char buffer[]="foobar"; I know that buffer is char* pointer to the first element so buffer==&buffer[0] but why &buffer==buffer? &buffer should give the memory address of the buffer char* and NOT the address of the first element?
Additionaly,What would happen when i do (int)buffer ?
buffer is the address of the first element and &buffer is indeed the address of the array itself. The array will be stored on the stack directly. That is why &buffer == buffer.
It is not a pointer but an array. If you had declared it as char*, it would not be &buffer == buffer
Think of it like this.
buffer is the address of the first element of the array. So it is the address of an integer.
&buffer is the address of the array. Hence &buffer will be in fact the same as buffer, but their behaviour will be different.
For e.g. if you do buffer+1, it will increment by the size of int, but &buffer+1 will increment by the size of the array, i.e size of one element * number of elements.
Edits. Initially I had written buffer++ instead of buffer+1. See comments section for the reason I edited it.
not doing buffer++. He said that it is an array, that decays to pointer. Hence, IMO, buffer++ is totally valid.
&bufferandbufferdo in fact have different datatypes, so you should at least get a warning if you actually code up the Expression&buffer == buffer. Try it.(int)buffer, the compiler will first turnbufferinto a pointer and then intepret the bits of that pointer as andint. This might also result in a warning if the pointer is bigger than the int, but I am not sure.bufferis not a pointer (char *) - it's an array (char []) - try doingbuffer++if you want to see one important difference. It will decay to a pointer is various cases though.