Programming Forums > Application Development > C++

Haz · Dec 10th, 2005, 9:43 PM

'lo,

I tried to create a loop which was governed by a character:

    while(1) {
	cout << "Large (L) or Small (S) computation?: ";
	cin >> losc; 
	cin.get();
	cout << "\n\n";

	if( losc != ('L' || 'S') ) {
		cout << "Invalid choice! Try again...\n\n";
		}
	else {
		break;
		}
	}

As you can see, if lost does not equal L or S, then it prints invalid choice and the loop reiterates, yet it doesn't work that way, if I insert ANYTHING, it still show "Invalid choice", is there a problem with the way I arranged the statement?

Thanks,

Haz

Master · Dec 10th, 2005, 9:58 PM

why don't you make it simpler by doing this

(Toggle Plain Text)

    while(1) {
	cout << "Large (L) or Small (S) computation?: ";
	cin >> losc; 
	cin.get();
	cout << "\n\n";

	if( losc != 'L' || losc != 'S' ) {
		cout << "Invalid choice! Try again...\n\n";
		}
	else {
		break;
		}
	}

Kilo · Dec 10th, 2005, 10:12 PM

or you could do,

(Toggle Plain Text)

if(losc!='L' && losc!='S')

Jessehk · Dec 10th, 2005, 10:46 PM

or, best of all, use a switch statemnt:

(Toggle Plain Text)

bool done = true;

while(done) {
	cout << "(L)arge or (S)mall computation?: ";
	cin >> losc;

	cout << endl << endl;
	switch(toupper(losc)) {
		case L:
		case S:
                        done = false;
                        break;
			
		default:
			cout << "invalid choice! Try again...\n\n";
			break;
	}
}

make sure include the cctype header file for the toupper() function

UnKnown X · Dec 11th, 2005, 4:59 AM

Shouldn't that be:

(Toggle Plain Text)

case 'L':
case 'S':
    done = false;
    break;

?

grumpy · Dec 11th, 2005, 5:05 AM

I would dispute the claim that a switch statement is 'best", but it is certainly an alternative in this case.

metsfan · Dec 11th, 2005, 5:17 AM

this is a typical problem with c++; no typechecking! the binary operator "or" does not operate like this. the syntax for the binary operator is that it takes in two booleans and returns a boolean. you are giving this binary operator two characters; hence this will never work. anyway, lets look at the statement in question:

if( losc != ('L' || 'S') )

this is not logically equivalent to saying "if losc is not equal to L or S". it is actually logically equvaletn to "if losc is not equal to the binary operation (L or S). because the binary operator "or" returns a boolean, you are asking the evaluator if a boolean is not equal to a character. this is of course true. so thats why you get the error.

otehrs posted the solution, but that is the explaination for the problem

if you tried to program such a statement in a typesafe language such as SML or OCAML this would not even compile

grumpy · Dec 11th, 2005, 7:51 AM

The issue has nothing to do type safety. The issue is related to the meaning of particular expressions.

The expression 'L' || 'S' (which is equivalent to 'L' or 'S') has a particular meaning in C++ and, in this case, yields a boolean result of true (which, in C/C++, is equivalent to an integer with a value of 1). The net effect is that the expression losc != ('L' || 'S') is testing if losc != 1, rather than testing if losc is not equal to one of 'L' or 'S'.

The reason such an expression would not compile in SML or OACML is that the meaning of the expression would be different (in fact, invalid) in those languages.

metsfan · Dec 11th, 2005, 8:03 AM

i was unaware of that...thank you for pointing that out. is there any particular reason why this is advantagous in c++? it seems as if allowing this expression to exist can only cause bugs and problems...

grumpy · Dec 11th, 2005, 8:15 AM

It's a side effect of other features of the language. While it does make for some problems like the one in this thread, it also offers some advantage. The advantage of it is that it allows chaining of expressions, and shortcuts.

(Toggle Plain Text)

    int x, y;
    // initialise x and y to something valid
    int x_nonzero = (x != 0);
    int y_nonzero = (y != 0);

    int x_or_y_nonzero = x || y;   // simpler than (x != 0) || (y != 0)

One characteristic of characters in C/C++ is that they have an integral value, which is why the last line above works if x and y are of type char.

It may seem that x || y is more cryptic than (x != 0) || (y != 0), despite the fact they have identifical meaning. But, if you think about it, the more wordy expression also takes more effort to interpret.

Consider the difference if you have several integer variables, and you wanted to find out if at least one of them is non-zero

Dec 10th, 2005, 9:43 PM	#1
Haz Programmer Join Date: Feb 2005 Location: England Posts: 37 Rep Power: 0	Problem with characters in if statement 'lo, I tried to create a loop which was governed by a character: (Toggle Plain Text) while(1) { cout << "Large (L) or Small (S) computation?: "; cin >> losc; cin.get(); cout << "\n\n"; if( losc != ('L' \|\| 'S') ) { cout << "Invalid choice! Try again...\n\n"; } else { break; } } while(1) { cout << "Large (L) or Small (S) computation?: "; cin >> losc; cin.get(); cout << "\n\n"; if( losc != ('L' \|\| 'S') ) { cout << "Invalid choice! Try again...\n\n"; } else { break; } } As you can see, if lost does not equal L or S, then it prints invalid choice and the loop reiterates, yet it doesn't work that way, if I insert ANYTHING, it still show "Invalid choice", is there a problem with the way I arranged the statement? Thanks, Haz __________________ MY BLOG :eek: AND WEBSITE :eek:

