I just run this c code, but when i enter A, the program not go to the printf fun
ID: 3586176 • Letter: I
Question
I just run this c code, but when i enter A, the program not go to the printf function, i don't know why. I think the value of in is A, and A belong to the case 'A', it should be do.
int main() char in, cis, r; printfA: Circle B : SquareinC: Triangle "); printf("Enter Which Shape to use: "); scanf("%c", ∈); switch('in') ( case 'A" printf("Enter P for solve for perimeter and A for solve for Area "); scanf("%c", &cis;): cis is choice printf("Enter circle radius: "); scanf("%lf", &r;); printf( "Circle perimeterr is:", perimeter( 'A', 2, 0, 0) return 0;Explanation / Answer
//problem is in switch statement
//switch('in')//this is wrong(not syntax wise, logically)
//you have to user like..
//switch(in)//variable name
#include<stdio.h>
int main()
{
char in,cis,r;
printf("A: Circle B: Square C: Triangle ");
printf("Enter Which Shape to user: ");
scanf("%c",&in);
switch(in)//switch('in')
{
case 'A':
printf("Enter P for solve for perimeter and A for solve for Area ");
scanf("%c",&cis);
if(cis == 'P')
{
printf("Enter circle radius: ");
scanf("%lf",&r);
printf("Circle perimeter is:",perimeter('A',2,0,0));
}
}
return 0;
}