函数以十六进制返回值

问题描述

我制作了一个函数，可以将数字的每隔一位数字加倍，但由于某种原因，它在返回之前将 in 转换为十六进制。我检查过，该十六进制准确到实际数字，但是如何让它停止返回十六进制？这是代码。

unsigned long long* Luhn_Algorigthem::double_every_other_value(unsigned long long int in) {
    unsigned long long int* out = new unsigned long long;
    int counter = 10;
    
    for (int i = 0; i < std::to_string(in).length(); i++) {
        if (i % 2 == 0) { //Is even
            counter * 10;
            out += (unsigned long long)((in % counter) * 2);
        }
        else { //Is odd
            out += (unsigned long long)(std::to_string(in).at(i));
        }
    }

    doubled_val = (unsigned long long)out;
    return (unsigned long long*)33;
    delete out;
}

解决方法

Unsigned 和 long 是类型修饰符（就像没有名词的形容词），您是否尝试过 unsigned long int 进行显式类型转换？

@TheUndedFish 帮我解决了这个问题，我最终摆脱了指针，并使用 doubled val 变量代替它而不为堆分配内存。这是更新后的代码：

unsigned long long int Luhn_Algorigthem::double_every_other_value(unsigned long long int in) {
    //unsigned long long int* out = new unsigned long long;
    int counter = 10;
    
    for (int i = 0; i < std::to_string(in).length(); i++) {
        if (i % 2 == 0) { //Is even
            counter * 10;
            doubled_val += (in % counter) * 2;
        }
        else { //Is odd
            doubled_val += std::to_string(in).at(i);
        }
    }
    //delete out;
    return doubled_val;
}

c++hex return