Passing char array to a function

Question

Passing an array to a function, do it need to use &?

In below example Case 1 or Case 2, which one is good ?

#define IP_ADDR_LEN  4 
char ip[4]={192,168,205,1};
char PeerIP[4];
int main()
{
   memcpy(PeerIP,ip,IP_ADDR_LEN)

   testip(PeerIP);   /* Case 1 */

   testip(&PeerIP);   /*Case 2*/
}

int testip(char *ip)
{
  /* IP check */

}

Note that you should declare (or define) testip() before you call it. If you do that, the compiler will be able to tell you that the testip(&PeerIP); call is passing the wrong type to the function — it's an int (*)[4] and not an int * that is being passed. Find out how to make your compiler insist on functions being defined or declared before being used, with strict prototypes, and then use it. — Jonathan Leffler
– Jonathan Leffler, Commented Jan 19, 2017 at 7:35

Sourav Ghosh · Accepted Answer · 2017-01-19 06:53:07Z

2

The first one.

  testip(PeerIP);

is the right way to do it, as an array name will decay to a pointer to the first element of the array, which matches the expected type for the called function parameter.

To elaborate, in your code, testip() expects an argument of type char * and by calling it like testip(PeerIP); you are passing a char *, so no issues.

In case you use the &PeerIP, the type will be [char (*) [4]] pointer to array, not pointer to the first element of the array [char *] and they are different type.

Quoting C11, chapter §6.3.2.1, Lvalues, arrays, and function designators, (emphasis mine)

Except when it is the operand of the sizeof operator, the _Alignof operator, or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue. [...]

edited Jan 19, 2017 at 6:53

answered Jan 19, 2017 at 6:42

Sourav Ghosh

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3 Comments

StoryTeller - Unslander Monica Over a year ago

Different type, but both point to the same address. Only pointer arithmetic will differ.

Sourav Ghosh Over a year ago

@StoryTeller right, but still the later would be wrong. Just because they point to same address doe not mean the expressions are interchangeable. Types matter. :)

StoryTeller - Unslander Monica Over a year ago

They do indeed matter very much. I just had my c++ background pop up. Where casting pointers to array references is done "correctly" thanks to that fact :)

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