谁帮我看看sml的C语言代码。给我注释啊。求啊
#ifndef _Simpletron_H_#define _Simpletron_H_
#include <stdio.h>
#include<string.h>
#ifdef __cplusplus
extern "C"
{
#endif
#define Read 10
#define Write 11
#define Loda 20
#define Store 21
#define Add 30
#define Subtract 31
#define Divide 32
#define Multiply 33
#define Branch 40
#define Branchneg 41
#define Branchzero 42
#define Halt 43
int accumulator;
short instructionCounter;
int instructionRegister;
short operationCode;
short operand;
int memory[100];
static void init();
void instruction()
{
printf("\n\n\n\n *** Welcome to Simpletron ***");
printf("\n\n It is Mrt Lin's works. Please don't copy!!\n\n\n\n ");
}
void dump()
{
char sign = '+';
int start = 9;
int i = 0;
int j = 0;
int skip = 7;
printf("\n\n\n\n\n");
printf("accumulator\t\t\t%+.4d\n\n", accumulator);
printf("instructionCounter\t\t %.2d\n\n", instructionCounter);
printf("instructionRegister\t\t%+.4d\n\n", instructionRegister);
printf("operationCode\t\t\t %.2d\n\n", operationCode);
printf("operand\t\t\t\t %.2d\n", operand);
printf("\n\n");
printf(" \t\tMemory:\n\n");
for (i=0 ;i<10 ;i++ )
{
if (i > 0)
printf("%7d", i);
else
printf("%9d", i);
}
printf("\n");
for (i=0 ;i<10 ;i++ )
{
printf("%.2d", i*10);
for (j=0 ;j<10 ;j++ )
{
printf(" %+.4d ", memory[i*10 + j]);
}
printf("\n");
}
}
void halt()
{
printf("\n\n *** Simpletron execution terminated ***\n\n");
dump();
exit(1);
}
void error(int errno)/*errno indicates error number*/
{
switch (errno)
{
case 1:
printf("\n\n[Instruction is negateive.]\n");break;
case 2:
printf("\n\n[Attempt to devide by zero.]\n");break;
case 3:
printf("\n\n[Result beyond -9999 ~ +9999.]\n");break;
}
printf("\n\n *** Simpletron execution abnormally terminated. ***\n");
dump();
exit(-1);
}
int read();
int write();
int load();
int store();
int add();
int subtract();
int divide();
int multiiply();
int branch();
int branchneg();
int branchzero();
void start();
void exec();
void exechelper();
#ifdef __cplusplus
}
#endif
#endif
#include <stdio.h>
#include <math.h>
int main(int argc, char *argv[])
{
init();
instruction();
start();
exec();
return 0;
}
static void init()
{
accumulator = 0;
instructionCounter = 0;
instructionRegister = 0;
operationCode = 0;
operand = 0;
}
void start()
{
int resultOfScanf = 0;
int scanfed = -1;
while (scanfed != -99999)
{
printf("%.2d <<<< ", instructionCounter);
resultOfScanf = scanf("%d", &scanfed);
if (0 == resultOfScanf || (scanfed!=-99999 && abs(scanfed)>9999))
{
printf("wrong instruction or data word.(-9999 ~ +9999)]\n");
continue;
}
if (scanfed == -99999)
{
instructionCounter--;
break;
}
memory[ instructionCounter++ ] = scanfed;
}
printf(" \n\n*** Program. Is completeing ***\n\n");
}
void exec()//判断是否超过一百个程序；
{
int i = 0;
init();
printf("*** Program. execution begins ***\n\n");
while (instructionCounter<100)
{
instructionRegister = memory[ instructionCounter++ ];
if (instructionRegister <= 0)
{
instructionCounter--;
error(1);
}
operationCode = instructionRegister/100;
operand = instructionRegister%100;
exechelper();
}
}
void exechelper()
{
switch (operationCode)
{
case Read:
read();break;
case Write:
write();break;
case Loda:
load();break;
case Store:
store();break;
case Add:
add();break;
case Subtract:
subtract();break;
case Divide:
divide();break;
case Multiply:
multiply();break;
case Branch:
branch();break;
case Branchneg:
branchneg();break;
case Branchzero:
branchzero();break;
case Halt:
halt();
}
}
int read()
{
int resultOfScanf = 0;
int scanfed = -1;
while (1)
{
printf(" \n\ndo your work!! \n");
printf("please input your munbers: ");
resultOfScanf = scanf("%d", &scanfed);
if (0 == resultOfScanf || abs(scanfed)>9999)
printf("[Please input a integer( -9999 ~ +9999 ).]\n");
memory[ operand ] = scanfed;
break;
}
return scanfed;
}
int write()
{
printf("%d", memory[ operand ]);
return 1;
}
int load()
{
accumulator = memory[ operand ];
return 1;
}
int store()
{
memory[ operand ] = accumulator;
return 1;
}
int add()
{
accumulator += memory[ operand ];
return 1;
}
int subtract()
{
accumulator -= memory[operand ];
return 1;
}
int divide()
{
int div = memory[ operand ];
if (div == 0)
{
error(2);
}
else
{
accumulator /= div;
}
return 1;
}
int multiply()
{
accumulator *= memory[ operand ];
if (abs(accumulator) > 9999)
error(3);
return 1;
}
int branch()
{
instructionCounter = operand;
return 1;
}
int branchneg()
{
if (accumulator < 0)
instructionCounter = operand;
return 1;
}
int branchzero()
{
if (accumulator == 0)
instructionCounter = operand;
return 1;
}
或者帮我来个简单的？
要求
这样就实现了Simpletron模拟器指令执行的过程；
1． 由键盘输入范例1的SML程序，执行并在屏幕上显示逐条指令执行过程中的“计算机k,hu信息
2． 将范例程序2（见附录B中）编写在文本文件fl2.txt中，Simpletron运行时从文件中读取程序再执行并最终将执行过程的“计算机转储”信息保存在另一个文本文件中。
3． 程序运行时，有键盘读入文本文件名，再打开指定的文件执行完成以下任务的SML程序：
1） 用标记控制的循环读取10个正数，计算并打印出它们的和；
2） 用计数器控制的循环读取7个数，其中既有正数，也有负数，计算并打印出它们的平均值。
3） 读取一组数，判断并打印出最大的数。读取的第一个数表示要处理的数的个数。
其它要求：
只能使用C/C++语言，源程序要有适当的注释，使程序容易阅读