Dec 10th, 2005, 9:58 PM	#2
Master Programmer Join Date: Sep 2005 Location: GA Posts: 99 Rep Power: 4	why don't you make it simpler by doing this (Toggle Plain Text) while(1) { cout << "Large (L) or Small (S) computation?: "; cin >> losc; cin.get(); cout << "\n\n"; if( losc != 'L' \|\| losc != 'S' ) { cout << "Invalid choice! Try again...\n\n"; } else { break; } } while(1) { cout << "Large (L) or Small (S) computation?: "; cin >> losc; cin.get(); cout << "\n\n"; if( losc != 'L' \|\| losc != 'S' ) { cout << "Invalid choice! Try again...\n\n"; } else { break; } }

Dec 10th, 2005, 10:12 PM	#3
Kilo Expert Programmer Join Date: Nov 2005 Location: In Pink Clam? Posts: 542 Rep Power: 0	or you could do, (Toggle Plain Text) if(losc!='L' && losc!='S') if(losc!='L' && losc!='S') __________________ "When in Rome, Do as the Romans Do" "Beauty is in the eye of the BEER holder" "Save your breath your going to need it for your blow up doll later" SearchLores.org

Dec 10th, 2005, 10:46 PM	#4
Jessehk The Oblivious One Join Date: May 2005 Location: Ontario, Canada Posts: 648 Rep Power: 4	or, best of all, use a switch statemnt: (Toggle Plain Text) bool done = true; while(done) { cout << "(L)arge or (S)mall computation?: "; cin >> losc; cout << endl << endl; switch(toupper(losc)) { case L: case S: done = false; break; default: cout << "invalid choice! Try again...\n\n"; break; } } bool done = true; while(done) { cout << "(L)arge or (S)mall computation?: "; cin >> losc; cout << endl << endl; switch(toupper(losc)) { case L: case S: done = false; break; default: cout << "invalid choice! Try again...\n\n"; break; } } make sure include the cctype header file for the toupper() function __________________ Dr. Zoidberg: [ecstatic] I'm going to a movie... with FRIENDS!

Dec 11th, 2005, 4:59 AM	#5
UnKnown X Hobbyist Programmer Join Date: Dec 2005 Location: Sandvika, Norway Posts: 114 Rep Power: 0	Shouldn't that be: (Toggle Plain Text) case 'L': case 'S': done = false; break; case 'L': case 'S': done = false; break; ?

Dec 11th, 2005, 5:05 AM	#6
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,260 Rep Power: 5	I would dispute the claim that a switch statement is 'best", but it is certainly an alternative in this case.

Dec 11th, 2005, 5:17 AM	#7
metsfan Programmer Join Date: Dec 2005 Posts: 57 Rep Power: 0	this is a typical problem with c++; no typechecking! the binary operator "or" does not operate like this. the syntax for the binary operator is that it takes in two booleans and returns a boolean. you are giving this binary operator two characters; hence this will never work. anyway, lets look at the statement in question: if( losc != ('L' \|\| 'S') ) this is not logically equivalent to saying "if losc is not equal to L or S". it is actually logically equvaletn to "if losc is not equal to the binary operation (L or S). because the binary operator "or" returns a boolean, you are asking the evaluator if a boolean is not equal to a character. this is of course true. so thats why you get the error. otehrs posted the solution, but that is the explaination for the problem if you tried to program such a statement in a typesafe language such as SML or OCAML this would not even compile

Dec 11th, 2005, 7:51 AM	#8
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,260 Rep Power: 5	The issue has nothing to do type safety. The issue is related to the meaning of particular expressions. The expression 'L' \|\| 'S' (which is equivalent to 'L' or 'S') has a particular meaning in C++ and, in this case, yields a boolean result of true (which, in C/C++, is equivalent to an integer with a value of 1). The net effect is that the expression losc != ('L' \|\| 'S') is testing if losc != 1, rather than testing if losc is not equal to one of 'L' or 'S'. The reason such an expression would not compile in SML or OACML is that the meaning of the expression would be different (in fact, invalid) in those languages.

Dec 11th, 2005, 8:03 AM	#9
metsfan Programmer Join Date: Dec 2005 Posts: 57 Rep Power: 0	i was unaware of that...thank you for pointing that out. is there any particular reason why this is advantagous in c++? it seems as if allowing this expression to exist can only cause bugs and problems...

Dec 11th, 2005, 8:15 AM	#10
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,260 Rep Power: 5	It's a side effect of other features of the language. While it does make for some problems like the one in this thread, it also offers some advantage. The advantage of it is that it allows chaining of expressions, and shortcuts. (Toggle Plain Text) int x, y; // initialise x and y to something valid int x_nonzero = (x != 0); int y_nonzero = (y != 0); int x_or_y_nonzero = x \|\| y; // simpler than (x != 0) \|\| (y != 0) int x, y; // initialise x and y to something valid int x_nonzero = (x != 0); int y_nonzero = (y != 0); int x_or_y_nonzero = x \|\| y; // simpler than (x != 0) \|\| (y != 0) One characteristic of characters in C/C++ is that they have an integral value, which is why the last line above works if x and y are of type char. It may seem that x \|\| y is more cryptic than (x != 0) \|\| (y != 0), despite the fact they have identifical meaning. But, if you think about it, the more wordy expression also takes more effort to interpret. Consider the difference if you have several integer variables, and you wanted to find out if at least one of them is non-zero

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Programming Forums > Application Development > C++

Problem with characters in if statement