How does falling through cases in c work? - c

Question

#include <stdio.h>
main() /* count digits, white space, others */
{
int c, i, nwhite, nother, ndigit[10];
nwhite = nother = 0;
for (i = 0; i < 10; i++)
ndigit[i] = 0;
while ((c = getchar()) != EOF) {
switch (c) {
case '0': case '1': case '2': case '3': case '4':
case '5': case '6': case '7': case '8': case '9':
ndigit[c-'0']++;
break;
case ' ':
case '\n':
case '\t':
nwhite++;
break;
default:
nother++;
break;
}
}
printf("digits =");
for (i = 0; i < 10; i++)
printf(" %d", ndigit[i]);
printf(", white space = %d, other = %d\n",
nwhite, nother);
return 0;
}
Question 1 - How does these cases work if c = 0 then does the control go to directly to the statements and break or does it remember that c = 0 and continue falling through until the statements?
#include <stdio.h>
int main()
{
int i = 0;
for(i = 0; i < 20; i++)
{
switch(i)
{
case 0: i+=5;
case 1: i+=2;
case 5: i+=5;
default: i+= 4;
break;
}
printf("%d ", i);
}
return 0;
}
Question 2 - In a program like this why does 2 add up in case 1? Where the current value of i will be 5? Does the control ignore all the remaining/underling case constant expressions until it finds a break if one case is true?

Answer 1

the switch in c jumps to the case with the matching value and then executes all code from there until it finds a break or switch ends. son in 2nd example for value 0 all other cases are evaluated too. if you don't want this you need a break.