help with a exercise of Accelerated C++

Hello,

I have this exercise :

The framing programm writes the mostly blank lines that seperates the borders from the greeting one character at the time. Change the programm so it writes all the spaces at one time.

So i thought this solution will work.

Code:

#include <iostream>
#include <string>

using namespace std;


int main()
{
	// ask for the person's name
	cout << "Please enter your first name: ";

	// read the name
	string name;
	cin >> name;

	// build the message that we intend to write
	const string greeting = "Hello, " + name + "!";

	// the number of blanks surrounding the greeting
	int pad ;
	cout << "hoeveel spaties tussen de rand en de tekst :";
	cin >> pad ;
	const int pad2 = 2 ;

	// the number of rows and columns to write
	const int rows = pad2 * 2 + 3;
	const string::size_type cols = greeting.size() + pad * 2 + 2;

	// write a blank line to separate the output from the input
	cout << endl;

	// write `rows' rows of output
	// invariant: we have written `r' rows so far
	for (int r = 0; r != rows; ++r) {

		string::size_type c = 0;

		// invariant: we have written `c' characters so far in the current row
		while (c != cols) {

			// is it time to write the greeting?
			if (r == pad2 + 1 && c == pad + 1) {
				cout << greeting;
				c += greeting.size();
			} else {

				// are we on the border?
				if (r == 0 || r == rows - 1 ||
				    c == 0 || c == cols - 1)
					cout << "*";
				else
					string z(pad," ");
					cout << z ;
                c=+pad ;
			}
		}

		cout << endl;
	}

	return 0;
}

But I see now that z is not declared in this scope.

What is my thinking error ?

Roelof

Within your loop, you have this

Code:

                                if (r == 0 || r == rows - 1 ||
				    c == 0 || c == cols - 1)
					cout << "*";
				else
					string z(pad," ");
					cout << z ;

The last line is not executed as part of the "else" clause, even though - from your indenting - you apparently think it is.

So, as part of the "else" a string named z is created, and immediately destroyed before you attempt to print it.

Oke,

I changed that part to :

Code:

else {
					string z(pad," ");
					cout << z ;
}

But now I get this message : C:\Users\wobben\Desktop\borland-source\chapter02\frame.cpp|53|error: invalid conversion from 'const char*' to 'char'|

Roelof

Then look at the types of your variables. You're doing something where the a pointer is expected, and supplying a char.

If you look up the constructors of the string type (or, more accurately, basic_string) there are none that accept an integer as the first argument and a pointer as the second argument. The string literal " " is being implicitly interpreted as a pointer to char.

Oke,

I thought this could work by this text :

string z(n, c)
Defines z as a type string that initially contains n copies of the caracter c.
Here C must be a char, nor a string or a string literal.

Roelof

Edit : I found it . If I change it to z(pad1, char(32)) it works.'

Hello,

I have not this code :

Code:

#include <iostream>
#include <string>

using namespace std;


int main()
{
	// ask for the person's name
	cout << "Please enter your first name: ";

	// read the name
	string name;
	cin >> name;

	// build the message that we intend to write
	const string greeting = "Hello, " + name + "!";

	// the number of blanks surrounding the greeting
	int pad ;
	cout << "hoeveel spaties tussen de rand en de tekst :";
	cin >> pad ;
	const int pad2 = 2 ;

	// the number of rows and columns to write
	const int rows = pad2 * 2 + 3;
	const string::size_type cols = greeting.size() + pad * 2 + 2;


	// write a blank line to separate the output from the input
	cout << endl;

	// write `rows' rows of output
	// invariant: we have written `r' rows so far
	for (int r = 0; r != rows; ++r) {

		string::size_type c = 0;

		// invariant: we have written `c' characters so far in the current row
		while (c != cols) {

			// is it time to write the greeting?
			if (r == pad2 + 1 && c == pad + 1) {
				cout << greeting;
				c += greeting.size();
			} else {

				// are we on the border?
				if (r == 0 || r == rows - 1 ||
				    c == 0 || c == cols - 1){
					cout << "*";
					c++ ; }
				else
				  if ( r==pad2-1 || r==pad2 ||
                       r==rows-1 || r==rows-2) // is this a blank line with only * and blanks ?
                     {
                         string z(cols-2, char(32));
                         cout << z ;
                         c+= cols -3 ;
                     }
                    else   // is this the line for the greeting. If so print the padding
				{
					string z(pad,char(32));
					cout << z ;
                    c +=pad ;
                    }
			}
		}

		cout << endl;
	}

	return 0;
}

But now I see only the blank lines.

