**When i=4 , j=1 , switch (4+1-1)=4, ( case 4: ) what will be the right output? **

Question

/* When i=4 , j=1 , switch (4+1-1)=4, ( case 4: ) which is not present on the code. So now it should go to default: (5+3)=8 But on the output I am getting 7 instead of 8 why? Can you please explain it? */

#include <stdio.h>
main()

{
int i, j, x = 0;

for (i = 0; i < 5; ++i)
for (j = 0; j < i; ++j) {

switch (i + j - 1) {

case -1:
case 0:
x += 1;
break;

case 1:
case 2:
case 3:

x += 2;
break;

default:

x += 3;
break;
}
printf("%d\n", x);
}
printf("\nx = %d", x);
}

Output:
1
3
5
7
9
12
14
17
20
23

x = 23Press any key to continue . . .
*/
https://code.sololearn.com/cYMktBgOzJ6h/?ref=app

Answer 1

Yes it goes to default but there x is already 14 so output 14+3=17

when i = 1, j= 0 , case = 0, x=x+1=1

when i = 2, j= 0 , case = 1, x=x+2=>x=3
when i = 2, j= 1 , case = 2, x=x+2=>x=5

when i = 3, j= 0 , case = 2, x=x+2=>x=7
when i = 3, j= 1 , case = 3, x=x+2=>x=9
when i = 3, j= 2 , case = 4, x=x+3=>x=9+3=12

when i = 4, j= 0 , case = 3, x=x+2=>x=14
when i = 4, j= 1 , case = 4, x=x+3=>14+3=17
when i = 4, j= 2 , case = 5, x=x+3=>x=20
when i = 4, j= 3 , case = 6, x=x+3=>x=23

RRR

Answer 2

Where is code!?

continue with last question.. instead of new!

don't change original post as it confuses readers...

edit: added.

Answer 3

Insert your code! For that you must have it saved in the "playground"

Using "printf" functions you can see the values ​​of j and i at each moment to understand what is happening, like this:

https://code.sololearn.com/cqa2gdb83AF5/?ref=app