Question

/* to find the age of individuals according to youngest to oldest */

#include <stdio.h>

int main(void){
    int age1, age2, age3, youngest, middle, oldest;

    printf ("Enter the age of the first individual: ");
    scanf  ("%d", &age1);
    printf ("Enter the age of the second individual: ");
    scanf  ("%d", &age2);
    printf ("Enter the age of the third individual: ");
    scanf  ("%d", &age3);

    if ( ( age1 == age2 ) && ( age2 == age3) ){
        printf("All individuals have the same age of %d", &age2);
    }

    else (age1 != age2) && (age1 != age3) && (age2 != age3);{
        youngest = age1;

        if (age1 > age2)
            youngest = age2;
        if (age2 > age3)
            youngest = age3;

        middle = age1;

        if (age1 > age2)
            middle = age2;
        if (age2 < age3)
            middle = age2;

        oldest = age1;

        if (age1 < age2)
            oldest = age2;
        if (age2 < age3)
            oldest = age3;

        printf("%d is the youngest.\n", youngest);
        printf("%d is the middle.\n", middle);
        printf("%d is the oldest.\n", oldest);
    }

    return 0;

}

您好我改变了我的代码，但是当我为每个人输入相同的年龄时，显示屏仍会显示一个奇怪的数字。我怎么做到这样，如果每个人都有相同的年龄，那么只有说所有人都具有相同年龄___的线。请帮我这个作为分级任务，我有这个

的问题

所有年龄相同的人都是63567321，是最年轻的。 1是中间。 1是最老的。

Answer 1

三个主要问题：

当年龄相等时，您将指针传递给整数而不是整数本身printf()
else声明没有`if＆＃39;
INDENTATION - 不会影响执行，但对于帮助您更轻松地查找代码中的问题至关重要

正确代码：

/* to find the age of individuals according to youngest to oldest */

#include <stdio.h>

int main(void)

{
    int age1, age2, age3, youngest, middle, oldest;
    printf ("Enter the age of the first individual: ");
    scanf  ("%d", &age1);
    printf ("Enter the age of the second individual: ");
    scanf  ("%d", &age2);
    printf ("Enter the age of the third individual: ");
    scanf  ("%d", &age3);

    if ( ( age1 == age2 ) && ( age2 == age3) )
    {
    printf("All individuals have the same age of %d", age2);
    }

    else if ((age1 != age2) && (age1 != age3) && (age2 != age3))
    {
        youngest = age1;

        if (age1 > age2)
        youngest = age2;
        if (age2 > age3)
        youngest = age3;

        middle = age1;

        if (age1 > age2)
        middle = age2;
        if (age2 < age3)
        middle = age2;

        oldest = age1;

        if (age1 < age2)
        oldest = age2;
        if (age2 < age3)
        oldest = age3;

        printf("%d is the youngest.\n", youngest);
        printf("%d is the middle.\n", middle);
        printf("%d is the oldest.\n", oldest);
    }

    return 0;

}

- 编辑

请注意，在上述解决方案中，如果其中两个年龄相等，则不会打印最旧和最年轻的年龄。要解决这个问题，我会删除else的if条件，并在打印中间值之前包含一个检查。新的其他声明：

    else
    {
        youngest = age1;

        if (age1 > age2)
        youngest = age2;
        if (age2 > age3)
        youngest = age3;

        middle = age1;

        if (age1 > age2)
        middle = age2;
        if (age2 < age3)
        middle = age2;

        oldest = age1;

        if (age1 < age2)
        oldest = age2;
        if (age2 < age3)
        oldest = age3;

        printf("%d is the youngest.\n", youngest);

        // If two ages are equivalent, do not print middle
        if ((age1 != age2) && (age1 != age3) && (age2 != age3))
            printf("%d is the middle.\n", middle);

        printf("%d is the oldest.\n", oldest);
    }

免责声明：虽然我已经提供了上述解决方案来保留原始代码，但我希望通过以下方式采用更通用的方法：

创建一个对整数数组进行排序的方法：sort(int *array, int size)
将输入年龄读入整数数组：int ages[3]
将年龄传递给排序功能：sort(ages, 3);
打印排序数组的结果：ages[0]，ages[1]，ages[2]

这样代码更具可读性和可重用性。

Answer 2

你为什么不做像

这样的事情

{{1}}

排序3个人的年龄

2 个答案: