C++ 打印素数的程序

代码中存在多个问题:

• 素数if (primeCheck(i, y) == 0)的测试是不正确的:第一个参数应该通过2，第二个参数应该通过i.
• 循环应该从i = y开始，在i >= x时运行，递减i以首先找到范围内最大的素数.
• 你应该储存素数并在找到2时停止.
• 素性测试函数应该修改为返回1 if i > j，以防止对某些输入(如primeCheck(2, 1))进行无限递归.
• 测试小于等于sqrt(j)的除数在primeCheck()中就足够了.您可以在j < i * i时停止递归，这可以在没有潜在溢出的情况下进行测试，如j / i < i.

以下是您的代码的修改版本:

#include <stdio.h>

int usersInput(int *x, int *y);
void swapping(int *x, int *y);
int primeCheck(int i, int j);

int main() {
    int x, y, i, num1, num2;
    
    if (!usersInput(&x, &y)) {
        return 1;
    }
    if (x > y) {
        swapping(&x, &y);
    }

    printf("The value of x is %d, and the value of y is %d.\n", x, y);

    for (i = y, num1 = num2 = 0; i >= x; i--) {
        if (primeCheck(2, i) == 0) {
            if (num2 == 0) {
                num2 = i;
            } else {
                num1 = i;
                break;
            }
        }
    }
    printf("The two largest prime numbers from %d to %d are: %.0d %.0d\n",
            x, y, num1, num2);
    return 0;
}

// read user input return zero on failure
int usersInput(int *x, int *y) {
    printf("Enter the value of x: ");
    if (scanf("%d", x) != 1)
       return 0;
    printf("Enter the value of y: ");
    return scanf("%d", y) == 1;
}

void swapping(int *x, int *y) {
    int temp = *x;
    *x = *y;
    *y = temp;
}

int primeCheck(int i, int j) {
    if (j < 2) {
        return 1;  // negative, 0 or 1: not prime
    } else if (j / i < i) {
        return 0;  // all factors tested up to sqrt(j): j is prime
    } else if (j % i == 0) {
        return 1;  // composite, not prime
    } else {
        return primeCheck(i + 1, j);
    }
}

输出:

Enter the value of x: 10
Enter the value of y: 3
The value of x is 3, and the value of y is 10.
The two largest prime numbers from 3 to 10 are: 5 7