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EOF Help NEEDED!

This is a discussion on EOF Help NEEDED! within the C Programming forums, part of the General Programming Boards category; Hi i have a program getting input from a user, I need to incorporate a way for the program to ...

  1. #1
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    EOF Help NEEDED!

    Hi i have a program getting input from a user, I need to incorporate a way for the program to end if the user presses CTRL + D while entering their input. I know you use EOF but i dont understand how to use it and where to use it. heres my code:

    Code:
    #include <stdio.h>
    
    void selection_sort(int a[], int size);
    
    int main(void) {
    
        int a[100], i, j, N;
    
        printf("How many numbers will you be entering: ");
        scanf("%d", &N);
    
        printf("Enter numbers to be sorted: ");
        for(i = 0; i < N; i++) {
            scanf("%d", &a[i]);
        }
        
        selection_sort(a, N);
        
        printf("Your sorted list of integers is : ");
        for(j = 0; j < N; j++) {
        printf("%4d ", a[j]);
        }
        printf("\n");
    
        return 0;
        }
    
    void selection_sort(int a[], int size) {
        int i, j, x;
    
        for(i = 1; i <= size - 1; i++) {
    
        j = i;
        x = a[i];
        
        while(a[j - 1] > x && j > 0) {
            a[j] = a[j-1];
            j = j - 1;
        }
        
        a[j] = x;
        }
    }

  2. #2
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    I'm helping by adding comments, so please read through your code below. I don't think their is a way to exit the program with CTRL + D, but anyone can close a program with CTRL + C

    Code:
    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43
    #include <stdio.h> void selection_sort(int a[], int size); int main(void) { int a[100], i, j, N; printf("How many numbers will you be entering: "); scanf("%d", &N); printf("Enter numbers to be sorted: "); //Its nice to space out your for loops with an empty line for(i = 0; i < N; i++) { //Because Arrays are pointers, you don't need the & also add an /n //scanf("%d /n", a[i[); scanf("%d", &a[i]); } selection_sort(a, N); printf("Your sorted list of integers is : "); for(j = 0; j < N; j++) { printf("%4d ", a[j]); } printf("\n"); return 0; } //While this is correct you should have a function as follows //void selection_sort(int *a, int size){ void selection_sort(int a[], int size) { int i, j, x; //remove the smaller than or equal to and you can just have //for(i = 1; i < size; i++){ for(i = 1; i <= size - 1; i++) { j = i; x = a[i]; //remember removing the smaller than or equal to //while(a[j] > x && j > 0){ while(a[j - 1] > x && j > 0) { a[j] = a[j-1]; j = j - 1; } a[j] = x; } }
    I might be missing some errors, I'm new to trying to help people with out giving them the full answer =/. Memory allocation and everything else is pretty much terminated when the process is closed, but its not nice to leave it like that... but the code you have, shouldn't raise any problems.

  3. #3
    and the hat of wrongness Salem's Avatar
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    Make your first loop this, then study the output.
    Code:
        for(i = 0; i < N; i++) {
            int result = scanf("%d", &a[i]);
            printf("Scan result = %d, is it EOF=%d\n", result, result == EOF );
        }
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  4. #4
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    Quote Originally Posted by 0x47617279 View Post
    Code:
    //Because Arrays are pointers, you don't need the & also add an /n
    //scanf("%d /n", a[i[);
    scanf("%d", &a[i]);
    clmoore3's scanf()-call is correct. a[i] is one element of the array (an int), thus the & is necessary.

    And please don't post code inside a table.

    Bye, Andreas

  5. #5
    TEIAM - problem solved
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    Quote Originally Posted by AndiPersti View Post
    clmoore3's scanf()-call is correct. a[i] is one element of the array (an int), thus the & is necessary.

    And please don't post code inside a table.

    Bye, Andreas
    Also, arrays are not pointers.
    AndiPersti likes this.
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