尝试在函数中传递地址时收到警告

Question

I get the warning:

expected 'const CHAR *' but argument is of type 'CHAR (*)[18]'

Why is the compiler giving me this warning? This is the line I am using to call to the api:

AG_AS(g_AgMgrCtx.audio_device.Profile_Type,&g_AgMgrCtx.SDevAddr);

The declaration of the function is

AG_AS(AG_DType d_type , CHAR *addr);
char SDevAddr[18];

What is the problem with passing the address? When I remove & the warning goes away.

What does the warning actually mean?

Answer 1

That means your call should be (without the & ):

   AG_AS(g_AgMgrCtx.audio_device.Profile_Type, g_AgMgrCtx.SDevAddr);

g_AgMgrCtx.SDevAddr is an array of 18 chars. So when you pass it to a function, it gets converted into a pointer to its first element (thus matching the type that AG_AS() expects).

Relevant: What is array decaying?

Answer 2

When you have an object like this

T obj;

where T is some type then expression &obj has type T * that is it is a pointer to an object of type T .

This array declaration

char SDevAddr[18];

can be transformed to the form shown above using a typedef. For example

typedef char T[18];
T SDevAddr; 

In this case expression &SDevAddr has type T * as it has been pointed above. however in this case T in turn an alias for the type char[18] . Thus &SDevAddr is a pointer to an object of type char[18] Its type looks like char ( * )[18] .

However as it is seen from the function declaration

AG_AS(AG_DType d_type , CHAR *addr);

the second parameter has type char * . If you will pass as an argument the array itself like SDevAddr then it will be implicitly converted to pointer to its first element and will have type char * that is required for the function call.