Trying to intuit how a language works tends to be problematic. To understand this operation, you need to look it up in the spec: http://www.open-std.org/jtc1/s

Trying to intuit how a language works tends to be problematic. To understand this operation, you need to look it up in the spec: http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1256.pdf http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1256.pdf Section 6.5.2.1 "Array Subscripting" states in item (2) under "Semantics" on page 70: 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). In other words: a[5] is equal to (*((a)+(5))) 5[a] is equal to (*((5)+(a))) So what are the rules for E1+E2? Refer to section 6.5.6 "Additive Operators", item (8) under "Semantics" on page 83: When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the pointer operand points to an element of an array object, and the array is large enough, the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression. In other words, if the expression P points to the i-th element of an array object, the expressions (P)+N (equivalently, N+(P)) and (P)-N (where N has the value n) point to, respectively, the i+n-th and i−n-th elements of the array object, provided they exist. In other words, if (P)+N is equal to N+(P), you need only substitute "a" for "P" and "5" for "N" and that proves a[5] == 5[a]. If I've missed something, corrections are welcomed.