The two buddies upstairs directly took X as an integer, but the landlord didn't say X was an integer, did he?
This is my program:
# include & ltstdio.h & gt
Master ()
{
Double x, y;
int a；
scanf("%lf "，& ampx)；
a =(int)x； /* Convert the value of x into an integer, that is, cut off the decimal point and assign it to a*/
Switch (a)
{
Case-4: y = x-1; Break;
Case-3: y = x-1; Break;
Case-2: y = x-1; Break;
case- 1:y = x- 1； Break;
/*********************
In fact, the above paragraph can be written like this.
Case 4:
Case 3:
Case 2:
Case-1:
y = x- 1； Break;
Simplify the process
The following paragraph is the same. ....
***********************/
Case 7: y = x+1; Break;
Case 6: y = x+1; Break;
Case 5: y = x+1; Break;
Case 4: y = x+1; Break;
Case 3: y = x+1; Break;
Case 2: y = x+1; Break;
Case1:y = x+1; Break;
Case 0: y = x; Break;
Default value: printf ("input error \ n"); Go to loop;
}
printf("x=%lf\ny=%lf\n "，x，y)；
Loop:;
getch()；
}
So it doesn't matter whether you enter an integer or a floating-point number. ....
Actually, the buddy's method on the first floor is super good. If int x, y is changed to double x, y will be perfect ... (this doesn't seem to be a problem ...)
I have tested this program and found no problems. (Hungry just means hungry. No problems were found. Whoever finds the problem will correct it quickly! ! ! )
====================================
Lipbb is right, the program I wrote above is really wrong ... I'm too lazy to change it ... this is another way I think:
Nested with switch, hunger test, no problem found. (Hungry just means hungry. No problems were found. Whoever finds the problem will correct it quickly! ! ! )
# include & ltstdio.h & gt
Master ()
{
Double x, y;
Printf ("Please enter x \ n");
scanf("%lf "，& ampx)；
Switch (-5 <; x & amp& ampx & lt0)
{
Case1:y = x-1; Break;
Case 0:
{
Switch (x==0)
{
Case1:y = x; Break;
Case 0:
{
Switch (0<x & amp& ampx & lt8)
{
Case1:y = x+1; Break;
Case 0:printf ("input error \ n"); Go to loop;
}
} broken;
}
} broken;
}
printf("x=%lf\ny=%lf "，x，y)；
Loop:;
getch()；
}
You can leave me a message if you don't understand.