用c++实现An object of class Bar contains n characters. By default, each character is a dot '.'

以下是用C++实现的类Bar，其默认包含n个字符，每个字符都是'.'。

#include <iostream>
#include <string>

class Bar {
public:
    Bar(int n) : m_chars(n, '.') {}
    std::string get_chars() const { return m_chars; }
private:
    std::string m_chars;
};

int main() {
    Bar b(5);
    std::cout << b.get_chars() << std::endl; // Output: .....
    return 0;
}

在上面的代码中，我们使用了std::string作为存储字符的容器，并使用类的构造函数初始化了这个容器。我们还提供了一个公共的方法get_chars()，以返回字符串。在主函数中，我们创建了一个名为b的对象，并打印出它包含的字符。

相关问题

Write a few lines of C++ code to show an example of an infinite loop. Explain why it is an infinite loop

Here's an example of an infinite loop in C++:

while(true){
    // code that never exits the loop
}

In this example, the `while` loop condition is always `true`, which means the loop will never exit. The loop will continue to execute the block of code inside the curly braces indefinitely.

This is an infinite loop because the loop condition is always true and there is no way for the loop to terminate on its own. The only way to exit the loop is to use a `break` statement or an external event (such as terminating the program).

用C++解决On an 8×8 grid of dots, a word consisting of lowercase Latin letters is written vertically in one column, from top to bottom. What is it? Input The input consists of multiple test cases. The first line of the input contains a single integer t (1≤t≤1000 ) — the number of test cases. Each test case consists of 8 lines, each containing 8 characters. Each character in the grid is either . (representing a dot) or a lowercase Latin letter (a –z ). The word lies entirely in a single column and is continuous from the beginning to the ending (without gaps). See the sample input for better understanding. Output For each test case, output a single line containing the word made up of lowercase Latin letters (a –z ) that is written vertically in one column from top to bottom. Example inputCopy 5 ........ ........ ........ ........ ...i.... ........ ........ ........ ........ .l...... .o...... .s...... .t...... ........ ........ ........ ........ ........ ........ ........ ......t. ......h. ......e. ........ ........ ........ ........ ........ .......g .......a .......m .......e a....... a....... a....... a....... a....... a....... a....... a....... outputCopy i lost the game aaaaaaaa

Sure! Here's a C++ code that solves the problem:

#include <iostream>
#include <vector>
#include <string>

using namespace std;

int main() {
    int t;
    cin >> t;

    while (t--) {
        vector<string> grid(8);
        for (int i = 0; i < 8; i++) {
            cin >> grid[i];
        }

        string word;
        for (int i = 0; i < 8; i++) {
            for (int j = 0; j < 8; j++) {
                if (grid[j][i] != '.') {
                    word += grid[j][i];
                }
            }
        }

        cout << word << endl;
    }

    return 0;
}

This code reads the number of test cases and then for each test case, it reads the grid of characters. It then iterates over each column and adds the non-dot characters to the word variable. Finally, it prints the word for each test case.

I hope this helps! Let me know if you have any further questions.