*& question

Posted on 2003-11-06
Last Modified: 2010-05-18
Hi! Experts,

When I work on ObjectArx which is a library by AutoDesk to program with AutoCad, I met a defination like this:

Acad::ErrorStatus getName(const char*& pName) const;

I don't understand what char*& mean. I did the following:

char name[20];
char* rname = name;
acutPrintf("\nThe block name is : %s", rname);

acutPrintf is similar to printf. It works.

But if I write

acutPrintf("\nThe block name is : %s", name);

It gives me junk chars.

If I write

char name[20];

The compiler gives me an error saying it can not convert char[20] to char*&.

What is the mystic *& mean?

Thanks for your help.

Question by:jzzhang
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LVL 86

Expert Comment

ID: 9694860
>>I don't understand what char*& mean

This means that the function takes a reference to a pointer. Technically speaking, the result is similar to using a 'char**'

>>if I write
>>acutPrintf("\nThe block name is : %s", name);
>>It gives me junk chars.

Yes, because 'pname' no longer points to 'name'.


strcpy ( name, pname);

before, and it will work.

LVL 86

Expert Comment

ID: 9694875
BTW, the behaviour would be the same if you used

void ChangeThePointer ( char*& p) {

   p = "ChangeThePointer";

char* p = "test";

printf ( "before: %s\n", p);

ChangeThePointer ( p);

printf ( "after: %s\n", p);

Expert Comment

ID: 9694911
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LVL 10

Accepted Solution

Sys_Prog earned 250 total points
ID: 9695530
Here's the explanation

In C, when u need to modify the contents of a variable, u pass the address of the variable and recieve it in a pointer variable of the same type. Using the pointer u can change the contents of your variable.

But if u need to change the contents of the pointer itself, i.e. make the pointer point to something else using a function, then u need to pass the address of the pointer variable, For this case the function would be taking a pointer to a pointer argument
void f ( int ** p ) ;
Inside f(), u would have use the cumbersome syntax for referring to the double pointer

Now C++ makes this easy by letting u pass a pointer by reference and that is what u asked for
getName(const char*& pName)

means pName is a reference to a pointer to a char const

Regarding ---------The compiler gives me an error saying it can not convert char[20] to char*&.====

Any statically declared array [e.g. char a[20]] is by definition a const pointer. Now Your function is trying to pass a reference to the pointer, thus it runs the risk of changing it i.e. making it point to something else, but this is not allowed for a const pointer OR an array.

That's the reason for the error


LVL 86

Expert Comment

ID: 9695551
>>Here's the explanation

Haven't I already posted that?

Author Comment

ID: 9696707

It looks like
getName(const char*& pName)
means pName is a reference to a pointer to a char const

I will test it.

Be back soon.


Author Comment

ID: 9698372

I tested it. It works. So char*& should be interpreted as a reference to a char pointer.

Here is the small test code:

bool AllocStr(char*& szStr)
      szStr = new char[15];
      lstrcpy(szStr, "This is a test");
      return 1;

int main()
      char* psz;

      wsprintf(g_szMsg, "Allocated string (%s) length is %d", psz, lstrlen(psz));
      delete psz;

Then, how to accept your answer?

LVL 10

Expert Comment

ID: 9698789
I guess when u open this question, against each of the answers, u get a button for accepting a particular answer.

U can also give grades to the answer.

read the help for Accepting a Answer
LVL 10

Expert Comment

ID: 9698802
Here's the link for the help

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