How to remove punctuation characters from a char array? - c++

Question

My program prompts a user for a phrase to check if its a palindrome, then it's supposed to print out the phrase without capitalization or special characters like " ' , ? etc. My problem is erasing those characters. I've gotten my program to ignore them I'm asking how should I erase them? I made a comment where I think the statement should go. Example output should be: "Madam I'm Adam" to "madamimadam"
#include <iostream>
#include <string>
#include <cctype>
using namespace std;
int main()
{
//Variables and arrays
int const index = 80;
char Phrase[index];
char NewPhrase[index];
int i, j, k, l;
bool test = true;
//Prompt user for the phrase/word
cout << "Please enter a sentence to be tested as a palindrome: ";
cin.getline(Phrase, 80);
//Make everything lowercase, delete spaces, and copy that to a new array 'NewPhrase'
for(k = 0, l = 0; k <= strlen(Phrase); k++)
{
if(Phrase[k] != ' ')
{
NewPhrase[l] = tolower(Phrase[k]);
l++;
}
}
//cout << "The Phrase without punctuation/extra characters: " << newPhrase[l];
int length = strlen(NewPhrase); //Get the length of the phrase
for(i = 0, j = length-1; i < j; i++, j--)
{
if(test) //Test to see if the phrase is a palindrome
{
if(NewPhrase[i] == NewPhrase[j])
{;}
else
{
test = false;
}
}
else
break;
}
if(test)
{
cout << endl << "Phrase/Word is a Palindrome." << endl << endl;
cout << "The Palindrome is: " << NewPhrase << endl << endl;
}
else
cout << endl << "Phrase/Word is not a Palindrome." << endl << endl;
system("Pause");
return 0;
}

Answer 1

Modify this line:
if(Phrase[k] != ' ')
To be:
if((phrase[k] != ' ') && (ispunct(phrase[k]) == false))
This means that we check for spaces and punctuation at the same time.
Also, consider rewriting this:
if(NewPhrase[i] == NewPhrase[j])
{;}
else
{
test = false;
}
As this:
if(NewPhrase[i] != NewPhrase[j])
test = false;

Answer 2

Here's suggestion:
Use an std::string
Use std::ispunct to determine whether a character in the string is a punctuation mark
Use the erase-remove idiom to remove punctuation
That is one line of code (plus one extra line for a convenience lambda):
std::string phrase = .....;
auto isPunct = [](char c) { return std::ispunct(static_cast<unsigned char>(c)); }
phrase.erase(std::remove_if(phrase.begin(), phrase.end(), isPunct),
phrase.end());
Next, for turning into lower case, from my answer to this recent question, another one-liner:
std::transform(phrase.begin(), phrase.end(), phrase.begin(),
[](char c)
{ return std::tolower(static_cast<unsigned char>(c));});