scanf("%d", int) good yesterday, not today. What is going on? - c

Question

I am getting comfortable with the foundations of C:-
There are two iterations of scanf() in this program. The first is fully functional.
The second appearance does nothing and instead quits the program and returns to the standard command prompt line.
Also, my second for loop only prints the first 4 numbers instead of 5.
Here is the code :
int main() {
int i;
int iNum[4]
int iMenu = 0;
printf("\n\n\tPlease enter five numbers: ");
for (i = 0; i < 4; i++) {
scanf("%d ", &iNum[i]);
}
printf("\n\tThank you. Here are your numbers: \n\t");
for (i = 0; i < 4; i++) {
printf("%d", iNum[i]);
}
printf("\n\tMenu:");
printf("\n1\tSort your numbers ascending.");
printf("\n2\tSort your numbers descending.");
printf("\n3\tQuit program");
printf("\n\nWhat would you like to do?\n\tEnter your option here: ");
scanf(" %d", &iMenu);
switch (iMenu) {
case 1:
printf("\nCase1testPrint");
sortAsc();
break;
case 2:
printf("\nCase2testPrint");
sortDesc();
break;
. . .//rest of code }
As soon as the program reaches the second scanf(), the program ends and returns to command prompt.
This is baffling, and I have tried many variations on what could be wrong. Please provide what you may.
Here is what it looks like if I run the program. Notice my entry "1" appears in the command line instead of the program:
C:\Users\Cortland\Documents\C projects>gcc arraysort.c
C:\Users\Cortland\Documents\C projects>a
Please enter five numbers: 3
6
5
4
5
Thank you. Here are your numbers:
3 6 5 4
Menu:
1 Sort your numbers ascending.
2 Sort your numbers descending.
3 Quit program
What would you like to do?
Enter your option here:
C:\Users\Cortland\Documents\C projects>1

Answer 1

Your for loop reads four numbers, not five.
You enter five numbers. The first four are read in the for loop. The fifth is read by
scanf(" %d", &iMenu);
So you've entered 3 6 5 4 as the four numbers, and then 5 as the choice of what to do. You haven't shown us the entire switch statement, but my guess is that it doesn't handle the value 5, and that it falls through and your program terminates.
(Also, you should always check the value returned by scanf(), and take some action if it indicates that the input operation failed.)

Answer 2

There are a couple of errors.
First off, the second line of your main() function contains a syntax error - there should be a semicolon at the end. This should stop your program from even compiling.
Second, even though you say you want the user to enter 5 numbers, your array and loops all use the number 4. The reason your program still allows you to enter five numbers, however, is because of how you've written the formatting string in your calls to scanf(). Instead of:
scanf("%d ", &iNum[i]);
use
scanf("%d", &iNum[i]);
Notice the lack of a space at the end of the formatting string? That's where your problems come in. Even though you are allowed to enter five numbers, the program only stores and checks four of them. The last number is stuck in the input stream - until, that is, your next call to scanf(). This fetches your last number, which is put through the switch/case statement (failing all cases, since you didn't input "1", "2" or "3"), and the program reaches the end of main() and exits. The "1" in the command line comes from trying to input something when the program has already ended.
Summarized:
1: Add the missing semicolon
2: Change the number used in the declaration of iNum and in the for loops from 4 to 5
3: Edit your scanf() calls, removing whitespace from the formatting strings.
This fixed the problem on my end.