Adding integers in a do-while statement - c

Question

I have to write a do while loop that reads integers and computes their sum. It continues to add the integers until the user enters -1, but the -1 should not be added to the sum.
do
{
printf("Enter a number. Enter -1 to stop\n");
scanf(" %d", &n);
sum = sum + n;
printf("The sum is %d\n", sum);
}
while (n >= 0);
{
return 0;
printf("Have a nice day!\n");
}
Where am I going wrong? How do I get the -1 to avoid being added to the sum?

Answer 1

You need to declare n, i before use. And you need a break when detecting input -1.
int sum = 0, n = 0;
do
{
printf("Enter a number. Enter -1 to stop\n");
scanf("%d", &n);
if( -1 == n )
break; // <-- break here to get out the loop
sum = sum + n;
printf("The sum is %d\n", sum);
} while (1); // <-- note while(1) should be used instead of while(n >= 0), this makes sure your loop only exits when detecting input -1
Also there's no point adding a printf after you return - ie this code can be removed:
{
return 0;
printf("Have a nice day!\n"); // <-- any code after `return` is meaningless
}

Answer 2

There are a couple of ways to avoid having -1 added to the sum. Let's look at your code, first:
do
{
printf("Enter a number. Enter -1 to stop\n");
scanf(" %d", &n);
sum = sum + n;
printf("The sum is %d\n", sum);
}
while (n >= 0);
The do/while loop will exit if n is less than zero. This is one way to catch the -1 as a sentinel value, so no problem there.
Unfortunately, your code reads the number, adds it to the total, and then checks if it should stop. With that order, you always add before checking.
The most obvious, brute-force way to avoid adding the number is to add an explicit check:
do {
printf ... scanf ... ;
if (n != -1)
sum = sum + n;
} while (n >= 0);
Another way is to cheat. Given that -1 always comes at the end of the list, just undo that at the end:
do {
printf... scanf...;
sum = sum + n;
} while (n >= 0);
sum = sum - n; // CHEAT - just subtract out the last number
printf("The sum is %d\n", sum);
Finally, there is a trick you can use. Sometimes in these situations, you can "shift" the loop, so that you actually do things slightly out of order. In this case, you would like to test before you add. So put the add at the top of the loop, and just set things up so that the very first add is harmless:
n = 0; // X + 0 is X, so adding 0 is harmless.
do {
sum = sum + n;
printf ... scanf ...;
} while (n >= 0);
This may make you uncomfortable for a minute, but it's an idiom you'll see a lot. The loop code has been "twisted" inside the loop, so that the top of the "natural" loop does not start at the top of the actual loop body. It's a good trick to know.

Answer 3

The behavior of your do-while loop will be clearer if you think of it in terms of the GOTOs the compiler sees it as:
// prior code before here...
START_OF_DO:
printf("Enter a number. Enter -1 to stop\n");
scanf(" %d", &n);
sum = sum + n;
printf("The sum is %d\n", sum);
if ( n >= 0 ) {
goto START_OF_DO; // this is what "while (n >= 0);" does
}
return 0;
printf("Have a nice day!\n"); // <-- note that this will never print, because you have already returned
Formed this way, you can see that your sum = sum + n happens before you check if ( n >= 0).
(note I am not suggesting that you ever use goto! But sometimes it's helpful to see these things in terms of the machine instructions the compiler emits.)

Answer 4

You're overwriting the value of n before it checks the condition to run the loop again.
Make these changes and it should work now.
int input = 0; // Change
int n = 0; // Change
do
{
printf("Enter a number. Enter -1 to stop\n");
scanf(" %d", &input); // Change
if(input >=0 ) // Change
{
n = input; // Change
sum = sum + n;
printf("The sum is %d\n", sum);
}
}
while (input != -1); // Change
{
return 0;
printf("Have a nice day!\n");
}