Question

我正在编写一个程序来确定一系列数字的素因子。在该计划中，该范围的startingNumber和endingNumber表示为std::pair。

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

const long long MAX = 100000000; // One hundred million
long long factor[MAX];


pair<const long long, const long long> get_pair() // Gets user start and end number to establish range in vector.
{
    long long largestNumber = 100000000;
    long long smallestNumber = 2;
    std::pair<long long, long long> p;
    cout << "Enter a start number: ";
    cin >> p.first;
    while (p.first < smallestNumber || p.first > largestNumber || cin.fail()) {
        cin.clear();
        cin.ignore(1000, '\n');
        cout << "That's not a valid choice. Please try again (" << smallestNumber << " - " << largestNumber << "): ";
        cin >> p.first;
    }
    cout << "Enter a end number: ";
    cin >> p.second;
    while (p.second < p.first || p.second > largestNumber || cin.fail()) {
        cin.clear();
        cin.ignore(1000, '\n');
        cout << "That's not a valid choice. Please try again (" << p.first << " - " << largestNumber << "): ";
        cin >> p.second;
    }
    cout << endl;

    return make_pair(p.first, p.second);
}

vector<long long> rangeToVector(const long long &startingNumber, const long long &endingNumber) // creates vector of numbers from std::pair.
{
    vector<long long> iterator;

    for (long long i = 1; i <= ((10 - 1) + 1); i += 1) iterator.push_back((1-1)+i);

    return iterator;
}
void precalculate() // Calculates all of the prime numbers between 1 and MAX. Stores prime numbers from smallest to largest.
{
    for (long long i = 1; i < MAX; i++) { // stores 1 through Max in factor.
        factor[i] = i;
    }

    for (long long i = 2; i * i < MAX; i++) if (factor[i] == i) { // If factor[i] is even.
        for (long long j = i + i; j < MAX; j += i) {
            factor[j] = i;
        }
    }
}

上述代码按预期工作。然后将代码输入到follow函数中，我不确定是否正常工作（因为我的函数在main中不会编译）。

vector<long long> get_factors(vector<long long> &vect) // performs calculation on rangeToVector[i] (determined by precalculate) until it is not divisible, then it returns those that were divided into the number as prime factors of the rangeToVector[i].
{
    vector<long long> factors;
    for (auto i : vect)
    {
        while (vect[i] > 1)
        {
            long long f = factor[vect[i]];
            factors.push_back(f);
            vect[i] /=f;
        }
    }

    sort(factors.begin(), factors.end()); // sorts the vector.
    factors.erase(unique(factors.begin(), factors.end()), factors.end()); // erases duplicates (requires sorted vector).

    return factors;
}

在主（第3行）中，我收到错误下划线rangeToVector，initial value of reference to non-const must be an lvalue。

int main()
{
    for (auto i : get_factors(rangeToVector(pairs.first, pairs.second))) // get_factors returns factors of numbers in vector rangeToVector
    {
        cout << i << " ";
    }
    system("pause");
    return 0;
}

我试图在rangeToVector常量中创建值，但这并没有解决问题。我的问题是，我的引用不正确吗？我应该以不同方式传递函数吗？或者我可以对代码做些什么来允许get_factors在第3行的主要内容中使用rangeToVector[i]的因子。我是c ++的新手，所以任何帮助都会非常感激。

Answer 1

你的问题在于：

std::vector<long long> get_factors(std::vector<long long> &vect) { ... }

变量vect是一个左值引用，这意味着它是一个无法绑定到临时变量的引用。

然而，你这样使用它：

get_factors(rangeToVector(...))

rangeToVector返回临时值。

解决方法就是让vect成为一个常量左值引用，它可以绑定到临时值：

//                                     const here  -------v
std::vector<long long> get_factors(std::vector<long long> const& vect) { ... }

您的代码中还有许多其他问题，我建议您阅读一个好的C ++并尝试不要假设它的语法。例如，您没有正确使用range for loops：

for (auto i : vect) {
    while (vect[i] > 1) {
        long long f = factor[vect[i]];
        factors.push_back(f);
        vect[i] /=f;
    }
}

正确的用法是：

 for (auto i : vect) {
    while (i > 1) {
        long long f = factor[i];
        factors.push_back(f);
        i /= f; // warning here!
    }
}

我发出警告的行是因为你正在修改一个局部变量，没有副作用。这条线的意图尚不清楚。

Answer 2

首先，我要感谢大家的意见和建议，这有助于我解决我的问题。正如Guillaume Racicot所说，我当然需要更多地研究载体。再次，谢谢。

我的问题出在下面的代码中：

vector<long long> get_factors(vector<long long> &vect) // Notice how I have it relying on "vector<long long> &vect".
{
    vector<long long> factors;
    for (auto i : vect)
    {
        while (vect[i] > 1)
        {
            long long f = factor[vect[i]];
            factors.push_back(f);
            vect[i] /=f;
        }
    }

    sort(factors.begin(), factors.end()); // sorts the vector.
    factors.erase(unique(factors.begin(), factors.end()), factors.end()); // erases duplicates (requires sorted vector).

    return factors;
}

以及我在main中调用它的方式：

 for (auto i : get_factors(rangeToVector(pairs.first, pairs.second))) // get_factors returns factors of numbers in vector rangeToVector
    {
        cout << i << " ";
    }

为了解决我的问题，我需要意识到，为了访问向量/数组，你必须创建一个for循环来收集它的值。因此，我的解决方案如下。

vector<long long> get_factors(long long x, vector<long long> vec )
    {
             vector<long long> factors;
        while (x > 1) { /* insert code*/ vec.push_back(f); } // Or use a for loop.

           return vec;
    }

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

    vector<long long> temp1;

    for (long long val : inputVector) // where initial data was stored (see original post)
            {
                  result1 = getPrimes(val, temp1); // iterate through vec as long long in function
            }

     return 0;
}

c ++如何解决：对非const的引用的初始值必须是左值

2 个答案: