switch-case statement without break

Question

According to this book I am reading：

Q What happens if I omit a break in a switch-case statement?

A The break statement enables program execution to exit the swi

Answer 1

Try yourself - Run the code using ideone available here.

#include 

void doA(int *i){
    printf("doA i = %d\n", *i);
    *i = 3;
}

void doB(int *i){
    printf("doB i = %d\n", *i);
    *i = 4;
}

void doC(int *i){
    printf("doC i = %d\n", *i);
    *i = 5;
}

int main(void) {
    int i = 1;
    switch(i){
        case 1:
            doA(&i);
        case 2:
            doB(&i);
        default:
            doC(&i);
            break;
    }
    return 0;
}

Output:

doA i = 1
doB i = 3
doC i = 4

Note:

• It will execute all the options from the selected case until it sees a break or the switch statement ends. So it might be that only C is executed, or B and then C, or A and B and C, but never A and C
• If you change the value of the variable analysed in switch inside the handle function (e.g doA), it does not affect the flow as describe above