我们可以扫描一个数组吗？

Question

我想输入一个整数数组，然后从输入的数字中打印出偶数..

example is if I input 2466688992,
it will output 24666882;

我的代码如下：

#include<stdio.h>
int main()
{
int a[5],i;
printf("Enter array of numbers: ");
scanf("%d",&a);

for(i=0; i<sizeof(a); i++){
if(a[i]%2==0)
printf("%d",a[i]);  
}
   getch();
   return 0;
   
}

结果变成垃圾：2468000075416640419940000004225568000

Answer 1

这是打印整数偶数的函数:

#include<stdio.h> 

int main(){

               int num,rem,even=0,digit;

               printf("  Enter an integer number: ");

               scanf("%d",&num);

               printf("\n  The even digits present in %d are \n",num);

               while(num>0){

               digit = num % 10;

               num = num / 10;

               rem = digit % 2;

               if(rem == 0)
               even++;
               printf("\n  %d.",digit);

               }

        return 0; 

    }

Answer 2

您应该将数组扫描为字符串（除非您想强加数组中的项目数），然后解析字符串以存储不同的数字：

long a[50];
char buf[1024];
printf("Enter array of numbers: ");
scanf("%s",buf);
int len = strlen(buf);
int j = 0;
for (int i = 0; i < len; ) {
    long sign = 1;
    long n = 0;
    if (buf[i] == '+') {
        ++i;
    }
    else if (buf[i] == '-') {
        sign = -1;
        ++i;
    }
    if (isdigit(buf[i])) {
        while (isdigit(buf[i])) {
            n = 10 * n + buf[i++] - '0';
        }
        a[j] = n * sign;
    }
    else
        i++;
}
for (int i = 0; i < j; i++)
   if (!(a[i] ℅ 2)) // true if even
      printf("%ld ", a[i]);

这会将所有数字存储在大小为 j 的数组 a 中。

编辑：如果您谈论的是数字，那么它更容易：

char buf[1024];
printf("Enter array of numbers: ");

scanf("%s",buf);
int len = strlen(buf);
for (int i = 0; i < len; i++)
   if (isdigit(buf[i]) && !((buf[i] - '0') ℅ 2)) // true if even, note that '0' equals 0x30 so there is no need to sub it to check for odd/even in reality.
      printf("%c ", buf[i]);

我们可以扫描一个数组吗？

Can we scanf an array?

arrays

c

integer

scanf

sizeof