[英]Enforce input greater than or equal to 0 and less than 24
#include<stdio.h>
int main ()
{
int n,a=0,b=24;
do
{
scanf("%d",n); //ask the user to enter a value of n less than 24
// but greater than 0.
} while(/*boolean expression logic*/)
if(a<n<b)
{
printf("%d\n",n);
}
return 0;
}
I need to evaluate: 我需要评估:
If the value of n is greater than or equal to 0 and less than 24 (less than or equal to 23) then 如果n的值大于或等于0且小于24(小于或等于23),则
.... go to the if statement and print the value of n ....转到if语句并打印n的值
otherwise 除此以外
... ask the user to input the value of n ie it should again go back in the loop. ...要求用户输入n的值,即应该再次返回循环。
You want the program to keep asking for values until n>=0 && n<24
; 您希望程序继续询问值,直到n>=0 && n<24
; in other words, you want to keep asking for values while !(n>=0 && n<24)
, which using De Morgan's law we can write as !(n>=0) || !(n<24)
换句话说,您想在!(n>=0 && n<24)
时继续询问值,根据De Morgan定律,我们可以写成!(n>=0) || !(n<24)
!(n>=0) || !(n<24)
, which can be reduced to n<0 || n>=24
!(n>=0) || !(n<24)
,可以减少为n<0 || n>=24
n<0 || n>=24
do
{
scanf("%d",n);
}
while(n<0 || n>=24)
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