Pass an unknown number of integers with scanf - c

Question

I need to create an array of ints of an unknown size and pass them all. My code looks like this:
int FillTable(int a[], int max){
int i;
int x = 0;
int m = 0;
for (i = 0; i < max; i++){
printf("Fill the table with integers: ");
scanf("%d", &m);
if (m != "" && m != NULL){
a[i] = m;
}else if (m != "" && m == NULL){
a[i] = 0;
}else{
break;
}
}
printf("\n");
return 0;
}
I know you can pass multiple ints separated by spaces with something like:
scanf("%d %d %d", &var1, &var2, &var3);
But I don't know how to pass a number of integers that I don't know how many will be there. Could I create a string with a bunch of %d and just repeat that for max times? I don't know, but right now, I just ask for ints until the array is full, and I need to be able to have the array be smaller than max if the user doesn't enter in enough values. Does anyone have any ideas as to how I would go about scanning for an unknown number of integers?

Answer 1

Does anyone have any ideas as to how I would go about scanning for an unknown number of integers?
This calls for Dynamic memory allocation!
One way of going with scanning unknown number of integers is, firstly allocate an integer array with size to hold max number of integers.
How to know whether user has ended his input?
If you are only scanning in positive integers from user at array entries then prompt him to end his input by inputting a negative number
or if you are sure about the range of input entries then break out of loop, when user enters input out of range
Example: (considering user inputs only positive numbers)
//creating a large enough array to store user inputs
int *array = malloc(sizeof(int) * max);
//check if memory was allocated or not
if(array == NULL)
{
printf("memory allocation problem!");
exit(1);
}
//integer to make note of size of array or you can use the sizeof() function instead
int size_of_array = 0;
for (i = 0; i < max; i++)
{
printf("Fill the table with integers: ");
if(scanf("%d", &m) != 1) //check for scanf return value
{
printf("wrong input, try again");
char consume; //to consume the character
scanf("%c", &consume);
i--;
continue;
}
if (m > 0) //if positive number, accept
{
array[i] = m;
size_of_array++;
}
else //else break out of scanning
{
break;
}
}
//do the program.....
//don't for get to free the memory at the end
free(array);
here's a working example: https://ideone.com/BHN4sk

Answer 2

You are trying to do something that is not necessary. To predict the size of the array and reallocate the appropriate exact size would be computationally more expensive (in terms of cpu time) so that benefit of saving the memory that you already had allocated is not enough.
The size of the array in c is stored somewhere that not necessarily has anything to do with the array itself. So you simply need to know how many of the array elements are interesting for the program and nothing else.
You could have something like
struct Array {
int *data;
size_t size;
size_t count;
};
where size is the total size of the array, count is the number of elements in the array and data are the elements. I use this pattern a lot and it's useful, specially if combined with realloc() as it saves from unecessarilly reallocating memory too many times which is expensive at the cost of using slightly more memory than actually needed.
But systems today have way more memory than can be used (except if you use Android Studio, which can use as much memory as your computer has).

Answer 3

First, m != "" && m != NULL probably does not do what you think it does. You're probably coming from a different language. What (I think) that statement does is compare the value in the integer variable m to the address of the string literal "" and then compare it to NULL (which evaluates to 0).
Second, scanf by default reads until either a space or a newline.
scanf returns a negative number on failure, so your code should look like this:
for (i = 0; i < max; i++){
printf("Fill the table with integers: ");
if(scanf("%d", &m) > 0) {
a[i] = m;
}
else {
break;
}
}
I left out the a[i] = 0 branch because I don't understand what you wanted there.
Also, you never use the variable x - unless there is more code that you left out.

Answer 4

your problem isn't understand for me properly,however i think this will be helped to you
int arraySize = 200; // Or whatever
int *array_ = malloc(arraySize * sizeof(int));
use this and the pass the *array_ as parameter,first defined array size or get array size as a user input,and run a for loop till size of array

Answer 5

You should decide how the user can stop his input (and include this info in your prompt). A quick-and-dirty way would be "enter anything that is not a number". I chose this way of terminating input, because it's easy to implement (hence, quick and dirty):
printf("Fill the table with integers; terminate with 'x':\n");
for (i = 0; i < max; i++)
{
int result = scanf("%d", &a[i]);
if (result != 1)
break;
}
Note:
The prompt tries to be user-friendly
The scanf function puts the number straight into the array, without using any intermediate variable
The scanf function returns how many numbers it read, which is normally 1; if it's not 1, then the user entered x or anything else
When the code finishes, i holds the number of iterations, which shows how many numbers were read from the user.
Your input function should return the size of the array:
return i; // instead of "return 0"
You might also want to clean the stdin buffer - discard anything that the user entered to terminate the array:
while (getchar() != '\n')
{
// do nothing - keep consuming input until end-of-line
}