Question

如果x> INT_MAX或x> INT_MIN，该函数将返回0 ...或这就是我正在尝试做的事情：）

在我的测试用例中，我传入的是INT_MAX +1 ... 2147483648 ...的值，以引入整数溢出来查看程序如何处理它。

我逐步完成...我的IDE调试器说，溢出后该值立即变为-2147483648，并且由于某种原因，程序将执行以下两个语句之外的操作：

如果（x> INT_MAX）如果（x

并保持崩溃int revInt = std :: stoi（strNum）; 说超出范围

一定很简单，但这让我感到困惑。给定x> INT_MAX，为什么程序在返回该std :: stoi（）之前不返回？任何帮助表示赞赏。谢谢！功能和测试平台的完整列表如下：（对不起，代码插入格式设置有误。）

#include <iostream> 
#include <algorithm>
#include <string> //using namespace std;
class Solution {
public: int reverse(int x)
{

    // check special cases for int and set flags:

    // is x > max int, need to return 0 now
    if(x > INT_MAX)
        return 0;

    // is x < min int, need to return 0 now
    if(x < INT_MIN)
        return 0;

    // is x < 0, need negative sign handled at end

    // does x end with 0, need to not start new int with 0 if it's ploy numeric and the functions used handle that for us

    // do conversion, reversal, output:

    // convert int to string
    std::string strNum = std::to_string(x);

    // reverse string
    std::reverse(strNum.begin(), strNum.end());

    // convert reversed string to int
    int revInt = std::stoi(strNum);

    // multiply by -1 if x was negative
    if (x < 0)
        revInt = revInt * -1;

    // output reversed integer
    return revInt;

}
};

主要：

#include <iostream>

int main(int argc, const char * argv[]) {


    // test cases
    // instance Solution and call it's method
    Solution sol;
    int answer = sol.reverse(0); // 0
    std::cout << "in " << 0 << ", out " << answer << "\n";

    answer = sol.reverse(-1); // -1
    std::cout << "in " << -1 << ", out " << answer << "\n";

    answer = sol.reverse(10); // 1
    std::cout << "in " << 10 << ", out " << answer << "\n";

    answer = sol.reverse(12); // 21
    std::cout << "in " << 12 << ", out " << answer << "\n";

    answer = sol.reverse(100); // 1
    std::cout << "in " << 100 << ", out " << answer << "\n";

    answer = sol.reverse(123); // 321
    std::cout << "in " << 123 << ", out " << answer << "\n";

    answer = sol.reverse(-123); // -321
    std::cout << "in " << -123 << ", out " << answer << "\n";

    answer = sol.reverse(1024); // 4201
    std::cout << "in " << 1024 << ", out " << answer << "\n";

    answer = sol.reverse(-1024); // -4201
    std::cout << "in " << -1024 << ", out " << answer << "\n";

    answer = sol.reverse(2147483648); // 0
    std::cout << "in " << 2147483648 << ", out " << answer << "\n";

    answer = sol.reverse(-2147483648); // 0
    std::cout << "in " << -2147483648 << ", out " << answer << "\n";

    return 0;

}

Answer 1

由于(x > INT_MAX)的值不能超过x，因此int类型为x的任何测试（如INT_MAX都将永远不会为真。

无论如何，即使2147483647是有效范围，其反向7463847412也不是。因此，我认为最好让stoi“尝试”转换值并“捕获”任何out_of_range-exception`。以下代码说明了这种方法：

int convert() {
    const char* num = "12345678890123424542";
    try {
        int x = std::stoi(num);
        return x;
    } catch (std::out_of_range  &e) {
        cout << "invalid." << endl;
        return 0;
    }
}

整数溢出和std :: stoi

1 个答案: