Can't convert from char*[] to char**

Question

When I try a test program with just these two lines

char array[256];
char** arrayPointer=&array;

I get the error

cannot convert from char*[256] to char**.

Yet if I do this:

char array[256];
char* temp=array;
char** arrayPointer=&temp;

I get no such complaint.

I figured that it was eclipse acting buggy (which my eclipase is acting funny right now) but when I tried to do a cast of the &array to char** for the function I ended up with unusual behavior and my debugger implying that the array isn't being modified as it should.

PS. all of this was written by hand, forgive typos.

Answer 1

In C++, arrays and pointers are not the same thing. Arrays in many cases can implicitly be converted to a pointer, but array types and pointer types are different.

In your case, the variable

char array[256];

has type char[256]. If you take its address by writing &array, you get a pointer to an array of 256 chars, which has type char (*)[256]. This is not the same a char**. This is actually a good thing. If you could do the conversion, what would happen if you did this?

char array[256];
char** ptrPtr = &array; // Not legal!
*ptrPtr = new char[256];

In this case, the third line would "reassign" array to point to a new array of 256 elements. However, array is not a pointer! This operation is meaningless.

The reason you got a weird debugger error when writing

char array[256];
char** ptrPtr = (char**) &array; // Legal, but Bad Times!
*ptrPtr = new char[256];

is that the cast you've put in results in undefined behavior. You're pretending that a pointer to an array of 256 actual char objects is really a pointer to a pointer to a char. This is a meaningless cast, so all bets are off when you do it.

On the other hand, if you explicitly introduce a char* variable like this:

char array[256];
char* ptr = array;
char** ptrPtr = &ptr;

Then everything is fine. In the second line, you create a pointer (actual type char*) that points to the first element of array. In the third line, you create a pointer to that new pointer. If you then write

*ptrPtr = new char[137];

Then nothing bad happens; you've just changed where ptr was pointing, and didn't destroy array.

Hope this helps!