Roelof

try using:
cout << greeting.c_str()

enjoy,
r

Oke,

I don't think that's the problem.
I have tested further and I think c+=-3 is the problem.
If I change it then on different numbers I see the problem.

I have now this : c += cols - pad2 ; and then it's going wrong if de padding is bigger then 3.

Roelof

[EDIT] This is incorrect. Evidently not enough coffee at this point in the day

Think:

Code:

			if (r == pad2 + 1 && c == pad + 1) {
				cout << greeting;
				c += greeting.size();
			} else {

				// are we on the border?
				if (r == 0 || r == rows - 1 ||
				    c == 0 || c == cols - 1){
					cout << "*";
					c++ ; }
				else
				  if ( r==pad2-1 || r==pad2 ||

You only have two real possibilities for each iteration, the first if and the first else, because the else has no condition. That means if the if is not true, the else will happen. So the other else/else if's you have can never happen.

Here's how an else-if sequence works:

Code:

if (condition1) ...
else if (condition2) ...
else if (condition3) ...
else if (condition4) ....
else {  // if none of the above conditions are met

What do you think would happen if I swapped line 5 for line 2 or line 3 here?

Do you still have
c += pad
inside the "else"-clause?

In your first code-snippet it wasn't inside...

r.

hello,

I don't follow your suggestion.

Code:

if (r == pad2 + 1 && c == pad + 1) {
				cout << greeting;
				c += greeting.size();
			} else {

				// are we on the border?
				if (r == 0 || r == rows - 1 ||
				    c == 0 || c == cols - 1){
					cout << "*";
					c++ ; }
				else
				  if ( r==pad2-1 || r==pad2 ||

If (r==pad2 and on is False then if (r=0) is carried out. If this is False for me if (r==pad2-1) is carried out.

Or im a wrong here.

Roelof

Hello,

I'm going to start all over again.
This is a big mess I believe.

Roelof

maybe you'd try something like this:

Code:

	std::vector< std::string *> screen;
	std::string vertical_line; vertical_line.resize(cols);
	std::string empty_line;	empty_line.resize(cols);
	std::string greet_line;	greet_line.resize(cols);

	std::fill( vertical_line.begin(), vertical_line.end(), '*' );
	std::fill( empty_line.begin(), empty_line.end(), ' ' );
	
	empty_line[0] = '*'; empty_line[ empty_line.size() - 1 ] = '*'; 

	greet_line.assign( empty_line );
	greet_line.replace( pad, greeting.size() , greeting );

	screen.push_back( &vertical_line );
	for (int i=0; i!=pad2; ++i)
	{
		screen.push_back ( &empty_line );
	}
	screen.push_back( &greet_line );
	for (int i=0; i!=pad2; ++i)
	{
		screen.push_back ( &empty_line );
	}
	screen.push_back( &vertical_line );

	for ( std::vector<std::string*>::const_iterator it = screen.begin(); it!=screen.end(); ++it )
	{
		cout << (**it) << endl;
	}

Have fun!

r.

Originally Posted by roelof

Or im a wrong here.

Roelof

No, you're not wrong. Sorry about, I must have been writing with my eyes closed.

oke,

I have one problem.

I trying to solve this and try to make the problem visible.

If I not mistaken.

If a user wants 2 lines between the border above and the greeting then rule nr. 2 is the rule with a * as start , then a lot of blanks and then a "*"

If a user wants 3 lines between the border and the greeting the rule nr. 2 and 3 are the rule with * , a lot a blanks and a *

So the nr. of rules with a *, a lot of spaces and a * are between 2 and the number the user inputs.

So i thought this would work.

Code:

for (int r = 0; r != rows; ++r) {

		string::size_type c = 0;

		// invariant: we have written `c' characters so far in the current row
		while (c != cols) {

			// is it time to write the greeting?
			if (r == rules + 1 && c == pad + 1) {
				cout << greeting;
				c += greeting.size();
			} else {

				// are we on the border?
				if (r == 0 || r == rows - 1 ||
				    c == 0 || c == cols - 1)
				    {
				       cout << "*";
                        c++ ;
				    }
				else
				    {
                                                                       for (teller=2; teller=rules; teller++
                                                                       { //write the spaces }
				    }


			}
		}

		cout << endl;
	}

	return 0;

But it don't work.
Where am i thinking wrong.

rvr : your solution can be right but you use things that I have to learn in chapters to come.
So no solution for me

Roelof