Pass an array to a function by value
Below is a snippet from the book C Programming Just the FAQs. Isn't this wrong as Arrays can never be passed by reference?
VIII.6: How can you pass an array to a function by value?
Answer: An array can be passed to a function by value by declaring in
the called function the array name
with square brackets (and[)]
attached to the end. When calling the
function, simply pass the address of
the array (that is, the array’s name)
to the called function. For instance,
the following program passes the array
to the function namedx[]
by value:byval_func()
Theparameter tells theint[]
compiler that thebyval_func()
function will take one argument—an
array of integers. When the
function is called, youbyval_func()
pass the address of the array to
:byval_func()
byval_func(x);
Because the array is being passed by
value, an exact copy of the array is
made and placed on the stack. The
called function then receives this
copy of the array and can print it.
Because the array passed to
is a copy of thebyval_func()
original array, modifying the array
within thefunction hasbyval_func()
no effect on the original array.
Because the array is being passed by value, an exact copy of the array is made and placed on the stack.
This is incorrect: the array itself is not being copied, only a copy of the pointer to its address is passed to the callee (placed on the stack). (Regardless of whether you declare the parameter as int[] or int*, it decays into a pointer.) This allows you to modify the contents of the array from within the called function. Thus, this
Because the array passed to byval_func() is a copy of the original array, modifying the array within the byval_func() function has no effect on the original array.
is plain wrong (kudos to @Jonathan Leffler for his comment below). However, reassigning the pointer inside the function will not change the pointer to the original array outside the function.