Once the expression into a suffix and then calculated on the expression of very simple. Symbols on the face of a pop-up from his recent two operands and then proceed to the designated operator of the operation ok on the pressure after the completion of the operation ... Stack continued to scan, followed by repeat ... do not wait until the operator. All operations will be over. ... The results will be out no longer have to determine what priority. What the brackets would also be omitted .. Shuangba.
In fact, in the suffix turn up on time can be calculated expression scanning, the completion of the calculation.
Yesterday, the direct source of revised copy .. ..
opt.push ('#');
int len = exp.length ();
for (int i = 0; i <len;)
(
char ch = exp.at (i);
if (isalnum (ch))
(
long num = 0;
while (isalnum (ch)) / / converted to an integer
(
num = (ch-'0 ') + num * 10;
if (+ + i> = len)
break;
ch = exp.at (i);
)
opd.push (num); / / operand Ruzhan
)
else / / operator on the judge and push
(
if (ch == '(') / / left parenthesis direct push
opt.push (ch);
else if (ch == ')') / / there are brackets on the stack to bomb the left encountered until the brackets
(
ch = opt.top (); / / Top-made operator
while (ch! = '(') / / pop up to left brackets
(
OP (ch);
opt.pop ();
ch = opt.top ();
)
opt.pop (); / / pop left in brackets
)
else
(
int thisPri = GetPri (ch); / / operator of the current priority
char prevOpt = opt.top (); / / one operator
int prevPri = GetPri (prevOpt); / / operators on a priority
while (thisPri <= prevPri)
(/ / Output in the stack operator face up to than the current operator of a lower priority
OP (prevOpt);
opt.pop (); / / output after the pop
prevOpt = opt.top ();
prevPri = GetPri (prevOpt);
)
opt.push (ch); / / push the current operator
)
i + +;
)
)
char ch = opt.top (); / / expression after scanning the stack in the remainder of the full output of the operator
while (ch! = '#')
(
OP (ch);
opt.pop ();
ch = opt.top ();
)
Yesterday and did not see the code almost no difference, just yesterday, the output of the local operations into the OP (const char & ch); completion of the operation
Need to understand that. Expression scanning, is likely to push the operator not to carry out operation (by the concrete expression of this decision), the operator should be the last stack empty shells, all of the completion of the operation .... Operand stack at this time is the only one of the few is the result of the operation.
Function OP. To achieve the following. Parameter is the operator ch
void Caculator:: OP (const char & ch)
(
long RightOperand = opd.top ();
opd.pop ();
long LeftOperand = opd.top (); / / These two operations can not be wrong in the order of a few
opd.pop ();
switch (ch)
(
case '+':
LeftOperand + = RightOperand;
break;
case '-':
LeftOperand -= RightOperand;
break;
case '*':
LeftOperand *= RightOperand;
break;
case '/':
if (RightOperand == 0) / / divisor 0
(
system ( "pause");
exit (1);
)
else
LeftOperand / = RightOperand;
default:
cout << "illegal operator !!----:"<< ch <<endl;
system ( "pause");
exit (1);
)
opd.push (LeftOperand); / / find the value of the pressure into the stack
)
opd is the operand stack, is a member of the class of variables. is the definition of stack <long> opd; / / can only carry out the whole operation
Simple ... |
