Need help in switch in c

Question

This is code:

#include<stdio.h>
int main() {
char inchar = 'A';
switch (inchar)
{
case 'A' :
printf ("choice A ") ;
case 'B' :
printf ("choice B ") ;
case 'C' :
case 'D' :
case 'E' :
default:
printf ("No Choice") ;
}
return 0;
}

when i compiled
answer is choice A choice B No Choice
why?? shouldn't answer be choice A

Answer 1

you have no use of break statement

Answer 2

It should be:

#include<stdio.h>
int main() {
char inchar = 'A';
switch (inchar)
{
case 'A' :
printf ("choice A ") ;
break;
case 'B' :
printf ("choice B ") ;
break;
case 'C' :
break;
case 'D' :
break;
case 'E' :
break;
default:
printf ("No Choice") ;
break;
}
return 0;
}

Answer 3

Here is my explanation to myself:
https://code.sololearn.com/cHpx2r85kn8A/#c

And here is how I figured it out, please someone, correct me if I am wrong.
In C, when the case is true all cases after that one, are also true or accounted for.
I was trying to work out why is the answer for a challenge question 17, this is the code for it:
int x
switch(x){
case 1: {x+=x;}
case 3: {x+=x;}
case 5: {x+=x;}
default: {x+=5;}
}
printf("%d", x);
I thought it was 11, case 3 and default, right?
But when I did this:
int x
switch(x){
case 1: {x+=x;}
case 2: {x+=x;}
case 3: {x+=x;}
default: {x+=5;}
}
printf("%d", x);

the result was 11

and after this

int x
switch(x){
case 1: {x+=x;}
case 3: {x+=x;}
case 5: {x+=x;}
case 6: {x+=x;}
default: {x+=5;}
}
printf("%d", x);

The result was 29.

same can be applied for your switch if you put case 'B' : before case 'A'... And

Yes because no breaks are applyed. :)

Answer 4

you forget to use break...
in the end of every case write break;