调试断言失败！表达式：result_pointer!=nullptr

Posted 2023-02-19

【中文标题】调试断言失败！表达式：result_pointer!=nullptr

【英文标题】：Debug Assertion Failed! Expression:result_pointer!=nullptr

【发布时间】：2018-03-23 11:21:18

【问题描述】：

我正在编写一个 C++ 程序来搜索数组中的给定整数，但是，当我尝试调试程序时，Visual Studio（我使用的是 vs 2015 专业版）抱怨调试断言失败： enter image description here

这是我的代码，非常简单：

int main() 
int searchArray[10] =  324,4567,6789,5421345,7,65,8965,12,342,485 ;
//use searchKey for the number to be found
//use location for the array index of the found value
int searchKey, location;

//write code to determine if integers entered by 
//the user are in searchArray
//initiate searchKey and location
searchKey = 0;
location = 0;
int n = sizeof(searchArray) / sizeof(searchArray[0]);
//let user define the search key, give -1 to quit
while (true)

    std::cout << "Enter an integer ('-1') to quit: ";
    scanf_s("%d", searchKey);
    std::cout << searchKey << "\n";
    if (searchKey == -1)
    
        break;
    
    for (location; location < n; location++)
    
        if (searchArray[location] == searchKey)
        
            break;
        
        location = -1;
    
    if (location != -1)
    
        std::cout << searchKey << " is at location " << location << " in the array.\n";
    
    else
    
        std::cout << searchKey << " is not in the array.\n";
    

return 0;

【问题讨论】：

您解决了这个问题吗？如果在代码行中的“searchKey”之前添加“&”，结果如何？

【参考方案1】：

每个参数都必须是指向一个类型变量的指针，该类型与格式中的类型说明符相对应。

只需将代码 "scanf_s("%d", searchKey)" 更改为：

scanf_s("%d", &searchKey);

它会很好用。