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解釋array與pointer關係

The name of an array usually evaluates to the address of the first element of the array, so arrayand &array have the same value (but different types, so array+1 and &array+1 will not be equal if the array is more than 1 element long).

There are two exceptions to this: when the array name is an operand of sizeof

 or unary &(address-of), the name refers to the array object itself. Thus sizeof array gives you the size in bytes of the entire array, not the size of a pointer.

For an array defined as T array[size], it will have type T *. When/if you increment it, you get to the next element in the array.

&array evaluates to the same address, but given the same definition, it creates a pointer of the type T(*)[size] -- i.e., it's a pointer to an array, not to a single element. If you increment this pointer, it'll add the size of the entire array, not the size of a single element. For example, with code like this:

char array[16];
printf("%p\t%p",(void*)&array,(void*)(&array+1));

We can expect the second pointer to be 16 greater than the first (because it's an array of 16 char's). Since %p typically converts pointers in hexadecimal, it might look something like:

0x123410000x12341010