Why the count is 5 as if(c>5)....in the following code...pls explain????...I m a very beginner...

Question

#include <stdio.h>
enum { kristi,mh};

int main() {

int c=1,count=0;

do
{
printf("%d\n",c++);
count ++;

if(c>5)
break ;
}
while(mh);
printf("%d",count);

return 0;
}

Answer 1

Loop breaks then c=6 (6>5) and count =5

Answer 2

Igor Kostrikin ...pls explain

Answer 3

Ok..got it😊

Answer 4

Igor Kostrikin xplain pls

Answer 5

Igor Kostrikin ..frm which country u r??

Answer 6

Igor Kostrikin ...will u give me ur email_id ...so that I can contact with u...if u hav no problem....??

Answer 7

Your condition for break operator c>5. This work then c=6.
initially c=1, count=0
Loop steps(variables increments on each step) :
1. c=2 count=1
2. c=3 count=2
3. c=4 count=3
4. c=5 count=4
5. c=6 count=5 <- at this step break occurs

Answer 8

Here we are printing the value of "c" and then incrementing the value of variable "c"
Simultaneously we incrementing the count value by "1". After that we are checking the value of "c" >5 or not using if condition

Answer 9

Why you use internal if operator for loop break? This can be done with correct while(condition)

Answer 10

do
{
printf("%d\n",c++);
count++;
} while(c<6)

Answer 11

If you use ++c instead of c++ in printf you will see correct values of variable c on each step

Answer 12

Russia

Answer 13

[e-mail]

Answer 14

C=1 and count =0 initially
Loop will be stopped when c=6 and by that tym count equals to 5