is it possible to write a function that takes an integer array as argument and returns the number of-CodePudding

I know it's possible to calculate the number of the elements of the array by sizeof(the_array)/sizeof(the_array[n]) but I dont know what to do with it if I had to write a function

CodePudding user response：

No, not in the sense you mean.

When an array is used as a function argument, it is automatically converted to a pointer to its first element. The C standard specifies only that this pointer points to the first element of the array, not that it contains any other information about the array. The C standard also does not specify there is any way to obtain the size of the array from the pointer or any other information passed to the function.

If you want a function to know the size of an array, pass the size as a separate argument.

(It is possible a C implementation could provide this information, by including it with the pointer, associating it with the pointer, or storing it before the beginning of the array. I am not aware of any that do, except that debugging features may provide this information.)

CodePudding user response：

It is impossible to pass an object of array type in the argument of a function call (unless the array is embedded as a member of a struct or union type), because the array "decays" to a pointer to its first element (C11 6.3.2.1/3).

Also, any parameter of array type in a function declaration will be automatically adjusted to a pointer type. For example, the declaration int foo(const int bar[5]) is automatically adjusted to int foo(const int *bar) (C11 6.7.5.3/7 or C17 6.7.6.3/7).

It is possible to define a macro to get the length of an array, but it is up to the programmer to ensure that the parameter really is an array and not a pointer:

#define ARRAYLEN(arr) (sizeof (arr) / sizeof *(arr))

For example, it cannot be used on the parameter of a function, even if the function's parameter type looks like an array, because the function's parameter type is really a pointer:

#define ARRAYLEN(arr) (sizeof (arr) / sizeof *(arr))

size_t foo(int bar[5])
{
    return ARRAYLEN(bar); // INCORRECT! bar is a pointer, not an array.
}