Attention ! I need help. I cant understand the following program.

Question

/*
A program that prints prime number in a given range.
*/

#include<stdio.h>
#include<math.h>
int main()
{
int start, end, count, flag,inum;
printf("Enter the range: ");
scanf("%d %d",&start,&end);

while(start<end)
{
inum=sqrt(start);
count=2;
flag=1;
while(count<=inum)
{
if(start%count==0)
{
flag=0;
break;
}
count++;
}
if(flag==1)
printf("%d ",start);
start++;
}

}

Answer 1

It is easy.
suppose the range is 4 to 6.
"while (start < end)" is to be executed 2 times, starting with 4.

First, "inum" is going to be equal to the square root of 4, which is 2. "count" is going to be equal to 2 and "flag" is going to be equal to 1.

Then "while (count <= inum)" is executed, since 2 is less than or equal to 2.

Then "if (start% count == 0)" is executed, which analyzes if the remainder of 4/2 is equal to 0, as this is true, the following lines are executed:

flag = 0;
break;

"break" will break "while (count <= inum)" and then execute "if (flag == 1)", as "flag" equals 0, this condition is false and the number is not printed on the screen, then run "start ++;" which changes the value of "start" from 4 to 5.

now we repeat the same procedure but with 5:

inum = 2.2360; (square root of 5)
count = 2;
flag = 1;

"while (count <= inum)" is executed since 2 is less than or equal to 2.2360

"if (start% count == 0)" is executed which evaluates if the remainder of 5/2 is equal to 0, as this is false "count ++" is executed which increases the value of "count" from 2 to 3.

Since count is 3 and is greater than "inum", "while (count <= inum)" is not rerun

Then "if (flag == 1)" is executed, which is true because "if (start% count == 0)" was not true, the following line is executed:

printf ("% d", start);

which prints the number 5 on the screen

then "start ++" increases the value of "start" from 5 to 6

As ( 6(start) < 6(end) ) is false the program ends giving as a result that 5 is a prime number.

I hope I've helped.