C語言 如何案ESC鍵回原問題
這是一道先乘除後加減的編程(按下 '= '鍵會算出答案)
題目:
在按下 '= '鍵之前 按下 'ESC ' 會回到原題目
在算出答案後 按任何建繼續 按 'ESC '回中程序
不知道哪為大大可以幫我改一下
以下是先乘除後加減的編程 要把它加上題目說的
#include <stdio.h>
#include <stdlib.h>
//prototypes
char GetInteger(int* x);
int main(int argc, char *argv[])
{
//declare variables
char state; //to the state of the state machine
int x, x1, x2, x3, result;
char op, op1;
int repeat;
//explain the program
printf( "Welcome to the single-digit calculator ");
printf( "with multiplication.\n\n ");
state = 'F '; //enter state F initially
repeat = 'y '; //enter the loop initially
while(repeat == 'y ' || repeat == 'Y ')
{
switch(state)
{
case 'F ':
//state F: prompt user for input
//and leave for state A unconditional
printf( "Please input the equation:\n ");
state = 'A '; //update the state
break;
case 'A ':
case 'B ':
//op,x <- input
//and leave for state C on op1=op; x1=x
//leave for state H on x1=x
//leave for state J on result=x
op = GetInteger(&x);
//make decision to update the state
if (op == '+ ' || op == '- ')
{
state = 'C ';
op1 = op;
x1 = x;
}
else if (op == '* ')
{
state = 'H ';
x1 = x;
}
else
{
state = 'J ';
result = x;
}
break;
case 'C ':
case 'D ':
case 'E ':
//op,x <- input
//and leave for state E on +,-; x1 <- x1(op1)x,op1=op
//leave for state G on *; x2=x
//leave for state J on =; result <- x1(op1)x
op = GetInteger(&x);
//make decision to update the state
if (op == '+ ' || op == '- ')
{
state = 'E ';
//update x1 based op1
if (op1 == '+ ')
{
x1 = x1 + x;
}
else
{
x1 = x1 - x ;
}
//update op1
op1 = op;
}
else if (op == '* ')
{
state = 'G ';
x2 = x;
}
else
{
state = 'J ';
//update result based op1
if (op1 == '+ ')
{
result = x1 + x;
}
else
{
result = x1 - x;
}
}
break;
case 'G ':
//op,x <- input
//leave for state D on +,-; x1=x1(op1)x2*x;op1=op
//leave for state G on *; x2=x2*x
//leave for state J on =; result=x1(op1)x2*x
op = GetInteger(&x);
//update the state based on op
if (op == '+ ' || op == '- ')
{
state = 'D ';
//update x1 based on op1
if (op1 == '+ ')
{
x1 = x1 + x2 * x;
}
else
{
x1 = x1 - x2 * x;
}
//update op1
op1 = op;
}
else if (op == '* ')
{
state = 'G ';
x2 = x2 * x;
}
else
{
state = 'J ';
//update result based on op1
if (op1 == '+ ')
{
result = x1 + x2 * x;
}
else
{
result = x1 - x2 * x;
}
}
break;
case 'H ':
case 'I ':
//op,x <- input
//leave for state C on +,-; x1 <- x1*x; op1=op
//or leave for state H on *; x1 <- x1*x
//or leave for state J on =; result <- x1*x
op = GetInteger(&x);
//make decision to update the state
if (op == '+ ' || op == '- ')
{
state = 'C ';
//update x1 and op1
x1 = x1 * x;
op1 = op;
}
else if (op == '* ')
{
state = 'H ';
//update x1
x1 = x1 * x;
}
else
{
state = 'J ';
result = x1 * x;
}
break;
case 'J ':
//print the result
//leave for state K unconditional
printf( " %d\n ", result);
//update state
state = 'K ';
break;
default:
case 'K ':
//prompt the user if repeat or not
//leave for state F on repeat
//otherwise break the loop and terminate the prog
printf( "Another calculation? (y/n) ");
repeat = getch(); //re-use op tempararily
printf( "\n ");
state = 'F '; //when repeat
break;
}
}
//prompt the user for completion
printf( "\nBye!\n\n ");
system( "PAUSE ");
return 0;
}
//Function definitions
char GetInteger(int* x)
{
char input;
*x = 0; //initialize the variable pointed to by x
while(1)
{
input = getch();
if (input > = '0 ' && input <= '9 ')
{
*x = 10 * (*x) + input - '0 '; //update *x
printf( "%c ", input); //print the digit
continue; //to go back the beginning of the loop
}
if (input == '= ' || input == '+ '
|| input == '- ' || input == '* ')
{
printf( " %c ",input);
break; //to terminate the loop
}
}
return input;
}
[解决办法]
帮顶
