Parsing Fully Parenthesized expression - c++

Question

I'm trying to Parse a Fully Parenthesized Exp for this grammar:
exp->(exp + exp)
|(exp - exp)
| num
num->[0-9]
...but I have a problem: when I enter "1+4" no error appears. How can I solve it ??

Answer 1

This is a classic problem of recursive descent parsers: your grammar does not require that the entire input is consumed. Once it finishes with the "while isdigit" loop, it considers its job done, ignoring the +4 portion of the input.
Add a check that the end of line is reached after the call of the top-level expression to address this problem:
void TopExp() {
Expr();
Match('\n');
}
You need to modify Match to allow matching \n when no additional input is available:
void Match(char c) {
int p = cin.peek();
if(p==c) {
cin>>c;
} else if (p == EOF && c == '\n') {
return
} else {
cout<<"Syntax Error! "<<(char)cin.peek()<<endl;
cin.ignore();
}
}

Answer 2

Try with this changed instruction: if (openParenthesis-closeParenthesis>0)
int Match(char c)
{
if(cin.peek()==c) {
cin>>c;
return 1;
} else
{
cout<<"Syntax Error! "<<(char)cin.peek()<<endl;
cin.ignore();
return 0;
}
}
void MatchOp()
{
if(cin.peek()=='+')
Match('+');
else if(cin.peek()=='-')
Match('-');
else
{
cout<<"invalid Operation: "<<(char)cin.peek()<<endl;
cin.ignore();
}
}
void Exp()
{ static int openParenthesis = 0;
static int closeParenthesis = 0;
if(cin.peek()!='\n')
if(cin.peek()=='(')
{
if (Match('(') == 1) {
openParenthesis += 1;
}
Exp();MatchOp();Exp();
if (Match(')') == 1) {
closeParenthesis -= 1;
}
}
else if(isdigit(cin.peek()))
{
if (openParenthesis-closeParenthesis>0) {
cout<<"Syntax Error! "<<(char)cin.peek()<<endl;
cin.ignore();
} else {
while(isdigit(cin.peek()))
{ cout<<(char)cin.peek();
Match(cin.peek());
}
}
}
else
{
cout<<"Syntax Error!"<<(char)cin.peek()<<endl;
cin.ignore();
}
}