Below is a snippet from the book C Programming Just the FAQs. Isn't this wrong as Arrays can never be passed by value?
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 arrayx[]
to the function namedbyval_func()
by value:The
int[]
parameter tells the compiler that thebyval_func()
function will take one argument—an array of integers. When thebyval_func()
function is called, you pass the address of the array tobyval_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
byval_func()
is a copy of the original array, modifying the array within thebyval_func()
function has 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 thebyval_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.