Arrays and pointers in C

Question

I have this piece of program and there are some parts that I don't understand...

‪#‎include‬ <stdio.h>
void increment( int *ptr) {++*ptr; }

int main(){
int a[] = { 5, 10 }, i = 0; // We have: vector a=[5,10] | i=0

increment(a); //This is the same as increment(&a) isn't it? 
              //So we have: a=[6,10] | i=0

increment(&i); // We increment i so: a=[6,10] | i=1

increment(&a[i]); //We increment the element at the position a[1]: a=[6,11] | i=1

increment(a+i); //OMG! What does it mean? a is an array isn't it? Then how can we
                 //make "a+1"? and how is this an address?

printf("\nResult: i= %d\n", i);
printf( "a[0] = %d\n" "a[1] = %d\n", a[0], a[1]);
return 0;

}

The printf s return:

 i=1
 a[0]=6
 a[1]=12

How can this be explained?

Answer 1

When you pass an array to a function, the array decays to a pointer to the first element. So in the function, doing eg *ptr is dereferencing the first element in the array.

Also, because arrays decays to pointers, doing eg a[i] is equal to *(a + i) .

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Unrelated side-note: Because of the commutative of addition, the expression *(a + i) can be written as *(i + a) which leads to i[a] actually being a valid expression.

Answer 2

Arrays and pointer arithmetic are the same in C. If you've not come across pointer arithmetic here's a tutorial on it: C Pointer Arithemtic

In essence:

a[3]

is the same as

*(a + 3)

Where a is a pointer.

Answer 3

When we give the name of array in the function call, the address of the arrya's first element is passed

ie; increment(a); is same as increment(&a[0]);

So while giving

incrementa(a+i);

since i and array a[] are both integers effectively that becomes

incrementa ( base address of array a (ie; &a[0]) + value of i )

i just acts as the offset to that array element

That it becomes the address of the second element which is &a[1]

so the 11 is modified to 12 while incrementing the value by the function