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.

Answer:

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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