错误"在int之前预期的primary-expression"

Question

我正在编写一个代码(希望)允许用户输入一个数字,并输出2和该数字(包括)之间的素数之和.但是,我在代码的倒数第二行遇到了一个问题.我已经查找了这个问题的其他解决方案,但它们似乎并不是由与我相同的错误引起的.这是代码:

#include <iostream>

using namespace std;

int Q;

int sum_primes(int N) {

    cout << "Enter a number and I will generate the sums of the primes up to (and including) that number: ";
    cin >> Q;

    int i, count, sum = 0;

    for(N = 1; N <= Q; N++) {    
         count = 0;

         for(i = 2; i <= N/2; i++) {
             if (N % i == 0) {
                 count++;
                 break;
             }
         }

         if (count == 0 && N != 1)
             sum = sum + N;

         return N = sum;
    }
}

int main() {
    cout << "The sum of these primes is: " << sum_primes(int N);

   return 0;
}

Answer 1

cout << "..." << sum_primes(int N);

替换int N为数字.您已经定义了该函数,现在需要为它提供一个参数.

或许你想N通过用户输入给出价值.然后使用它代替:

int N;

cin >> N;

cout << "The sum of these primes is: " << sum_primes(N);

此外,正如GigaWatt指出的那样,你做过的路线:

return N = sum;

没必要.简单地返回sum也会起作用.

这是完整的代码:

#include <iostream>
#include <cmath>

bool isPrime(int x) {

    if (x == 1) return false;
    if (x == 2) return true;

    bool prime = true;
    for (int i = 2; i <= sqrt(x); i++) {
        if (x % i == 0) { prime = false; break; }
    }
    return prime;
}

int sum_primes(unsigned int N) {

    int sum = 0;

    for ( int i = 1; i <= N; i++ ) {

        if (isPrime(i)) sum += i;

    }
    return sum == 0 ? 1 : sum;
}

int main() {

    int Q;

    std::cin >> Q;

    std::cout << "Sum of primes " << sum_primes(Q);

}