Thread: First Program

you can use the newline character \n to create a new line.

Well here is a look at the corrected code.

Code:

#include <iostream>
using namespace std;

int main()
{	
	int convert;
	char letter;

	cout << "Enter a Character: ";
	cin >> letter; 
	cin.ignore();      // Takes the \n character from the stream
	convert = (int) letter;	
	cout << "\n" << letter << " in ASCII is: " << convert;  // Newline at the beginning
	cin.get();                                              // of this output
	return 0;
}

The \n character is good where I put it, keep in mind this could be replaced with an endl if you wish, though with more output to come, I don't really see the point. The reason cin.get() wasn't working for you is because you still had a newline character in the input stream and it was picking that up and closing. I added a cin.ignore() after the input to take that away for you. You could also have used 2 cin.get() calls at the end if you wish.

This should work fine. I wish you explained in detail why \n wasn't working for you, I'm just assuming you didn't have it in the right place, though.

thankyou so much for your replies, and thankyou SlyMaelstrom for re-writing the program, it now works just the way i wanted it to

Please could you explain a bit more about why cin.ignore() was required and cin.get() didn't work because i dont understand where i had the newline character in the input stream or what it causes

I would also like to know about how program code is read by the computer (program\source code flow) so that i can understand how to write code in the correct order because you were right about me not putting the "\n" character in the right place.

Just a quick note:

Code:

convert = static_cast<int>( letter );

is usually used with C++, whereas ( int )letter is a C-Style cast.

- SirCrono6

Please could you explain a bit more about why cin.ignore() was required and cin.get() didn't work because i dont understand where i had the newline character in the input stream or what it causes

A line like this:

cin >>num

waits for the user to enter some input. Then the user can type something in and hit Return. However, when the user hits Return, an invisible \n or "newline" character is entered after the input ('\n' is considered a single character). So, if the user types in 10 and hits Return, the input actually looks like this:

10\n

Now, let's examine how that input is handled. cin>> is programmed to skip all leading whitespace(e.g. spaces, tabs, and newlines), read in data, and stop reading in data when the first whitespace character is encountered. And, very importantly, the terminating whitespace character is not read in. So, with this input:

10\n

1) cin>> skips any leading whitespace(there's no whitespace in front of 10)
2) cin>> reads in 10, and
3) cin>> stops reading when it encounters the whitespace character \n
4) cin>> does not read in the \n, and leaves the \n in the input stream.

Sometime thereafter, if you have a line like this:

cin.get();

that is an instruction to read in more input. Specifically, cin.get() says to read in one character. In this case, there is still a \n left in the input stream, and as far as cin.get() is concerned that is good input, so it doesn't need to wait for the user to enter data--cin.get() goes ahead and reads in the \n. So, execution continues on without waiting for the user to enter something.

The line:

cin.ignore(n);

will cause n characters in the input stream to be skipped. Since 1 is the default value, you can skip one character by writing:

cin.ignore();

And what about program\source code flow? how does C++ read the source code when it is compiled

Sequentially, from the top of the file to the bottom of the file and from left to right.

That code should work fine, Swatto. Are you sure it doesn't create a newline for you?