1
我正在用C编写一个虚拟机，并且我具有所有各种功能，但是我无法将它们放在一起。具体来说，我遇到了问题，我需要一种方法来增加程序计数器，而不会干扰改变PC所指向的指令，如JMP，JPC，CAL和RET。当我尝试采取措施取消pC++时，如PCvalueAfterJmp - 1或if声明不会在这些情况下递增，它会突然进入无限循环，似乎会重复执行指令。模拟虚拟机，在增加电脑和跳转时遇到问题
这个程序读取输入文件并打印到屏幕上正在处理什么指令和堆栈
int main(int argc, char* argv[]){
int running = 1;
int numInstructions = 0;
int lineRun;
int arcntr = 0;
//Memory
int stack[MAX_STACK_HEIGHT];
instruction code[MAX_CODE_LENGTH];
int arlist[MAX_STACK_HEIGHT];
//Registers
int sp=0;
int bp=1;
int pc=0;
instruction ir;
//Initializing ir
ir.op = 0;
ir.l = 0;
ir.m = 0;
//Initializing stack
stack[1] = 0;
stack[2] = 0;
stack[3] = 0;
//Reading the input file
numInstructions = readFile(argc, argv, code);
if(numInstructions < 0) //Exit with error if readFile returns invalid
return 1;
//show input code
printFile(code, numInstructions);
//setup and labeling
printState(-1, ir, pc, bp, sp, stack, arlist);
//Execution loop
while(running)
{
lineRun = pc;
//Fetch cycle
ir = code[pc];
//Execution cycle returns a nonzero to keep program running until end
if(!execOp(&sp, &bp, &pc, ir, code, stack, arlist, &arcntr))
running = 0;
//if statement didn't work
printState(lineRun, ir, pc, bp, sp, stack, arlist);
//if (!(ir.op == 5 || ir.op == 7 || ir.op == 8 || (ir.op == 2 && ir.m == 0)))
pc++;
}
return 0;
}
这里的当前状态是我的执行周期
int execOp(int* sp, int* bp, int* pc, instruction ir, instruction code[],
int stack[], int arlist[], int* arcntr){
switch((opcode)ir.op){
case LIT:
stack[++(*sp)] = ir.m;
break;
case OPR: //Operators
switch((operator)ir.m){
case RET:
if(*bp == 1) //Kill the simulation if we're at the base level
return 0;
arlist[--(*arcntr)] = 0;
*sp = *bp - 1;
*pc = stack[*sp+3];
*bp = stack[*sp+2];
break;
case NEG:
stack[*sp] = -stack[*sp];
break;
case ADD:
(*sp)--;
stack[*sp] = stack[*sp] + stack[*sp+1];
break;
case SUB:
(*sp)--;
stack[*sp] = stack[*sp] - stack[*sp+1];
break;
case MUL:
(*sp)--;
stack[*sp] = stack[*sp] * stack[*sp+1];
break;
case DIV:
(*sp)--;
stack[*sp] = stack[*sp]/stack[*sp+1];
break;
case ODD:
stack[*sp] = stack[*sp] % 2;
break;
case MOD:
(*sp)--;
stack[*sp] = stack[*sp] % stack[(*sp)+1];
break;
case EQL:
(*sp)--;
stack[*sp] = stack[*sp] == stack[*sp+1];
break;
case NEQ:
(*sp)--;
stack[*sp] = stack[*sp] != stack[*sp+1];
break;
case LSS:
(*sp)--;
stack[*sp] = stack[*sp] < stack[*sp+1];
break;
case LEQ:
(*sp)--;
stack[*sp] = stack[*sp] <= stack[*sp+1];
break;
case GTR:
(*sp)--;
stack[*sp] = stack[*sp] > stack[*sp+1];
break;
case GEQ:
(*sp)--;
stack[*sp] = stack[*sp] >= stack[*sp+1];
break;
}
break;
case LOD:
stack[++*sp] = stack[base(ir.l, *bp, stack) + ir.m];
break;
case STO:
stack[base(ir.l, *bp, stack) + ir.m] = stack[(*sp)--];
break;
case CAL:
arlist[(*arcntr)++] = *sp + 1;
stack[*sp + 1] = base(ir.l, *bp, stack);
stack[*sp + 2] = *bp;
stack[*sp + 3] = *pc - 1;
*bp = *sp + 1;
*pc = ir.m;
break;
case INC:
*sp = *sp + ir.m;
break;
case JMP:
*pc = ir.m;
break;
case JPC:
if(!stack[(*sp)--])
*pc = ir.m;
break;
case SOI:
printf("%d\n", stack[(*sp)--]);
break;
case SIO:
scanf("%d", &stack[++(*sp)]);
break;
}
return 1; //A non-zero return value keeps the machine running
}
增量计算机并调用'execOp'后更新堆栈指针。或者在'execOp'之前存储pc值，如果改变，请不要增加... –
考虑在指令提取之后但在解释指令之前递增PC。改变PC的指令将写入递增的值 – infixed
将PC递增到'execOp'。对于大多数指令，它会执行'（* pC++）'，但是对于跳转指令，它会执行其他操作。 – Barmar