C++ Basic while loop with keystroke to exit

Question

I am new to programming and this is my second assignment. It is supposed to accept input from user until they enter the | to exit the program. What have I done wrong?

int main()

    int i = 0;
    char a = 0;
        while ( a != "|" ){  //has also told me this is an
                             // invalid operator
    int numeric;
    cout << "Give character: ";
    cin >> a ;
    cout << "Its ascii value is: " << (int) a << endl;
    ++i;

    }
}

Here is the error:

2   IntelliSense: operand types are incompatible ("char" and "const char *")

Enclosing stuff inside "s results in a string literal, which is of type const char *. This is a pointer to an array of characters. You can't compare a single char to an array of characters. Use |. — BWG
– BWG, Commented Feb 12, 2015 at 1:05
@BWG: "string literal, which is of type const char *" - a string literal is actually an array of const char, not a const char*; the confusion's not uncommon given the standard conversion to const char* kicks in quite readily. The difference is significant: e.g. for sizeof and when passing it to template <size_t N> void f(char (&s)[N]);. — Tony Delroy
– Tony Delroy, Commented Feb 12, 2015 at 1:36

edtheprogrammerguy · Accepted Answer · 2015-02-12 01:10:05Z

3

Don't use "|". Put '|' instead. You want to compare a char to a char, NOT compare a char to a char*.

edited Feb 12, 2015 at 1:10

answered Feb 12, 2015 at 1:04

edtheprogrammerguy

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1 Comment

Tony Delroy Over a year ago

"NOT compare a char to a char*" - my comment to BWG under the question is relevant here too.

BigTailWolf · Accepted Answer · 2015-02-12 01:06:08Z

1

You shall use while ( a != '|' ) instead

'|' is a character and "|" is a string which has only 1 character

answered Feb 12, 2015 at 1:06

BigTailWolf

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2 Comments

Tony Delroy Over a year ago

""|" is a string which has only 1 character" - that's debatable... "|" is a string literal composed of '|' and NUL / '\0', which is '|' with a terminating sentinel in ASCIIZ encoding....

BigTailWolf Over a year ago

@TonyD You are right. LOL. Let's say "one valid character"

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