a function that works with array elements - c

Question

I need to write a function that subtracts digits.
If user inputs 2345, the output should be 111 (5-4, 4-3, 3-2); another example would be 683, where the output should be 25 (3-8(abs value is taken), 8-6).
I have wrote the following code which works only when the size of the array is declared.
int subtraction(int arr[], int size) {
int sub = 0;
for (int i = 0; i < size-1; i++) {
sub = sub * 10 + abs(arr[i] - arr[i+1]);
}
return sub;
}
However, the number that the user inputs is random and can have various digits, so I don't know what limit to put in the for loop.
For example:
int arr[] = {1, 2, 55, 56, 65, 135}, i;
subtraction(arr, 6);
for (i=0; i<6; i++)
printf("%d ", arr[i]);
expected output: 0 0 0 1 1 22
The function is supposed to subtract the second-to-last digit from the last one, by the way , / from right to left / from a random number that the user inputs ; for example if the input is 5789, the output is supposed to be 211 (9-8, 8-7, 7-5); if user inputs a negative number, the program should take it's absolute value and then do the subtracting. If user input is a one digit number the result should be 0.
The function I wrote only works when the size of the array is declared. I don't know how to make it work when the size is undeclared (pointers and malloc are required I believe, as that's what I managed to find out by googling for ages, but unfortunately, I don't know how to do it).
please help?

Answer 1

You are not actually changing any values, here is the line you need to look at.
sub = sub * 10 + abs(arr[i] - arr[i+1]);
As you are printing the array you actually need to store the calculated value in the array again.

Answer 2

#include <stdio.h>
#include <stdlib.h>
int subtract(int n)
{
int factor = 1;
int total = 0;
int lastPlace = n%10;
n /= 10;
while (n>0)
{
total += (factor * abs((n%10) - lastPlace));
factor *= 10;
lastPlace = n%10;
n /= 10;
}
return total;
}
void subtractArray(int* arr, unsigned int size)
{
for (int i=0; i<size; ++i)
{
if (arr[i] < 0)
arr[i] = abs(arr[i]);
arr[i] = subtract(arr[i]);
}
}
int main()
{
int arr[] = {1, 2, 55, 56, 65, 135};
int size = sizeof(arr)/ sizeof(arr[0]);
subtractArray(arr, size);
for (int i=0; i<size; ++i)
{
printf("%d ", arr[i]);
}
return 0;
}

Answer 3

Here is a simple code that solve your problem :)
#include <stdio.h>
#include <stdlib.h>
int *subtraction(int arr[], int size)
{
int *sub = calloc(sizeof(int*) , size), i = 0, rev; //allocating memory
for (i = 0; i < size; i++)
{
rev = 0;
arr[i] = abs(arr[i]);
for (int a = 0; arr[i] != 0; arr[i] /= 10)
rev = (rev * 10) + (arr[i] % 10);
for (i; (rev / 10) != 0; rev /= 10) //the loop ends when rev = 0
sub[i] = ((sub[i] * 10) + abs( (rev % 10) - ((rev / 10) % 10) )); //easy math => for example rev = 21 > sub[i] = (0 * 10) + ( (21 % 10) - ((21 / 10) %10)) = abs(1 - 2) = 1;
}
return sub;
}
int main()
{
int arr[] = {-9533, 7, -19173}, i;
int len = sizeof(arr)/sizeof(arr[0]); //size of arr
int *sub = subtraction(arr, len);
for(int i = 0; i < len; i++) //for test
printf("%d ", sub[i]);
return 0;
}
output for {1, 2, 55, 56, 65, 135}:
0 0 0 1 1 22
output for {987654321, 123456789, 111111111} :
11111111 11111111 0
output for {38279}:
5652
output for {-9533, 7, -19173}:
420 0 8864

Answer 4

Well as for the array of undefined size. What you probably want is a dynamically allocated array.
Here we get the number of array elements based on user input, within limits, of course.
first we're gonna get the number from the user using fgets() which will give us a string, then we'll use strtol() to convert the number part to scalar (int). you can use scanf("%d", &n) if you want.
Then we can count the digits from that number, and that value will be the number of elements of our array.
#include <stdio.h>
#include <stdlib.h> //for strtol(), malloc() and NULL guaranteed
//you may also want to add
#include <limits.h>
#include <errno.h>
#define MAX_STRLEN 12 // can hold all digits of INT_MAX plus '\0' and a posible, AND very likely, '\n'
#define DEC 10 // for strtol base argument
/*
* I'm lending you my old get_n_dits() function that I used to count decimal digits.
*/
int get_n_dits(int dnum) {
unsigned char stop_flag = 0; //we'll use to signal we're done with the loop.
int num_dits = 1, dpos_mult = 1; //num_dits start initialized as 1, cause we're pretty sure that we're getting a number with at least one digit
//dpos_mult stands for digital position multiplier.
int check_zresult; //we'll check if integer division yields zero.
/**
* Here we'll iterate everytime (dnum / dpost_mult) results in a non-zero value, we don't care for the remainder though, at least for this use.
* every iteration elevates dpost_mult to the next power of ten and every iteration yielding a non-zero result increments n_dits, once we get
* the zero result, we increment stop_flag, thus the loop condition is no longer true and we break from the loop.
*/
while(!stop_flag) {
dpos_mult *= 10;
check_zresult = dnum / dpos_mult;
(check_zresult) ? num_dits++ : stop_flag++;
}
return num_dits;
}
int main(void) {
int num, ndits; //we'll still using int as per your code. you can check against INT_MAX if you want (defined in limits.h)
int *num_array = NULL; //let's not unintentionally play with an unitialized pointer.
char *num_str = malloc(MAX_STRLEN); //or malloc(sizeof(char) * MAX_STRLEN); if there's any indication that (sizeof(char) != 1)
printf("please enter a number... please be reasonable... or ELSE!\n");
printf(">>> ");
if(!fgets(num_str, MAX_STRLEN, stdin)) {
fprintf(stderr, "Error while reading from STDIN stream.\n");
return -1;
}
num = (int)strtol(num_str, NULL, DEC); //convert the string from user input to scalar.
if(!num) {
fprintf(stderr, "Error: no number found on input.\n");
return -1;
}
ndits = get_n_dits(num);
if(ndits <= 0) {
fprintf(stderr, "Aw, crap!\n");
return -1;
}
printf("number of digits: %d\n", ndits);
num_array = malloc(sizeof(int) * ndits); //now we have our dynamically allocated array.
return 0;
}