Char Array Augmented Grammar For LR0 Parser Question

[sev908]

Greetings,

I am creating a program that will find the canonical LR0 sets for a given grammar.
To start I have 2 char arrays, production, and production2. production is simply the grammar imported from a text file. production2 is the augmented grammar with a "." shifted throughout each production: see expected output.

My problem occurs when iterating through production to create the augmented production 2 starting at line 66 to line 117.

I believe I screwed up my logic somewhere as the loop does not process the correct output.

I will try to solve the problem on my own, but hopefully it is a simple solution. Any assistance would be greatly appreciated.

My code below is what I currently have for augmenting my grammar:
See line 66 for the loop in question.

int main() 
{ 
    char terminals[10][10] = {}; 
    char production[10][10] = {};
    char production2[10][10] = {};
    vector <char> v;

    bool product = false;
    ifstream inFile;
    string strFilename;
    string strword;
    cout << "enter file name" << endl;
    cin >> strFilename;
    
    while (inFile.fail()) //While the file does not open display error message and asks for another file name and location
	{
		cout << "Input file error!\n";
		cout << "Please enter the data file name (with location): ";
		cin >> strFilename;
	}

	inFile.open(strFilename.c_str()); //Opens file
	int i = 0;
	int j = 0;
	while (inFile >> strword) //While the file is copying into the word
	{
	    if (strword == "$")//if a dollar sign is present then we start saving the cahracters in our production array
	    {
	        product = true;
	    }
	    if (product == false)//else we save the poroductions in the terminal array
	    {
	        strcpy(terminals[i], strword.c_str());
	        i++;
	    }
	    else if (product == true && strword != "$")
	    {
	        
	        strcpy(production[j], strword.c_str());
	        j++;
	    }
    
	}
	inFile.close(); //Close the input file
	
    for(int i = 0; i < 10; i++)//loop to insert . at the begining of each rpoduction: E->.TR
	{
	    //int j = 0;
	    if (production[i][0] == '\0')
        {}
	    for(int j = 0; j<10; j++)
	    {
	        if (production[i][0] == '\0')
            {}
	        else if(production[i][j] == '>')
	        {
	            for(int a=9; a>j; a--)
	            {
		            production[i][a]=production[i][a-1];
	            }
	            production[i][j+1]='.';
	        }
	    }
	}
		
	int a = 0;
	for(int i = 0; i < 9; i++)//begin creation of augmented grammar
	{
		for(int j = 3; j <10; j++)
		{
			if(production[i][j-1] == '.')
			{
				for(int b=0; b < 10; b++)
	        	{
		    		production2[a][b]=production[i][b];
	        	}
	    		a++;

				int num = j;
				for(int b = 1; b <= 3; b++)
				{
					if(production2[b-1][num] == '\0')
					{
						break;
					}
					else
					{
						for(int a = 0; a < 10; a++)
						{
							if(a == num-1)
							{
								a++;
							}
							if(a==num+1)
							{
								v.push_back('.');
							}
							char x =production2[b-1][a];
							if(a>=0)
							{
								v.push_back(x);
							}
						}
						for (int c = 0; c < v.size(); c++)
						{
							production2[b][c] = v[c];
						}
						num++;
						a++;
						v.clear();
					}

				}
			}
		}
	}
	
	
	for (int i = 0; i < 10; i++) 
    {
        if (production[i][0] == '\0')
        {}
        else
        {
            for (int j = 0; j < 10; j++)
            {
                cout << production[i][j];//print all of the productions
            }
            cout << endl;
        }

    }
	
	cout << endl;

	
	for (int i = 0; i < 10; i++) 
    {
        if (production2[i][0] == '\0')
        {}
        else
        {
            for (int j = 0; j < 10; j++)
            {
                cout << production2[i][j];//print all of the augmented productions
            }
            cout << endl;
        }

    }
    cout << "end";
    return 0; 
}

output:

E->.TR
R->.+TR
T->.FY
Y->.*FY
F->.(E)

E->.TR
E->T.R
E->TR.
R->.+TR
T->.FY
Y->.*FY
F->.(E)

expected output:

E->.TR
R->.+TR
T->.FY
Y->.*FY
F->.(E)

E->.TR
E->T.R
E->TR.
R->.+TR
R->+.TR
R->+T.R
R->+TR.
T->.FY
T->F.Y
T->FY.
Y->.*FY
Y->*.FY
Y->*F.Y
Y->*FY.
F->.(E)
F->(.E)
F->(E.)
F->(E).

[ne555]

char production[10][10];
for (int i = 0; i < 15; i++) 
   for (int j = 0; j < 10; j++)
      production[i][j+1];

out of bounds.

[sev908]

Corrected the out of bound errors. Thank you for pointing them out. Original problem still persists.

[ne555]

provide your input file

[sev908]

apologies. input is a text file

input.txt

+
*
(
)
$
E->TR
R->+TR
T->FY
Y->*FY
F->(E)
F->i
$

[ne555]

				int num = j;
				for(int b = 1; b <= 3; b++) //¿why range [1:3]?
				{
					if(production2[b-1][num] == '\0')
					{
						break;
					}
					else
					{
						for(int a = 0; a < 10; a++) //hiding `a' that gived you production2.size
						{
							if(a == num-1)
							{
								a++;
							}
							if(a==num+1)
							{
								v.push_back('.');
							}
							char x =production2[b-1][a];
							if(a>=0) //tautology
							{
								v.push_back(x);
							}
						}
						for (int c = 0; c < v.size(); c++)
						{
							production2[b][c] = v[c];
						}
						num++;
						a++;
						v.clear();
					}
				}

you are only operating on production2[0], production2[1] and production2[2] over and over again.
the logic is unnecessarily complicated for just moving a dot.
say that s is a std::string that has a dot

int position = s.find('.');
ss = s;
augmented_grammar.push_back(ss);
while(position < s.size()-1) {
	next = ss;
	swap(next[position], next[position+1]); //move the '.' to the right
	++position;
	ss = next;
	augmented_grammar.push_back(ss);
}

for (int b = 0; b < 10; b++) {
  production2[a][b] = production[i][b];
}

`a' reaches 12, out of bounds.
production2 is not big enough to store the augmented grammar

[sev908]

Thenk you for pointing out the production 2 error. After talking to some other people i was told to use a vector of strings and just use the swap. Thank you again.