Finding Array Length with sizeof in C: The Trick That Only Works in the Declaring Scope

sizeof(arr) / sizeof(arr[0]) computes the number of elements in an array. It works because sizeof(arr) gives the total byte size of the array and sizeof(arr[0]) gives the byte size of one element. The division yields the count. But this only works in the scope where arr was declared as an array. Pass arr to a function and it decays to a pointer. sizeof gives the pointer size. The element count is lost. This is the most repeated mistake in C array code.

In main, sizeof(data) is 28 (7 ints × 4 bytes). The count is 7. Inside the function, sizeof(arr) is 8 (one pointer on a 64-bit system). The count is 8/4 = 2, which is wrong. The array decayed to a pointer at the function boundary. The size information is gone. The only way to know the length inside the function is to pass it as a separate parameter.

#The Macro Guard#link

The macro works for any array type. The parentheses around (arr)[0] prevent operator precedence issues if the argument is an expression. The macro only works in the declaring scope. If you pass the array to a function, the macro silently gives the wrong answer. Some codebases add a compile-time check using _Static_assert to detect decay, but the simplest defense is discipline: compute the length at the call site and pass it.

The sizeof trick is convenient and correct in one place: the scope where the array was declared. Everywhere else, pass the length. This is not a limitation to resent. It is the C memory model. Arrays are raw memory. You track the size. Next: if statement best practices, including the Yoda condition debate and brace discipline.