c++如何求解:引用非常量的初始值必须是左值

c++ How to solve : Initial value of reference to non-const must be an lvalue

本文关键字:何求解 引用 常量 非常 c++      更新时间:2023-10-16

我正在编写一个程序来确定一系列数字的质因数。在程序中,范围的startingNumberendingNumber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;
        }
    }
}

上面的代码按预期工作。然后将代码输入到以下函数中,我不确定 or 是否正常工作(因为我在 main 中的函数将无法编译(。

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 main(第 3 行(中,我收到错误下划线rangeToVectorinitial 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常量中生成值,但这并没有解决问题。我的问题是,我的引用不正确吗?我应该以不同的方式传递函数吗?或者我可以对代码做些什么,以允许get_factors在第 3 行的 main 中获取rangeToVector[i]的因素。我是 c++ 的新手,所以任何帮助将不胜感激。

你的问题就在这里:

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

变量vect是一个左值引用,这意味着它是一个不能绑定到临时变量的引用。

然而,你这样使用它:

get_factors(rangeToVector(...))

其中rangeToVector返回临时。

解决方案是简单地将vect设置为常量左值引用,该引用可以绑定到临时

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

你的代码中还有许多其他问题,我建议你阅读一篇好C++,尽量不要假设它的语法。例如,您没有正确使用循环的范围:

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;
    }
}

首先,我要感谢大家的意见和建议,它帮助我解决了我的问题。正如纪尧姆·拉西科特(Guillaume Racicot(所提到的,我肯定需要更多地研究载体。再次,谢谢。

我的问题在下面的代码中:

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 循环来收集其值。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;
}