CODEWITHSUNDEEP
c++visual-studio-2008recursionbacktrackingoutofrangeexception
forest.peterson
forest.peterson

Reputation: 763

recursive function has out_of_range exception

Given a global vector list of ASCII codes and corresponding number values and a string, like 000.00-000.0.0.0, this function takes an input token strings 2-char or 3-char long and replaces it with a single ASCII symbol that represents the number value between 0 and 184, then returns the shortened string without deliminator as out. Also, in reverse (direction 1) given the ASCII symbol it converts back to the number string and returns.

//looks for input string in vector and returns output, 'c' is check row, 'r' is return row
string vectorSearch(string &check, int n, int c, int r) 
{
    if (check.length() <= 1) 
        return check;
    if (list[n][c] == check || list[n][c] == ('0'+check)) //adds leading zero if 2char long
        return list[n][r];
    else 
        return vectorSearch (check, ++n, c, r);
}

//this function takes an ontology and either changes from single char 
//to string or takes strings and converts to char representation
string Lexicon::convertOntology(string input, int direction, string out, string temp) 
{
    if (input == "" && temp == "") 
        return out; //check for completed conversion
    else {
        if (input[0] == '.' || input[0] == '-' || input == "") { //found deliniator or endk
            if (input != "") return convertOntology(input.substr(1),direction, 
                 out+=vectorSearch(temp, 0, direction, 1-direction), "");
            else return convertOntology("", direction, 
                 out+=vectorSearch(temp, 0, direction, 1-direction), "");
        } else 
            return convertOntology(input.substr(1), direction, out, temp+=input[0]); //increment and check
    }
}

These functions work fine except for on output after the last char is parsed. With a break on the line return convertOntology(input.substr(1), direction, out+=add, temp); there is an error when input == "" and temp == "0" - the last pass through vectorSearch() should clear the temp and add the temp char to the out string, since the temp is == 1char then it should be returned from vectorSearch() as it is. Then clear the convertOntology() return check of input and temp == "". But, it never reaches a break on the first line of vectorSearch() and there is an

Unhandled exception at 0x77bc15de exception: std::out_of_range at memory location 0x0035cf1c

What is going on? Is this an issue with recursion backtracking through returns and I am missing a return somewhere to break the recursion loop?

Upvotes: 1

Views: 251

Answers (2)

Bo Persson
Bo Persson

Reputation: 92371

Even if you don't get to the else part,

input.substr(1)

will throw an exception when the input string is exactly one character long.

Seems it doesn't - input.substr(input.size()) is allowed, and returns an empty string.

You will later likely have a similar problem in VectorSearch. If there is no match, you will increment n until it gets out of range.

Upvotes: 1

yuri kilochek
yuri kilochek

Reputation: 13589

for the case where temp == "" and input != "" you call input.substr(1) which is, well, out of range.

Upvotes: 2

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