3 ms·
Your link points out that this decays to a raw pointer in all cases and is not enforced by the compiler.
by senkora 3y ago
Your link points out that this decays to a raw pointer in all cases and is not enforced by the compiler.
- joshuaissac 3y agoIt does not. The decay is for the arr[size] form. > using the int array1[2]; like this: arraytest(array1); causes array1 to automatically decay into an int . > HOWEVER, if you take the address of array1 instead and call arraytest(&array1), you get completely different behavior! > Now, it does NOT decay into an int