hi i have written a calculator that first converts from
infix to postfix.
generally it works, but the problem is that if my infix calculations are too complicated, the conversion no longer works.
for this expression (3+2)+1+3*((4*4+6)/2+3)*8+(5+4)/(3+1),
my program prints out -62, whereas C++ and the Windows calculator print out -190.
here is my source code for the conversion:
int precedence(char op)
{
if (op == ''+'' ) return 0;
if (op == ''-'') return 0;
if (op == ''*'' ) return 1;
if (op == ''/'') return 1;
}
void infixToPostfix()
{
char str[255] = "(3-2)+1-3*((4*4+6)/2-3)*8+(5-4)/(3-1)";
char ch;
stack<char> st;
char op;
int n = 0;
while ((ch = str[n++]) != ''\0'')
{
if (isdigit(ch))
postfix[x++] = ch;
if ( ch == ''('' )
st.push(ch);
else if (ch == '')'')
{
char op = st.pop();
while ( op != ''('' )
{
if (op != ''('') postfix[x++] = op;
if (!st.count()) break;
op = st.pop();
}
}
else if ( ch == ''+'' || ch == ''-'' || ch == ''*'' || ch == ''/'' )
{
while (st.count() && ( precedence(st.top()) >= precedence(ch) ))
{
char op = st.pop();
if (op != ''('')
postfix[x++] = op;
}
st.push(ch);
}
}
while ( st.count() )
{
op = st.pop();
if (op != ''('') postfix[x++] = op;
}
postfix[x++] = ''\0'';
}
And here is the source code for my expression evaluation:
int main()
{
printf("%d\n", (3-2)+1-3*((4*4+6)/2-3)*8+(5-4)/(3-1));
stack<int> expr;
int n = 0;
char ch;
infixToPostfix();
printf(postfix);
while ((ch = postfix[n++]) != ''\0'')
{
if (isdigit(ch))
expr.push(ch - ''0'');
if (ch == ''+'' )
expr.push(expr.pop() + expr.pop());
if (ch == ''*'' )
expr.push(expr.pop() * expr.pop());
if (ch == ''-'' )
{
int a = expr.pop();
expr.push(expr.pop() - a);
}
if (ch == ''/'' )
{
int a = expr.pop();
expr.push(expr.pop() / a);
}
}
printf("\n");
printf("Ergebnis: %d\n", expr.pop());
return 0;
}
can someone help me on this problem?
thanks
Gammastrahler
