提供的示例代码返回一个随机数，即使在抛出异常之后（提供的代码） [英] The sample code provided returns a random number, even after throwing an exception (code provided)

问题描述

我有一个示例代码来修改和抛出异常处理。问题是甚至在我抛出异常后，代码仍然返回一个随机0.我花了一些时间，试图找出为什么我还有一个0返回，但我找不到答案。有没有人知道为什么代码的行为像这样？

I have a sample code to modify and throw exception handling. The problem is even after I threw an exception, the code still returns a random 0. I have spent some time trying to figure out why I still have a 0 returned but I could not find the answer. Does anyone have an idea why the code behaves like this?

#include <stdexcept>
#include <iostream>
#include <string>

using namespace std;


struct myException_Product_Not_Found : exception 
{
     virtual const char* what() const throw() {
        return "Product not found";
     }
} myExcept_Prod_Not_Found;  

int getProductID(int ids[], string names[], int numProducts, string target) {
    for (int i=0; i<numProducts; i++)  {
       if(names[i] == target)
            return ids[i];          
    } 
    try {
       throw myExcept_Prod_Not_Found;   
    }
    catch (exception& e) {
       cout<<e.what()<<endl;     
    }                                       
}

// Sample code to test the getProductID function
int main() {
    int    productIds[] = {4,5,8,10,13};
    string products[]   = {"computer","flash drive","mouse","printer","camera"};

    cout << getProductID(productIds, products, 5, "computer") << endl;
    cout << getProductID(productIds, products, 5, "laptop") << endl;
    cout << getProductID(productIds, products, 5, "printer") << endl;

    return 0;
} 

推荐答案

getProductID 不会抛出异常。你捕获在 getProductID 有机会抛出的异常。因此，你回来...好吧，没有。函数结束后不调用 return 。

getProductID doesn't throw an exception. You catch the exception you do throw before getProductID has a chance to throw it. As such, you return ... well, nothing. The functions ends without you calling return.

如果你打开了编译器的警告* do），编译器应该使用控制到非结束函数的消息来警告。 g ++ 在此实例中显示为返回零，但返回零可能是未定义的行为。

If you had turned on your compiler's warnings* (as should should be doing), the compiler should warn with a message like control reaches end of non-void function. g++ appears to return zero in this instance, but returning zero is probably undefined behaviour.

抛出异常，不捕捉你抛出的异常在函数内部。将捕获移到外面。

If you want a function to throw an exception, don't catch the exception you've thrown inside of the function. Move the catch to the outside.

int getProductID(...) {
   ...
   throw myExcept_Prod_Not_Found;
}

string product = "computer";
try {
   cout << getProductID(productIds, products, 5, product) << endl;
} catch (exception& e) {
   cout << "Can't find product id for " << product << ": " << e.what() << endl;
}

*—要打开 g ++ 中的警告， -Wall 是一个很好的起点。 @Tomalak Geret'kal建议 -Wall -Wextra -std = c ++ 98 -pedantic 或 -Wall -Wextra -std = c ++ 0x -pedantic 。

* — To turn on warnings in g++, -Wall is a good starting point. @Tomalak Geret'kal suggests -Wall -Wextra -std=c++98 -pedantic or -Wall -Wextra -std=c++0x -pedantic.

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