How to print array name in function from argument?

Question

I have a function like this:

foo(int a[], int b);

and I want to print the name of the array in the function. If I call

foo(cards, 5);

I want to print this: array name:cards;array size:5. How should I do this?

possible duplicate of http://stackoverflow.com/questions/15349761/get-called-function-name-as-string — JLT, Apr 30 '16 at 13:30
No, it's not - that deals with getting the enclosing function's name. — Isiah Meadows, Apr 30 '16 at 16:30

score 3 · Accepted Answer · answered Apr 30 '16 at 13:29

3

You can't. By the time your program is executing, the name "cards" that was used in the source code is no longer available. But you can do:

void foo(int *a, int b, const char *name);
...
foo(cards, 5, "cards");

answered Apr 30 '16 at 13:29

William Pursell

score 1 · Answer 2 · answered Apr 30 '16 at 14:30

1

To create a wrapper macro.

#define STR(v) #v

#define FOO(name, value) do{ fprintf(stderr, "array name:%s;array size:%d\n", STR(name), value);foo(name, value); }while(0)

Use FOO(cards, 5); instead of.

answered Apr 30 '16 at 14:30

BLUEPIXY

2 Answers2