How do Array Pointers behave? - c

Question

I was trying to write the code for storing multiplication table in array.
First I wrote the following code.
#include<stdio.h>
void Multiplication(int arr[][10], int num,int size);
int main(){
int arr[3][10];
Multiplication(&arr[0],2,3);
Multiplication(&arr[1],7,3);
Multiplication(&arr[2],9,3);
for(int i=0;i<3;i++){
printf("\n\nMultiplication Table of %d\n\n",arr[i][0]);
for(int j=0;j<10;j++){
printf("%d * %d = %d\n",arr[i][0],j,arr[i][j]);
}
}
return 0;
}
void Multiplication(int arr[][10], int num, int size){
for(int j=0;j<size;j++){
for(int i=0;i<10;i++){
arr[j][i] = num * (i+1);
}
}
}
When I was checking whether there is any reduntant part, I removed a little bit of pieces.
#include<stdio.h>
void Multiplication(int *arr, int num,int size);
int main(){
int arr[3][10];
Multiplication(arr[0],2,3);
Multiplication(arr[1],7,3);
Multiplication(arr[2],9,3);
for(int i=0;i<3;i++){
printf("\n\nMultiplication Table of %d\n\n",arr[i][0]);
for(int j=0;j<10;j++){
printf("%d * %d = %d\n",arr[i][0],j,arr[i][j]);
}
}
return 0;
}
void Multiplication(int *arr, int num,int size){
for(int i=0;i<10;i++){
arr[i] = num * (i+1);
}
}
Now, I noticed that &arr[0] and arr[0] are behaving same? I wanna know why is it so?

Answer 1

The following is the question asked:
I noticed that &arr[0] and arr[0] are behaving same? I wanna know why is it so?
We have
int arr[3][10];
arr[0] is a int[10], which is to say an array of 10 int values.
But it's being treated as a pointer. When an array is treated as a pointer, it produces a pointer to the first element of the array. So this means arr[0] is automatically converted to a int* pointing to the first element of the first row of arr when necessary.
&arr[0] means &(arr[0]). It produces a int(*)[10] (a pointer to an array of 10 int values) pointing to the first row of arr.
To recap,
arr[0] is a int[10] which degrades to a int*.
It points to the first element of the first row of arr.
&arr[0] is a int(*)[10].
It points to the first row of arr.
So, arr[0] is NOT equivalent to &arr[0], contrary to your assertion.
And this lack of equivalence is why you had to change from &arr[0] when you when you had int arr[][10] (which means int (*arr)[10] a parameter), to arr[10] when you had int *arr as the parameter.

Answer 2

The both programs are incorrect.
You declared a two dimensional array
int arr[3][10];
So expressions like this arr[0], arr[1], arr[2] have the array type int[10]. That is they produce the first "row", the second "row" and the third "row" of the two-dimensional array.
In the first program you are calling the function Multiplication like
Multiplication(&arr[0],2,3);
Multiplication(&arr[1],7,3);
Multiplication(&arr[2],9,3);
and within the function there are two nested loops
void Multiplication(int arr[][10], int num, int size){
for(int j=0;j<size;j++){
for(int i=0;i<10;i++){
arr[j][i] = num * (i+1);
}
}
}
according to the argument specified for the parameter size the function deals with a two dimensional array with three elements (rows) of the array type int[10] using a pointer to the first element of the array. As a result for the argument expressions &arr[1] and &arr[2] there is an access to the memory beyond the original two dimensional array.
You may consider these calls
Multiplication(&arr[0],2,3);
Multiplication(&arr[1],7,3);
Multiplication(&arr[2],9,3);
like
Multiplication( arr, 2, 3 );
Multiplication( arr + 1, 7, 3 );
Multiplication( arr + 2, 9, 3 );
The array designator arr is implicitly converted to a pointer to its first element (row) of the type int ( * )[10].
As you are setting only one "row" within the array then the function can be declared and defined the following way
void Multiplication(int arr[][10], int num ){
for(int j=0;j < 10;j++){
( *arr )[j]= num * (j+1);
}
}
and called like
Multiplication( arr, 2 );
Multiplication( arr + 1, 7 );
Multiplication( arr +2, 9 );
Of you want to set elements of a sub-array then the function can be defined the way it is defined initially but called like
Multiplication( arr, 2, 3 );
Multiplication( arr + 1, 7, 2 );
Multiplication( arr +2, 9, 1 );
In the second program the function is declared like
void Multiplication(int *arr, int num,int size){
for(int i=0;i<10;i++){
arr[i] = num * (i+1);
}
}
and you are calling it like
Multiplication(arr[0],2,3);
Multiplication(arr[1],7,3);
Multiplication(arr[2],9,3);
But within the function the parameter size is not used. There is used the magic number 10. And the passed argument 3 does not make a sense.
You should declare the function like
void Multiplication(int *arr, int num,int size){
for(int i=0;i<size;i++){
arr[i] = num * (i+1);
}
}
and call it like
Multiplication(arr[0],2,10);
Multiplication(arr[1],7,10);
Multiplication(arr[2],9,10);