C++ 当我试图在三个给定年龄中查找最小的年龄时，为什么我的 C 程序没有在终端中显示任何输出

发布于06月04日

终端未显示任何输出(空白)

// if the ages of 3 people are put telling which one is youngest//

#include<stdio.h>

int main()
{
    int a1,a2,a3;

    printf("enter the age of ram,shyam and ajay");
    scanf("%d,%d,%d",&a1,&a2,&a3);

    if(a1>a2>a3)
        printf("%d,%d,%d",a1,a2,a3);
    if(a3>a2>a1)
        printf("%d,%d,%d",a3,a2,a1);
    if(a2>a1>a3)
        printf("%d,%d,%d",a2,a1,a3);
    if(a2>a3>a1)
        printf("%d,%d,%d",a2,a3,a1);
    if(a1>a3>a2)
        printf("%d,%d,%d",a1,a3,a2);
    if(a3>a1>a2)
        printf("%d,%d,%d",a3,a1,a2);

    return 0;
}

我不确定您是否应该在1&>2&>3这样的if语句中编写3个表达式如果我说错了，请纠正我

推荐答案

根据C标准(6.5.8关系运算符)

6运算符<；(小于)、>；(大于)、<；=(小于或等于)，以及^gt=(大于或等于)shall yield 1 if the specified relation is true and 0 if it is false.其结果是键入int.

并且从左到右对运算符进行判断，

所以如果像这样的声明

if(a1>a2>a3)

相当于

if( ( a1 > a2 ) > a3 )

根据a1是否大于a2，你可以 Select

if( ( 1 ) > a3 )

或

if( ( 0 ) > a3 )

Instead you need to include in the expression logical AND operat或 like

if( a1 > a2 && a1 > a3 )

但是，如果您要这样做，如果至少有两个年龄相等，程序将不会输出任何内容.

也是在这个scanf的呼唤中

scanf( "%d,%d,%d", &a1, &a2, &a3 );

it is better to remove commas between the conversion specifiers. Let the user will enter ages separated by spaces like f或 example

20 22 21

或

20
22
21

If you want to output the ages in the descending 或der you could write f或 example

#include <stdio.h>

int main( void )
{
    int a1, a2, a3;

    printf( "enter the age of ram,shyam and ajay: " );
    scanf( "%d %d %d", &a1, &a2, &a3 );

    if ( a1 < a2 )
    {
        int tmp = a1;
        a1 = a2;
        a2 = tmp;
    }

    if ( a2 < a3 )
    {
        int tmp = a2;
        a2 = a3;
        a3 = tmp;
    }
 
    if ( a1 < a2 )
    {
        int tmp = a1;
        a1 = a2;
        a2 = tmp;
    }

    printf( "%d, %d, %d\n", a1, a2, a3 );
}