C ++从'char'到'const char *'的无效转换[-fpermissive]

Question

WARNING: Extremely limited knowledge of C++ and coding in general. Please refrain from advanced terminology.

for ( i = 0; i < answer.size(); ++i) {
            if (guess == answer.at(i)) {                            //to display correct letters in answerDisplay
                answerDisplay.replace( (2 * i), 1, answer.at(i) );
                correctGuesses += 1;
            }

Given: answerDisplay and answer are strings.

When I run my program there is a compile-time error at the third line of what I've posted saying:

invalid conversion from ‘char’ to ‘const char*’ [-fpermissive]

What's the problem? How can I fix it? All other posts with this error talked about pointer characters but I don't know what those are.

Answer 1

Pointer characters are they way plain strings are implemented in C and C++. In C++ you have the nice class std::string , but string literals are still array of characters. And arrays in C and C++ can be seen as pointers.

For example, "hello" is of type const char[6] (5 characters plus the ending NUL), but it can be trivially converted to const char * , and that in turn can be converted to std::string .

In line 3, the only relevant code is a call to the member function std::string::replace() . There are a lot of overrides of this function (different sets of parameters to be used), but the one the compiler is trying to use is this one:

string& replace (size_t pos,  size_t len,  const char* s);

As you can see, it takes two numbers and a const char * (an old-string/char-array). But you are passing as third parameter answer.at(i) that is of type char . Hence the error:

invalid conversion from ‘char’ to ‘const char*’

Solution? You can build a string from that char:

answerDisplay.replace( (2 * i), 1, std::string(1, answer.at(i))

Or you can get a substring of the original string instead of a plain character.

answerDisplay.replace( (2 * i), 1, answer.substr(i, 1))