array goes above limit when taking user input - c

Question

I am taking a couple of numbers as input from the user, storing them into a file and then reading from that file, sorting the numbers and displaying them to the user. The program works except for 1 issue. When taking input from the user it asks for an extra elememnt than the specifies limit. This extra element doesn't get stored anywhere. I have tried reducing the limit value by one but that results in loss of 1 element(i.e. 1 element becomes 0). I understand that this may be a very newbie issue but I coudn't find any direct answer to this issue. I wrote the program in Visual Studio 2010 as a C program. here's the code:
#include <stdio.h>
int arr[50];
int n;
int writefile()
{
FILE *ptr;
ptr = fopen("Sort.txt","w");
if(ptr==NULL)
{
printf("No such file exists\n");
fclose(ptr);
return 0;
}
else
{
int i;
printf("Enter number of elements in array\n");
scanf("%d", &n);
printf("Enter %d elements\n", n);
for(i=0; i<n; i++) //The issue is here
{
scanf("%d\n",&arr[i]); //say the user enter n=3, then after adding 1,2,3 it will ask for another 4th element but only 1,2,3 will get stored.
}
fwrite(arr, sizeof(int), n, ptr);
fclose(ptr);
printf("done\n");
}
}
void sortarr(int arr1[]);
int readfile()
{
FILE *ptr;
ptr = fopen("Sort.txt","r");
if(ptr==NULL)
{
printf("No such file exists\n");
fclose(ptr);
return 0;
}
else
{
int i;
int arrb[50];
fread(arrb, sizeof(int), n, ptr);
printf("Before sorting data\n");
for(i=0; i<n; i++)
{
printf("%d",arrb[i]);
}
printf("\n");
sortarr(arrb);
printf("After Sorting data\n");
for(i=0; i<n; i++)
{
printf("%d",arrb[i]);
}
printf("\n");
}
}
void sortarr(int arr1[])
{
int i,j,temp;
for (i = 0; i < n; i++) {
for (j = i + 1; j < n; j++) {
if (arr1[i] > arr1[j]) {
temp = arr1[i];
arr1[i] = arr1[j];
arr1[j] = temp;
}
}
}
}
int main()
{
writefile();
readfile();
return 0;
}

Answer 1

scanf("%d\n",&arr[i]);
Should be:
scanf("%d",&arr[i]);
If you ask it to ignore white space after the number, it will have to keep reading until it reads some non-whitespace in order for it to ensure it has ignored all the whitespace. You definitely don't want scanf to try to ignore whitespace after it has read the number, you want it to terminate.
The %d format specifier already ignores whitespace before the number.

Answer 2

Let's just narrow this down to the following code:
scanf("%d", &n);
for (i = 0; i < n; i++)
{
scanf("%d\n", &arr[i]);
}
Notice the difference in the scanf calls? The first call doesn't have a \n in its format string. The second call does.
Remove the \n from the scanf format string (in the second call). Things should work more like you're expecting then.
An online manual page for scanf says:
The format string consists of a sequence of directives which describe how to process the sequence of input characters... A directive is one of the following: A sequence of white-space characters (space, tab, newline, etc.; see isspace(3)). This directive matches any amount of white space, including none, in the input.
So the looped over scanf (with the extra \n in it) was effectively reading the number, the newline, the next number, then realizing it's read all of the white space it could. scanf then returned with the first number assigned to your array entry and the next number ready for the next call to reading from standard input. This left things effectively offset appearing like you saw and requiring the extra number before ending (in order that it could detect that the white space had ended).