Question

如何检测用户输入int是否介于两个数字之间？

我试过这样做：

int x=3;
printf("Enter the size of the triangle: ");
scanf("%d", &size);
odd=size%2;

for(x=3;x<21;x++)
{
  if(x==size&&odd==1)
  {
     break;
  }
  else
  {
 printf("The size must be an odd number and be between\n3 and 21, inclusive, please try again\n\n");
     printf("Enter the size of the triangle: ");
     scanf("%d",&size);
     odd=size%2;
     x=3;
  } 
}

但我唯一可以使用的输入是3。

Answer 1

您已经在代码中拥有了解决方案的所有内容：

if (size >= 3 && size <= 21) {
   // size is between 3 and 21 inclusive
} else {
   // size is less than 3 or more than 21
}

如果您还想确保它是奇数，您可以添加条件：

if ((size >= 3) && (size <= 21) && (size % 2 == 1)) {
   // size is between 3 and 21 inclusive, and odd
} else {
   // size is less than 3 or more than 21 or even
}

Answer 2

如果您需要不断询问用户输入的号码，直到他们输入以下号码：

奇;和
介于3和21之间，

你可以使用类似的东西：

printf ("Enter the size of the triangle: ");
scanf ("%d", &size);
while ((x < 3) || (x > 21) || (x % 2 == 0)) {
    printf ("The size must be an odd number and be between\n"
        "3 and 21, inclusive, please try again.\n\n");
    printf ("Enter the size of the triangle: ");
    scanf ("%d", &size);
}

这可能是最简单的形式。它得到数字然后进入while循环，直到它有效。

你可以将printf/scanf对重构为一个单独的函数，但在这样的小片段中这可能不那么重要。

Answer 3

怎么样：

if(x >= 3 && x <= 21 && x%2) {
    ...
}

Answer 4

printf("Enter the size of the triangle");
scanf("%d", &size);

if (3 <= size && size <= 21 && size % 2) {
    // size is between 3 and 21 (inclusive) and odd
}

如何检测数字是否在C中的其他两个数字之间？

4 个答案: