# Why in the C language is array[6] equal to 6[array]?

## Question:

Note: Question I saw on the SO in English , but I thought it was interesting to put it here (because we still don't have many C questions):

Why in the C language, this code prints "true"?

``````#include <stdio.h>

int main(void) {
int reais[10];
for(int i=0; i<10; ++i) reais[i] = i;
int num = reais[6];
int num2 = 6[reais];
if(num == num2) printf("verdadeiro"); else printf("falso");
return 0;
}
``````

Nota: Compilei isso com C99, ele NÃO COMPILA no C89 (porque não pode declarar variável dentro do for() )

Quem quiser pode copiar e colar isso no ideone.com para ver compilar e rodar.

Referenciando o standard do C:

# 6.5.2.1 Array subscripting

Constraints
One of the expressions shall have type “pointer to complete object type“, the other
expression shall have integer type, and the result has type “type“.

Semantics
A postfix expression followed by an expression in square brackets `[]` is a subscripted
designation of an element of an array object. The definition of the subscript operator `[]`
is that `E1[E2]` is identical to `(*((E1)+(E2)))`. Because of the conversion rules that
apply to the binary + operator, if `E1` is an array object (equivalently, a pointer to the
initial element of an array object) and `E2` is an integer, `E1[E2]` designates the `E2`-th
element of `E1` (counting from zero).

O relevante é que ao escrever `a[b]`, um dos valores deve ser um ponteiro (ou uma array que decaia a um ponteiro) e outro deve ser um tipo integral. Não é especificado nenhuma restrição sobre qual deve ser o que. O segundo ponto é que `a[b]` é equivalente a `*(a+b)`. Assim temos:

``````a[b] = *(a+b) = *(b+a) = b[a]
``````

The same can also be applied to literal strings, as they are `const char[]` :

``````char M = 4["Ola Mundo"];
``````

While this syntax is perfectly valid and legal, it's not good practice to access arrays by reversing the operands like this. Reading is quite unintuitive and there are no advantages.

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