Ignoring puncuations in input stream

hello, this is my first post, nice to meet everyone

I am taking input from the keyboard, and I want to ignore
whitespace, punctuation and the case of letters, i.e. 'a' = 'A'

I figured out the whitespace, but I can't figure out letter case and
punctuation. Here is my code (my main function, which uses a stack
and queue objects). The program is meant to determine whether or
not a statement is a palindrome.

Code:

---

#include "stack.h"
#include "queue.h"
#include <iostream>

using namespace std;

int main()
{
    bool isPalindrome = true;
    char ch, ch1, ch2;
    StackType s;
    QueueType q;

    cout << "Enter a statement: ";

    cin >> ch;
    while(!s.IsFull() && ch != '\n')
    {
        s.Push(ch);
        q.Enqueue(ch);
        if(cin.peek() != '\n')
            cin >> ch;
        else
            cin.get(ch);
    }

    while(isPalindrome && !s.IsEmpty())
    {
        q.Dequeue(ch1);
        ch2 = s.Top();
        s.Pop();
        isPalindrome = ch1 == ch2;
    }

    cout << endl;

    if(isPalindrome)
        cout << "Entered statement is a Palindrome\n";
    else
        cout << "Entered statement is not a Palindrome\n";
    return 0;
}

---

thanks!

Some useful things from working a bit with assembly:

• To guarantee a char will be lowercase you can OR it with 0x20.
• To guarantee a char will be uppercase, you can AND with 0xDF.

You can't handle case and punctuation with the input stream, you have to read in the character first and then handle them.

You can use tolower and/or toupper to change the case of a character. You can use ispunct to check to see if the character is punctuation and skip it if it is.

>> Some useful things from working a bit with assembly
Those might be useful for a code obfuscation contest, but for a real program I would use tolower and toupper.

thank you, everything works, but for the tolower function,
it returns the ASCII value for the char. Is there any way I can
change that value back to the char instead of an ASCII int?

thanks again

Simply assign it to a char variable and it will be converted correctly.

If you want to avoid a warning, just use a cast. The C++ style cast is:

Code:

---

ch = static_cast<char>(::tolower(ch));

---

What's the difference between "::tolower" and "tolower"?
Are they located in different libraries or namespaces?

I just tried them out and they give different outputs for the following lines of code:

Code:

---

char c;
...
cout << hex << tolower(c) << endl;
cout << hex << ::tolower(c) << endl;

---

The upper output is hexadecimal, the lower is plain decimal. Why is that?

Quote:

---

What's the difference between "::tolower" and "tolower"?

---

Depends on how your header files and namespace stuff is set up:

Code:

---

#include <ctype.h>

// both are the same because tolower() is in the global namespace
cout << hex << tolower(c) << endl;
cout << hex << ::tolower(c) << endl;

---

Code:

---

#include <cctype>

// both are wrong because tolower() is in the std namespace; might not compile
cout << hex << tolower(c) << endl;
cout << hex << ::tolower(c) << endl;

---

Code:

---

#include <cctype>

using namespace std;

// this one is right because it's defaulted to the std namespace
cout << hex << tolower(c) << endl;

// this one is not because tolower() is in the std namespace; might not compile
cout << hex << ::tolower(c) << endl;

---

What's the difference being "defaulted" to the std namespace and being "in" the namespace?

In other words, what does "::" do, when it doesn't have a scope modifier (as in class A::B)?

Quote:

---

What's the difference being "defaulted" to the std namespace and being "in" the namespace?

---

Nothing. "defaulted" means that you said 'using namespace std' or 'using std::tolower' and from then on every tolower() becomes std::tolower() by default.

Quote:

---

In other words, what does "::" do, when it doesn't have a scope modifier (as in class A::B)?

---

It's explicitly saying that the thing is in the global namespace.