Edit: Just noticing that this answer probably goes far beyond the scope of the task you have been given (just filling in some skeleton that has been provided by your teacher).
I'll still leave it here to illustrate what could be done with arbitrary input. Maybe you want to experiment a little further than what you have been asked...
bool within_width(...)
Pretty simple: string.length() <= max – just wait a second, you need to consider asterisks and spaces at beginning and end of output, so: max - 4
But you can do better, you can split the string, best at word boundaries. That's a bit difficult more difficult, though:
std::vector<std::string> lines;
// we'll be starting with an initially empty line:
auto lineBegin = text.begin();
auto lineEnd = text.begin();
for(auto i = text.begin(); i != text.end(); ++)
// stop condition empty: we'll stop from inside the loop...
{
// ok, we need to find next whitespace...
// we might try using text.find_first_of("..."), but then we
// need to know any whitespace characters ourselves, so I personally
// would rather iterate manually and use isspace function to determine;
// advantage: we can do other checks at the same time, too
auto distance = std::distance(lineBegin, i);
if(std::distance(lineBegin, i) > maxLineLength)
{
if(lineEnd == lineBegin)
{
// OK, now we have a problem: the word itself is too long
// decide yourself, do you want to cut the word somewhere in the
// middle (you even might implement syllable division...)
// or just refuse to print (i. e. throw an exception you catch
// elsewhere) - decide yourself...
}
else
{
lines.emplace_back(lineBegin, lineEnd);
lineBegin = lineEnd; // start next line...
}
}
// OK, now handle current character appropriately
// note: no else: we need to handle the character in ANY case,
// if we terminated the previous line or not
if(std::isspace(static_cast<unsigned char>(*i)))
{
lineEnd = i;
}
// otherwise, we're inside a word and just go on
}
// last line hasn't been added!
lines.emplace_back(lineBegin, lineEnd);
Now you can calculate maximum length over all the strings contained. Best: Do this right when adding a new line to the vector, then you don't need a separate loop...
You might have noticed that I didn't remove whitespace at the end of the strings, so you wouldn't need to add you own one, apart, possibly, from the very last string (so you might add a lines.back() += ' ';).
The ugly part, so far, is that I left multiple subsequent whitespace. Best is removing before splitting into lines, but be aware that you need to leave at least one. So:
auto end = text.begin();
bool isInWord = false; // will remove leading whitespace, if there is
for(auto c : text)
{
if(std::isspace(static_cast<unsigned char>(c)))
{
if(isInWord)
{
*end++ = ' '; // add a single space
isInWord = false;
}
}
else
{
*end++ = c;
isInWord = true;
}
}
This would have moved all words towards the beginning of the string, but we yet to drop the surplus part of the string yet contained:
text.erase(end, text.end());
Fine, the rest is pretty simple:
- iterate over maximum length, printing a single asterisk in every loop
- iterate over all of your strings in the vector:
std::cout << "* " << line << "*\n";
- repeat the initial loop to print second line of asterisks
Finally: You introduced a fix line limit of 80 characters. If console is larger, you just won't be using the entire available width, which yet might be acceptable, if it is smaller, you will get lines broken at the wrong places.
You now could (but that's optional) try to detect the width of the console – which has been asked before, so I won't go any deeper into.
Final note: The code presented above is untested, so no guarantee to be bugfree!
