How to allow user to exit while loop from terminal?

scanf returns information in two different ways: in the variable i, and as its return value. The content of the variable i is the number that scanf reads, if it is able to return a number. The return value from scanf indicates whether it was able to read a number.

Your test i != EOF is fundamentally a type error: you're comparing the error indicator value EOF to a variable designed to hold a floating-point number. The compiler doesn't complain because that is accidentally valid C code: EOF is encoded as an integer value, and that value is converted to a floating-point value to perform the comparison. In fact, you'll notice that if you enter -1 at the prompt, the loop will terminate. -1 is the value of the EOF constant (on most implementations).

You should store the return value of scanf, and store it into a separate variable. If the return value is EOF, terminate the loop. If the return value is 1, you have successfully read a floating-point value.

If the return value is 0, the user typed something that couldn't be parsed. You need to handle this case appropriately: if you do nothing, the user's input is not discarded and your program will loop forever. Two choices that make sense are to discard one character, or the whole line (I'll do the latter).

double i;
double array[42];
int x = 0;
int r = 0;
while (r != EOF) {
    printf("type in a number: \n");
    r = scanf("%f", &i);
    if (r == 1) {
        /* Read a number successfully */
        array[x] = i;
        x++;
    } else if (r == 0) {
       printf("Invalid number, try again.\n");
       scanf("%*[^\n]"); /* Discard all characters until the next newline */
    }
}

You should also check that x doesn't overflow the bounds of the array. I am leaving this as an exercise.

I want to do it by typing in something that's not a number

Then get the input as a string, and exit if it cannot be converted to a number.

char buf[0x80];
do {
    fgets(buf, sizeof(buf), stdin);
    if (isdigit(buf[0])) {
        array[x++] = strtod(buf);
    }
} while(isdigit(buf[0]);

In case of no input scanf() does not set i to EOF but rather can return EOF. So you should analyze scanf() return code. By the way you can receive 0 as result which actually means there is no EOF but number cannot be read.

Here is example for you:

#include <stdio.h>
#define MAX_SIZE 5

int main()
{
    int array[MAX_SIZE];
    int x = 0;
    int r = 0;
    while (x < MAX_SIZE)
    {
        int i = 0;
        printf("type in a number: \n");
        r = scanf("%d",&i);
        if (r == 0)
        {
            printf("ERROR!\n");
            break;
        }
        if (r == EOF)
        {
            printf("EOF!\n");
            break;
        }
        array[x]=i;
        x++;
    }
}

You cannot write 'EOF'.. since you are reading into a number... EOF equals -1.. so if he enterd that, the loop would stop

You can test for the return value of the scanf function. It returns EOF on matching failure or encountering an EOF character.

printf("type in a number:" \n);    
while(scanf("%f",&i)!=EOF){
    array[x]=i;
    x++;
    printf("type in a number:" \n);  
}