错误：在抛出 'std::out_of_range 的实例后调用终止 [重复]

Question

当我输入这段代码时

#include <iostream>
#include <string>

class binary
{
    std::string s;

public:
    void read();
    void check_format();
};
void binary::read()
{
    std::cout << "Enter a number\n";
    std::cin >> s;
}
void binary ::check_format()
{
    for (int i = 1; i <= s.length(); i++)
    {
        if (s.at(i) != '0' && s.at(i) != '1')
        {
            std::cout << "Incorrect format\n";
            exit(0);
        }
    }
};
int main()
{
    binary num;
    num.read();
    num.check_format();
    return 0;
}

对于其中没有“1”和“0”的数字，例如 44，我得到了正确的输出，但是对于其中带有“1”和“0”的数字，我得到了这个错误

terminate called after throwing an instance of 'std::out_of_range'
  what():  basic_string::at: __n (which is 2) >= this->size() (which is 2)

请帮忙解决这个问题。

Answer 1

C++ 字符串索引是从零开始的。这意味着如果字符串的大小为n。它的索引是从0 到n - 1。

例如：

#include <iostream>
#include <string>

int main()
{
    std::string s {"Hello World!"} // s has a size of 12.
    std::cout << s.size() << '\n'; // as shown
    std::cout << s.at(0); << '\n' // print the first character, 'H'
    std::cout << s.at(11); << '\n' // print the last character, '!'
}

但是您正在从1 循环到n。当i == n、s.at(i) 越界时，会抛出std::out_of_range 错误

将循环更改为：

void binary::check_format()
{
    for (int i = 0; i < s.length(); i++)
    {
        if (s[i] != '0' && s[i] != '1') // at() does bounds checking, [] is faster
        {
            std::cout << "Incorrect format\n";
            exit(0);
        }
    }
}

或者更好：

void binary::check_format()
{
    for(const auto &i : s)
    {
        if (i != '0' && i != '1')
        {
            std::cout << "Incorrect format\n";
            exit(0);
        }
    }
}