String literals as char arrays

Question

Is it considered bad practice to pass string literals as char*, when an array and not a c-string is expected by the function? From my code:

char ConsoleIO::GetNextChar(string prompt, char* choices, int numChoices) {
 // Loop until valid input, or 10 times.
 for (int i = 0; i < 10; i++) {
 Write(prompt);
 char* buffer = new char[200];
 cin.getline(buffer, 200);
 char choice = tolower(buffer[0]);
 if (in(choice, choices, numChoices)) {
 return choice;
 }
 WriteLine("Invalid choice.");
 }
 // Just return the first choice as the default.
 return choices[0];
}
// In main:
char userChoice = ConsoleIO::GetNextChar("Make a selection: ", "234", 3);

It compiles, but is it bad practice?

• nwp: How does this even compile? "234" is a const char[] which cannot be converted to a char *. You should at least get a warning.
• Java Matrix: Nope, I don't even get a warning on this. My guess is that since it's not by-ref the constness can be ignored, and char[] and char* are really the same thing in c++, it allows me to pass a const char[] as a char*.

Answer 1

When a C or C-related programmer sees char* they're going to naturally think of passing in a C-string. In my opinion, it's absolutely fine to do that. If you want to invoke an array of characters in C, I'd do it like this:

char ConsoleIO::GetNextChar(string prompt, char[] choices, int numChoices)

But, you're in C++ so you should use a std::vector<char> and then don't pass in numChoices, like this:

char ConsoleIO::GetNextChar(string prompt, std::vector<char> const& choices)

Also, you have a bug. You aren't ever deleteing buffer, which means it's leaking.

• Java Matrix: I appreciate the input, especially on the memory leak. Thanks!
• R Sahu: I suggest use of std::vector<char> const& choices.
• user1118321: @RSahu Excellent point! I'll update it.