Use of a function on a 2d array - c

Question

I have written code which allows me to modify the elements of a 1D array within my function by passing the element of the array:
I print the original array
I pass each element of the array to the function
Within the function I add the value 50 to each element of the array
I then call the function, and print out to screen the modified element value (i.e the value of each element +50)
I have been able to do this for a 1D array, with example values in the array being (10,20,30) and the valued printed after modification being (60,70,80).
What I am hoping to do is adapt that code to work for 2D arrays, you will see my attempt at doing this below. This code focuses on the use of int, but once I understand how to achieve this I am hoping to adapt for the use of a 2D string as well.
With the code below:
My objective is
Print to screen the original 2D array
Pass each element of the 2D array to the function
Within the function add the value 50 to each element of the array
Then call the function, and print out the modified element values to the screen(expected result displayed on screen 60,61,etc,.)
So far I have been able to print the original 2D array to the screen. It is the function I think I am messing up and would appreciate any advice. Thank you.
#include <stdio.h>
#include <string.h>
#define M 4
#define N 2
int function(int **arr);
int main() {
int i, a;
int arr[N][M] = {10, 11, 12, 13, 14, 15, 16, 17};
// the int array first
for(i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
// Accessing each variable
printf("value of arr[%d] is %d\n", i, arr[i][j]);
}
}
printf("\n ***values after modification***\n");
a = function(&arr[i][0]);
// int array print results
for(int i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
printf("value of arr %d\n", arr[i][j]);
}
}
return 0;
}
int function(int **arr) {
int i;
int j;
for(int i = 0; i < 3; i++) {
for(size_t j = 0; j < 5; j++) {
arr[i][j] = arr[i][j] + 50;
}
}
}
My apologies in advance for silly mistakes I am very new to C.
Thank you in advance.

Answer 1

The function int function(int **arr) does not return an int so make it void.
When you call it, a = function(&arr[i][0]);, you do not use a after the assignment. I suggest that you remove a from the program completely since it's not used anywhere.
The call to the function, function(&arr[i][0]);, should simply be function(arr);
The function signature needs to include the extent of all but the outermost dimension:
void function(int arr[][M])
Inside the function, you use 3 and 5 instead of N and M. That accesses the array out of bounds.
In function, the i and j you declare at the start of the function are unused. Remove them.
arr[i][j] = arr[i][j] + 50; is better written as arr[i][j] += 50;
When initializing a multidimensional array, use braces to make it simpler to read the code:
int arr[N][M] = {{10, 11, 12, 13}, {14, 15, 16, 17}};
In main you mix int and size_t for the indexing variables. I suggest you settle for one type.
Remove unused header files (string.h)
Example:
#include <stdio.h>
#define N 2
#define M 4
void function(int arr[][M]) {
for(int i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
arr[i][j] += 50;
}
}
}
int main() {
int arr[N][M] = {{10, 11, 12, 13}, {14, 15, 16, 17}};
for(size_t i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
printf("value of arr[%zu][%zu] is %d\n", i, j, arr[i][j]);
}
}
printf("\n ***values after modification***\n");
function(arr);
// int array print results
for(size_t i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
printf("value of arr[%zu][%zu] is %d\n", i, j, arr[i][j]);
}
}
}
Since you print the array more than once, you could also add a function to do so to not have to repeat that code in main:
void print(int arr[][M]) {
for(size_t i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
printf("value of arr[%zu][%zu] is %d\n", i, j, arr[i][j]);
}
}
}

