Can't understand this way to calculate the square of a number

c arrays pointers c99 variable-length-array
Emanuel picture Emanuel · Jan 7, 2015 · Viewed 9.2k times · Source

I have found a function that calculates square of a number:

int p(int n) {
    int a[n]; //works on C99 and above
    return (&a)[n] - a;
}

It returns value of n2. Question is, how does it do that? After a little testing, I found that between (&a)[k] and (&a)[k+1] is sizeof(a)/sizeof(int). Why is that?

Answer

Mark Lakata picture Mark Lakata · Jan 7, 2015

Obviously a hack... but a way of squaring a number without using the * operator (this was a coding contest requirement). 

(&a)[n]

is equivalent to a pointer to int at location

(a + sizeof(a[n])*n)

and thus the entire expression is

  (&a)[n] -a 

= (a + sizeof(a[n])*n -a) /sizeof(int)

= sizeof(a[n])*n / sizeof(int)
= sizeof(int) * n * n / sizeof(int)
= n * n

Related questions

Literal string initializer for a character array

In the following rules for the case when array decays to pointer: An lvalue [see question 2.5] of type array-of-T which appears in an expression decays (with three exceptions) into a pointer to its first element; the type of the resultant …

Read More
How to find the 'sizeof' (a pointer pointing to an array)?

First off, here is some code: int main() { int days[] = {1,2,3,4,5}; int *ptr = days; printf("%u\n", sizeof(days)); printf("%u\n", sizeof(ptr)); return 0; } Is there a way to find out the size of the array that ptr is pointing …

Read More
C pointer to array/array of pointers disambiguation

What is the difference between the following declarations: int* arr1[8]; int (*arr2)[8]; int *(arr3[8]); What is the general rule for understanding more complex declarations?

Read More