If the name of a file to open is in the string variable name fileName, which of the following will correctly open the file for output? out_file.open(\"fileName\"); out_file.open(fileName.c_str()); fileName.open(outfile); out_file.open(fileName);

In C++, when you have the name of a file stored in a `std::string` variable (e.g., `fileName`), you need to pass it correctly to the `open` function of an output file stream (`std::ofstream`). Let's analyze each option: 1. **`out_file.open("fileName");`** - **Explanation:** This line attempts to open a file literally named `"fileName"`, not the value stored in the `fileName` variable. - **Correct:** ❌ 2. **`out_file.open(fileName.c_str());`** - **Explanation:** The `c_str()` function converts the `std::string` to a C-style string (`const char*`), which is a valid argument for the `open` function. - **Correct:** ✅ 3. **`fileName.open(outfile);`** - **Explanation:** This is incorrect because `fileName` is a `std::string`, not a file stream. You cannot call `open` on a `std::string`. - **Correct:** ❌ 4. **`out_file.open(fileName);`** - **Explanation:** Starting from C++11, the `open` function of `std::ofstream` has an overload that accepts a `std::string` directly. - **Correct:** ✅ **Conclusion:** The correct ways to open the file for output using the `fileName` variable are: - `out_file.open(fileName.c_str());` - `out_file.open(fileName);` Both of these options properly use the value stored in the `fileName` string to open the desired file. **Example Usage:** ```cpp #include #include int main() { std::string fileName = "output.txt"; std::ofstream out_file; // Both of these are correct out_file.open(fileName.c_str()); // or out_file.open(fileName); if (out_file.is_open()) { out_file << "Hello, World!"; out_file.close(); } return 0; } ```