Reputation: 51
I'm getting this error on my program and I don't understand why. The code essentially has to check for tags stored in a set which is declared as a global variable. If it's a valid tag it stores it in the stack if not returns a error message. Then it checks (if its a valid tag)if the closing tags are in order. This is all for the is_well_formed method. For the print_well_formed_file method it essentially checks if the given file is well formed if it is it'll display the file.:
terminate called after throwing an instance of 'std::out_of_range'
what(): basic_string::substr
What can I do to fix this error ? This is part of the code:
bool is_well_formed(ifstream& ifs, string& error_msg) {
// your code goes here
string fname, line;
Token tok;
Lexer lexer;
tags.insert("blue");
tags.insert("red");
tags.insert("cyan");
tags.insert("white");
tags.insert("yellow");
tags.insert("magenta");
tags.insert("dim");
tags.insert("underline");
tags.insert("bold");
while (getline(cin, fname)) {
// tries to open the file whose name is in string fname
string name = fname.substr(1, fname.length() - 2);
cout << "Name" + name;
ifs.open(name.c_str());
if (ifs.fail()) {
cerr << "ERROR: Failed to open file " << fname << endl;
ifs.clear();
} else {
while (getline(ifs, line)) {
lexer.set_input(line);
while (lexer.has_more_token()) {
tok = lexer.next_token();
string tmpTok = tok.value;
switch (tok.type) {
case TAG:
// If it has /, remove / from tmpTok
if (tok.value[0] == '/') {
tmpTok = tmpTok.substr(1, tmpTok.length() - 1);
}
if (tags.find(tmpTok) == tags.end()) {
// Check whether the encountered tag is valid
error_return("Tag " + tmpTok + " is invalid!");
} else {
// Valid Tag encountered
stack < string > tagstack;
tagstack.push(tmpTok);
// Check if the tags are formed properly
if (tok.value[0] == '/') {
// Remove / from tmpTok
string closingTag = tmpTok;
string openingTag = tagstack.top();
tagstack.pop();
if (closingTag.compare(openingTag) != 0) {
error_return(
closingTag + "doesn't match"
+ openingTag);
} //else
// return true; // if the file is well formed
}
}
break;
case IDENT:
cout << "IDENT: " << tok.value << endl;
break;
case ERRTOK:
error_return("Syntax error on this line\n");
//cout << "Syntax error on this line\n";
break;
case ENDTOK:
break;
}
}
}
}
}
return true; // if the file is well-formed
}
void print_well_formed_file(ifstream& ifs) {
//Check if file is well formed.
string line;
Lexer command;
if (is_well_formed(ifs, line)) { //if well formed display
command.set_input(line);
display(command);
}
}
void display(Lexer cmd_lexer) {
string file_name;
if (!parse_input(cmd_lexer, file_name)) {
error_return("Syntax error: display <filename>");
return;
}
ifstream ifs(file_name.c_str());
string error_msg;
if (ifs) {
if (!is_well_formed(ifs, error_msg)) {
error_return(error_msg);
} else {
ifs.clear(); // clear EOF flag
ifs.seekg(0, ios::beg); // go back to the very beginning
print_well_formed_file(ifs);
}
} else {
error_return("Can't open " + file_name + " for reading");
}
ifs.close();
}
Example of user input:
validate <file name>
display <file name>
exit
Upvotes: 0
Views: 214
Reputation: 3089
string name = fname.substr(1, fname.length() - 2);
Will throw this kind of an exception if fname
's length is <= 1 . I bet, this is the case. The simplest(not the best) solution is to skip such lines.
Upvotes: 1