EOF Help NEEDED!

[clmoore3]

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:

#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;
    }
}

[0x47617279]

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

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#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.

[Salem]

Make your first loop this, then study the output.

    for(i = 0; i < N; i++) {
        int result = scanf("%d", &a[i]);
        printf("Scan result = %d, is it EOF=%d\n", result, result == EOF );
    }

[AndiPersti]

//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

[Click_here]

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.