Why is func(buf) equal func(&buf)

Question

Please, can somebody explain why line 15 and 16 are equal in result (and also in assembler)? Why does it compile only as long as the function parameter is void* ?

Beside this, don’t blame me ‘bout the code quality. It is legacy code and shortened to show the essential part.
https://code.sololearn.com/cwE4r7D69AHa/?ref=app

Answer 1

It is a very interesting thing. While buf and &buf have different types, physical address value is the same for both. The address of the first byte is the same as the address of the whole string.
Thanks for a good question.

Answer 2

In the C tutorial they explain that an array (buf, in this case) is actually syntactic sugar for a pointer to the first element in the array, so &buf points to the first character, and so does buf