while loop



Problems such as Hw#3 question #1 are a bit complex to come up with the algorithm if you are not used to the concept of while loop. First let's take a look at the syntax of while:

syntax 
while (condition)
    statement;

or

while (condition) {
    statement;
    statement;
    statement;
    etc...;
}
while will repeat the stętements as long as the condition is true. If while is used without { }, it will repeat only one statement right after. If while is used with { }, it will repeat all the statements surrouneded by { }. For example: 
#include < stdio.h >
main() {
    int i = 0;
    while ( i < 10 )
        i++;
    printf ("i = %d\n", i);
}
The program above will repeat i++; until i becomes 10, and then print the value of i. The flow fo the program is as follows. 
int i = 0;        /* initialize i to 0   */
while ( i < 10 )  /* i < 10 ( i = 0 ),
                         thus keep going */
i++;              /* i = 1 now           */
while ( i < 10 )  /* i < 10 ( i = 1 ),
                         thus keep going */
i++;              /* i = 2 now           */
while ( i < 10 )  /* i < 10 ( i = 2 ), 
                         thus keep going */
i++;              /* i = 3 now           */
while ( i < 10 )  /* i < 10 ( i = 3 ), 
                         thus keep going */
...
...
i++;              /* i = 8 now           */
while ( i < 10 )  /* i < 10 ( i = 8 ),
                         thus keep going */
i++;              /* i = 9 now           */
while ( i < 10 )  /* i < 10 ( i = 9 ), 
                         thus keep going */
i++;              /* i = 10 now          */
while ( i < 10 )  /* i is no longer < 10 
                          ( i = 10),
                     thus stop the loop! */
printf ( "i = %d\n", i );  
            /* prints out the value of i
                            which is 10  */
Notice that the it skips the i++; part does not get executed after the last while condition check..

Then, how would you write a program that will print the sum of number 1 through 10 ? First, you need to think about which part of your program needs to be repeated. Adding 1 to 10 can be written as
sum = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10;
Once again, this can be written as 
i = 1;
sum += i;
i = 2;
sum += i;
i = 3;
sum += i;
i = 4;
sum += i;
i = 5;
sum += i;
i = 6;
sum += i;
i = 7;
sum += i;
i = 8;
sum += i;
i = 9;
sum += i;
i = 10;
sum += i;
Notice that i gets incremented by one, and then added to sum. Thus, it can be written as: 
i = 1;
sum += i;
i++;
sum += i;
i++;
sum += i;
...
...
i++;
sum += i;
Notice that the i++; and sum +=i; will be repeated 10 times.. 
i = 1
sum += i;  /* REPEAT 10 TIMES */   
i++;       /* REPEAT 10 TIMES */
Unfortunately, computer won't understand REPEAT 10 TIMES.. =(
so you have tp tell the compiler to repeat the part 10 times by using while. 
i = 1;
while ( some_condition ) {
    sum +=i;
    i++;
}
Now.. how would you set the some_condition so sum += i; i++; will repeat exactly 10 times, not 9 or 11 times? You have to think a bit carefully here because often times condition is the tricky part. Let's try i < 10 like we did in the previous example.

i = 1;
sum = 0;
while ( i < 10 ) {
    sum += i;
    i++;
}
printf ("sum = %d\n", sum);
printf ("i = %d\n", i);
Although it seems perfect at first, this program does not perform as you might have guessed. Let's examine by line by line to see what happens in this program.

i = 1;/* initialize i with 1 */
while ( i < 10 )/* i < 10, (i = 1) thus keep going*/
sum += i;/* add i to sum. sum = 1 now */
i++;/* increment i, i = 2 now*/
while ( i < 10 )/* i < 10, ( i = 2) thus keep going*/
sum += i;/* add i to sum. sum = 3 now */
..........
i++;/* increment i, i = 8 now*/
while ( i < 10 )/* i < 10, (i = 8) thus keep going*/
sum += i;/* add i to sum, sum = 36 now*/
i++;/* increment i, i = 9 now */
while ( i < 10 )/* i < 10. ( i = 9 ) thus, keep going*/
sum += i;/* add i to sum, sum = 45 now*/
i++;/* increment i, i = 10 now
while ( i < 10 )/* i is not < 10, ( i == 10) thus, finish the loop */
printf ("sum = %d\n", sum);/* print sum to screen. 45 gets printed to screen*/
printf ("i = %d\n", i);/* print i to screen. 10 gets printed to screen*/


Wait a sec!!!! it didn't add 10 to the sum! If you look at the code, sum += i; i++; part won't be executed when i == 10 because the condition was set as i < 10.
To fix this problem, we need to use i <= 10 instead. Thus, the complete code will look like: 
/*
 * Name    : Ji Lee
 * Login   : cs5a6
 * Descrip : this program will add
 *           1 to 10 and print the sum
 *           to screen.
 */
#include < stdio.h >
void main()
{
    /* initializing variables */
    int i = 1, sum = 0;

    /* getting the sum of 1 + 2 + .. 10 */
    while ( i <= 10) {
        sum += i;
        i++;
    }

    /* print out the result */
    printf ("sum = %d\n", sum);
    printf ("i = %d\n", i);
}
This program will run as follows:
i = 1;/* initialize i with 1 */
while ( i <= 10 )/* i <= 10, (i = 1) thus keep going*/
sum += i;/* add i to sum. sum = 1 now */
i++;/* increment i, i = 2 now*/
while ( i <= 10 )/* i <= 10, ( i = 2) thus keep going*/
sum += i;/* add i to sum. sum = 3 now */
...........
i++;/* increment i, i = 8 now*/
while ( i <= 10 )/* i <= 10, (i = 8) thus keep going*/
sum += i;/* add i to sum, sum = 36 now*/
i++;/* increment i, i = 9 now */
while ( i <= 10 )/* i <= 10. ( i = 9 ) thus, keep going*/
sum += i;/* add i to sum, sum = 45 now*/
i++;/* increment i, i = 10 now
while ( i <=10 )/* i <= 10, ( i = 10), thus keep going*/
sum += i;/* add i to sum, sum = 55 now*/
i++;/* increment i, i = 11 now
while ( i <= 10 )/* i is not <= 10, ( i == 11) thus, finish the loop */
printf ("sum = %d\n", sum);/* print sum to screen. 55 gets printed to screen*/
printf ("i = %d\n", i);/* print i to screen. 11 gets printed to screen*/
thus, this program will print:
sum = 45
i = 11
to screen. Notice the use of indent within while loop for readability. It is a convention which most programmers, if not all, follow..

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