如何在代码中识别const char *和const char []的类型？

Question

const char* s1   = "teststirg";  
const char  s2[] = "teststirg";

I want a method tell me that s1 is "char*" and s2 is "char[]",how to write the method?

Answer 1

Use templates:

template<typename T, unsigned int SIZE>
bool IsArray (T (&a)[SIZE]) { return true; }

template<typename T>
bool IsArray (T *p) { return false; }

This will evaluate at runtime.
Usage:

if(IsArray(s1))
...

if(IsArray(s2))
...

If interested, you can use some advance techniques, which will tell you this as compile time.

Edit :

typedef char (&yes)[2];

template<typename T, unsigned int SIZE>
yes IsArray (T (&a)[SIZE]);

template<typename T>
char IsArray (T *p);

Usage:

if(sizeof(IsArray(s1)) == sizeof(yes))
...
if(sizeof(IsArray(s2)) == sizeof(yes))
...

Answer 2

If you have access to the original definition, then typeid can be used (but what for, I don't know). If you don't have access to the original definition... There's no way of knowing whether a char* was initialized from another char* , or from an array.

Answer 3

In the above context ( that is in the same method where we have the declaration),

   /*1*/ s1[0]='\0';
   /*2*/ s2=s1;
   /*3 Only This is valid*/  s1=s2;
   /*4*/  s2[0]='\0';

Your compiler wouldn't allow step 1,2,4 to pass, while step 3 would succeed. This clearly indicates the nature of the variables. Now, as regards the method (function call) to determine that, you will have to have the definition in the method signature anyways, so I dont see any purpose/utility/possiblity of this method.

determiner (const char* s1,const char *const s2)

You already have the definition in the signature.You need to bypass compiler, to get a use case for this. I apologise , If I haven't got your requirement correct.