程序挂起较小的整数

Question

我遇到了下面粘贴的代码问题,它只是在运行时挂起.VS2010没有给我任何警告或错误.

#include <stdio.h>
#include <stdlib.h>
#include <time.h>


void clear_buffer(void)
{
    while(getchar() != '\n');
}

int validate(int low, int high) {
    int num;

    scanf("%d", &num);
    while(num < low || num > high) 
    {
        clear_buffer();
        printf("INVALID! Must enter value between %d and %d: ", low, high);
        scanf("%d", &num);
    }
    return num;
}

int getRand(int max) {
    int number;
    number = rand() % max + 1;
    return number;
}

int validatePick(int pick, int one, int two, int three, int four, int five) {
    int valid = 0;

    if (pick != one && pick != two && pick != three && pick != four && pick != five) {
        valid = 1;
    } else {
        valid = 0;
    }
    return valid;
}
void prnt(int qty, int one, int two, int three, int four, int five, int six) {
    int i = 0, flag = 0;
    while (flag != 2) {
        flag = 0;
        if (sort2(&one, &two) == 0 && sort2(&three, &four) == 0 && sort2(&five, &six) == 0)
            flag = 1;
        if (sort2(&two, &three) == 0 && sort2(&four, &five) == 0)
            flag = 1;
        flag += flag;
    }

    printf("Picks: %d, ", one);
    while (i <= qty){
        if (i== 2 && qty == 2)
            printf("%d\n", two);
        else if (i == 2 && qty != 2) 
            printf("%d, ", two);

        if (i == 3 && qty == 3) 
            printf("%d\n", three);
        else if (i == 3 && qty != 3) 
            printf("%d, ", three);

        if (i == 4 && qty == 4)
            printf("%d\n", four);
        else if (i == 4  && qty != 4)
            printf("%d, ", four);

        if (i == 5 && qty == 5)
            printf("%d\n", five);
        else if (i == 5  && qty != 5) 
            printf("%d, ", five);

        if (i == 6 && qty == 6) 
            printf("%d\n", six);
        i++;
    }
}



int sort2(int *n1, int *n2) {
    int tmp, valid = 0;

    if (*n1 > *n2)
    {
        tmp = *n2;
        *n2 = *n1;
        *n1 = tmp;
        valid = 1;
    }
    return valid;
}

int main () {
    int num1, num2;
    int pick, one = 0, two = 0, three = 0, four = 0, five = 0, six = 0;

    srand(time(NULL));

    printf("LOTTERY GENERATOR\n");
    printf("Enter the maximum value between 1 and 100: ");
    num1 = validate(2,100);
    printf("Enter quantity of numbers to pick, between 1 and 6: ");
    num2 = validate(1, 6);

    one = getRand(num1);
    while (two == 0 || three == 0 || four == 0 || five == 0 || six == 0) {
        pick = getRand(num1);
        if (validatePick(pick, one, two, three, four, five) == 1 && two == 0)
            two = pick;
        else if (validatePick(pick, one, two, three, four, five) == 1 && three == 0)
            three = pick;
        else if (validatePick(pick, one, two, three, four, five) == 1 && four == 0)
            four = pick;
        else if (validatePick(pick, one, two, three, four, five) == 1 && five == 0)
            five = pick;
        else  if (validatePick(pick, one, two, three, four, five) == 1)
            six = pick;
    }
    prnt(num2, one, two, three, four, five, six);
}

如果我输入说3 Enter the maximum value between 1 and 100然后2,程序就会挂起.我不明白为什么会这样.我没有在代码中看到错误.有任何想法吗？

Answer 1

我认为问题在于你的逻辑.如果我理解正确的程序,您要挑选6张不同的数,在范围内随机生成的,从1到num1.

现在问题在于:while只有当所有六个变量(one,two.. six)都没有时,你的循环才会终止0.这些变量中的一个可以被设置为一个非零值的唯一方法是,如果validatePick回报率1,并认为只有在产生的随机数尚未分配到一个发生one,two... six.

这归结为类似反向鸽笼问题; 你试图用少于六羽的鸽子填补六个鸽笼,这是一项不可能完成的任务.

如果num1小于6,那么你不可能满足你的while循环的终止条件,你的程序似乎会挂起.

您可以通过else在while循环中放置一个case并打印生成的随机数来验证这一点,并在每次迭代时打印每个变量的值.

请注意,您的第二个输入,num2在while循环之后才会以任何方式引用,因此您输入的值将无法限制程序尝试生成的唯一随机值的数量.