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In With arrays, why is it the case that a[5] == 5[a]? it is explained that the [] operator in a[5] is defined as *(a + 5) and because + is commutative, 5[a] means *(5 + a) and so the two expressions refer to the same memory location. Fine.

However, C also defines in 6.4.2.1 that an identifier cannot start with a digit. In 5[a] the array identifier is 5 which is not a valid identifier. Why does 5[a] not generate an error about an invalid identifier?

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Paul Ogilvie
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1 Answers1

10

5 is not an identifier, it is an integer literal.

The C standard literally state that 5[a] is just syntactic sugar that must be equivalent with *(5 + a). There is no requirement in C that the first operand of the + operator is an identifier, so the code works just fine.

6.5.6, emphasis mine:

For addition, either both operands shall have arithmetic type, or one operand shall be a pointer to a complete object type and the other shall have integer type.

Lundin
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