRMOVL, IPOPL, IRET ,ILEAVE };

## Set write control signal
bool mem_write = icode in { IRMMOVL, IPUSHL, ICALL };

## Select memory address
int mem_addr = [
	icode in { IRMMOVL, IPUSHL, ICALL, IMRMOVL } : valE;
	icode in { IPOPL, IRET ,ILEAVE} : valA;
	# Other instructions don't need address
];

## Select memory input data
int mem_data = [
	# Value from register
	icode in { IRMMOVL, IPUSHL } : valA;
	# Return PC
	icode == ICALL : valP;
	# Default: Don't write anything
];

## Determine instruction status
int Stat = [
	imem_error || dmem_error : SADR;
	!instr_valid: SINS;
	icode == IHALT : SHLT;
	1 : SAOK;
];

################ Program Counter Update ############################

## What address should instruction be fetched at

int new_pc = [
	# Call.  Use instruction constant
	icode == ICALL : valC;
	# Taken branch.  Use instruction constant
	icode == IJXX && Cnd : valC;
	# Completion of RET instruction.  Use value from stack
	icode == IRET : valM;
	# Default: Use incremented PC
	1 : valP;
];

6.33

先计算出s、b、t，将地址解析成指引高速缓存的位置

6.34

组5意味着地址第4到第2位为“101”，观察4路组相联高速缓存索引5中 V 表示为“1”的行，里面的数据块即是可能命中的数据，8个存储器地址为 0x1314 0x1315 0x1316 0x1317 0x1794 0x1795 0x1796 0x1797