Answer 2

Two-Dimensional arrays in C (and C++) are actually one-dimensional arrays whose elements are one-dimensional arrays. The indexing operator [] has left-to-right semantics, so for a type arr[N][M] the first index (with N elements) is evaluated first. The resulting expression, e.g. arr[0], the first element in arr, is a one-dimensional array with M elements. Of course that array can be indexed again , e.g. arr[0][1], resulting in the second int in the first sub-array.
One of the quirks in the C language is that if you use an array as a function argument, what the function sees is a pointer to the first element. An array used as an argument "decays" or, as the standard says, is "adjusted" that way. This is no different for two-dimensional arrays, except that the elements of a two-dimensional array are themselves arrays. Therefore, what the receiving function gets is a pointer to int arr[M].
Consider: If you want to pass a simple integer array, say intArr[3], to a function, what the function sees is a pointer to the first element. Such a function declaration might look like void f(int *intPtr) and for this example is simply called with f(intArr). An alternative way to write this is void f(int intPtr[]). It means exactly the same: The parameter is a pointer to an int, not an array. It is pointing to the first — maybe even only — element in a succession of ints.
The logic with 2-dimensional arrays is exactly the same — except that the elements, as discussed, have the type "array of M ints", e.g. int subArr[M]. A pointer argument to such a type can be written in two ways, like with the simple int array: As a pointer like void f(int (*subArrPtr)[M]) or in array notation with the number of top-level elements unknown, like void f(int arr[][M]). Like with the simple int array the two parameter notations are entirely equivalent and interchangeable. Both actually declare a pointer, so (*subArrPtr)[M] is, so to speak, more to the point(er) but perhaps more obscure.
The reason for the funny parentheses in (*subArrPtr)is that we must dereference the pointer first in order to obtain the actual array, and only then index that. Without the parentheses the indexing operator [] would have precedence. You can look up precedences in this table. [] is in group 1 with the highest priority while the dereferencing operator * (not the multiplication!) is in group 2. Without the parentheses we would index first and only then dereference the array element (which must therefore be a pointer), that is, we would declare an array of pointers instead of a pointer to an array.
The two possible, interchangeable signatures for your function therefore are
void function( int (*arrArg)[M] ); // pointer notation
void function( int arrArg[][M] ); // "array" notation (but actually a pointer)
The entire program, also correcting the problems Ted mentioned, and without printing the original values (we know them, after all), is below. I have also adapted the initialization of the two-dimensional array so that the sub-arrays become visible. C is very lenient with initializing structures and arrays; it simply lets you write consecutive values and fills the elements of nested subobjects as the come. But I think showing the structure helps understanding the code and also reveals mistakes, like having the wrong number of elements in the subarrays. I have declared the function one way and defined it the other way to show that the function signatures are equivalent. I also changed the names of the defines and of the function to give them more meaning.
#include<stdio.h>
#define NUM_ELEMS_SUBARRAY 4
#define NUM_ELEMS_ARRAY 2
/// #arrArg Is a pointer to the first in a row of one-dimensional
/// arrays with NUM_ELEMS_SUBARRAY ints each.
void add50ToElems(int arrArg[][NUM_ELEMS_SUBARRAY]);
int main()
{
// Show the nested structure of the 2-dimensional array.
int arr[NUM_ELEMS_ARRAY][NUM_ELEMS_SUBARRAY] =
{
{10, 11, 12, 13},
{14, 15, 16, 17}
};
// Modify the array
add50ToElems(arr);
// print results
for (int i = 0; i < NUM_ELEMS_ARRAY; i++) {
for (int j = 0; j < NUM_ELEMS_SUBARRAY; j++)
{
printf("value of arr[%d][%d]: %d\n", i, j, arr[i][j]);
}
}
return 0;
}
// Equivalent to declaration above
void add50ToElems(int (*arrArg)[NUM_ELEMS_SUBARRAY])
{
for (int i = 0; i < NUM_ELEMS_ARRAY; i++)
{
for (size_t j = 0; j < NUM_ELEMS_SUBARRAY; j++)
{
//arrArg[i][j] = arrArg[i][j] + 50;
arrArg[i][j] += 50; // more idiomatic
}
}
}
Why is it wrong to pass a two-dimensional array to a function expecting a pointer-to-pointer? Let's consider what void f(int *p) means. It receives a pointer to an int which often is the beginning of an array, that is, of a succession of ints lying one after the other in memory. For example
void f(int *p) { for(int i=0; i<3; ++i) { printf("%d ", p[i]); }
may be called with a pointer to the first element of an array:
static int arr[3];
void g() { f(arr); }
Of course this minimal example is unsafe (how does f know there are three ints?) but it serves the purpose.
So what would void f(int **p); mean? Analogously it is a pointer, pointing to the first in a succession of pointers which are lying one after the other in memory. We see already why this will spell disaster if we pass the address of a 2-dimensional array: The objects there are not pointers, but all ints! Consider:
int arr1[2] = { 1,2 };
int arr2[2] = { 2,3 };
int arr3[2] = { 3,4 };
// This array contains addresses which point
// to the first element in each of the above arrays.
int *arrOfPtrToStartOfArrays[3] // The array of pointers
= { arr1, arr2, arr3 }; // arrays decay to pointers
int **ptrToArrOfPtrs = arrOfPtrToStartOfArrays;
void f(int **pp)
{
for(int pi=0; pi<3; pi++) // iterate the pointers in the array
{
int *p = pp[pi]; // pp element is a pointer
// iterate through the ints starting at each address
// pointed to by pp[pi]
for(int i=0; i<2; i++) // two ints in each arr
{
printf("%d ", pp[pi][i]); // show double indexing of array of pointers
// Since pp[pi] is now p, we can also say:
printf("%d\n", p[i]); // index int pointer
}
}
}
int main()
{
f(ptrToArrOfPtrs);
}
f iterates through an array of pointers. It thinks that the value at that address, and at the subsequent addresses, are pointers! That is what the declaration int **pp means.
Now if we pass the address of an array full of ints instead, f will still think that the memory there is full of pointers. An expression like int *p = pp[i]; above will read an integer number (e.g., 1) and think it is an address. p[i] in the printf call will then attempt to access the memory at address 1.
Let's end with a discussion of why the idea that one should pass a 2-dimensional array as a pointer to a pointer is so common. One reason is that while declaring a 2-dimensional array argument as void f(int **arr); is dead wrong, you can access the first (but only the first) element of it with e.g. int i = **arr. The reason this works is that the first dereferencing gives you the first sub-array, to which you can in turn apply the dereferencing operator, yielding its first element. But if you pass the array as an argument to a function it does not decay to a pointer to a pointer, but instead, as discussed, to a pointer to its first element.
The second source of confusion is that accessing elements the array-of-pointers uses the same double-indexing as accessing elements in a true two-dimensional array: pp[pi][i] vs. arr[i][j]. But the code produced by these expressions is entirely different and spells disaster if the wrong type is passed. Your compiler warns about that, by the way.