如何将一个字符串复制到另一个字符串？

Question

I am writing a function that removes vowels from a string (users name), and will then pass that function into another function that prints out the name without vowels. In order to copy the input[i] to the outputNoVowels , would this be the correct way of doing it?

string removeVowels(string input) {
    string outputNoVowels;
    
    for (int i = 0; i < input.length() - 1; i++) {
        switch (input[i]) {
        case 'a':
            break;
        case 'e':
            break;
        case 'i':
            break;
        case 'o':
            break;
        case 'u':
            break;
        case 'A':
            break;
        case 'E':
            break;
        case 'I':
            break;
        case 'O':
            break;
        case 'U':
            break;
        default:
            outputNoVowels = input[i];

        }
    }
    return outputNoVowels;
}

How come, when the program runs, it only prints out one letter in the string?

Provided Algorithm:

This is what is printing in the console window:

---

UPDATE : how would I do the same for char[] strings?

void removeVowels(const char input[], char output[]) {
    for (int i = 0; i < strlen(input) - 1; i++) {
        switch (input[i]) {
        case 'a':
            break;
        case 'e':
            break;
        case 'i':
            break;
        case 'o':
            break;
        case 'u':
            break;
        case 'A':
            break;
        case 'E':
            break;
        case 'I':
            break;
        case 'O':
            break;
        case 'U':
            break;
        default:
            strcpy(output, input);
            strcat(output, input);
        }
    }
}

Provided Algorithm:

Answer 1

You are getting 1 letter in the output because you are assigning each letter to outputNoVowels via its = assignment operator, wiping out its current content each time, so there is at most only 1 letter in outputNoVowels upon return .

You need to instead append each letter to outputNoVowels , via its += operator (or its push_back() or append() method), thus preserving any existing content.

Also, your loop is ignoring the last character in the input string. There is a typo in the instructions.

Try this instead:

string removeVowels(const string &input) {
    string outputNoVowels;
    const size_t len = input.length();
    for (size_t i = 0; i < len; ++i) {
        switch (input[i]) {
            case 'a':
            case 'e':
            case 'i':
            case 'o':
            case 'u':
            case 'A':
            case 'E':
            case 'I':
            case 'O':
            case 'U':
                break;
            default:
                outputNoVowels += input[i];
                // or: outputNoVowels.push_back(input[i]);
                // or: outputNoVowels.append(1, input[i]);
                break;
        }
    }
    return outputNoVowels;
}

That being said, you can eliminate the switch altogether by using the standard std::copy_if() algorithm instead (with std::back_inserter as the destination), eg:

string removeVowels(const string &input) {    
    static const string vowels = "aeiouAEIOU";
    string outputNoVowels;
    copy_if(input.begin(), input.end(),
        back_inserter(outputNoVowels),
        [&](char ch){ return vowels.find(ch) == string::npos; }
    );
    return outputNoVowels;
}

Or, by using the std::remove_if() algorithm, eg:

string removeVowels(string input) {    
    static const string vowels = "aeiouAEIOU";
    input.erase(
        remove_if(input.begin(), input.end(),
            [&](char ch){ return vowels.find(ch) != string::npos; }
        ),
        input.end()
    );
    return input;
}

Or, by using the std::erase_if() algorithm (C++20 and later), eg:

string removeVowels(string input) {    
    static const string vowels = "aeiouAEIOU";
    erase_if(input,
        [&](char ch){ return vowels.find(ch) != string::npos; }
    );
    return input;
}

---

UPDATE : in your char[] version, you have the opposite problem. You are appending too many letters to the output string. For each letter you want to keep, you are assigning the entire input string to the output, and then appending the entire input string to the end, rather than just the 1 letter.

You can use techniques similar to above to solve the char[] version, eg:

void removeVowels(const char input[], char output[]) {
    int idx = 0;
    const int len = strlen(input);
    for (int i = 0; i < len; ++i) {
        switch (input[i]) {
            case 'a':
            case 'e':
            case 'i':
            case 'o':
            case 'u':
            case 'A':
            case 'E':
            case 'I':
            case 'O':
            case 'U':
                break;
            default:
                output[idx++] = input[i];
                break;
        }
    }
    output[idx] = '\0';
}

Or, using the std::copy_if() algorithm:

void removeVowels(const char input[], char output[]) {    
    static const char[] vowels = "aeiouAEIOU";
    static const char*  vowels_end = vowels + strlen(vowels);
    char* output_end = copy_if(input, input + strlen(input),
        output,
        [&](char ch){ return find(vowels, vowels_end, ch) == vowels_end; }
    );
    *output_end = '\0';
}