how to understand this typedef

[evenstar]

Hi,

  typedef const char* CCARRY[10];
        char * p="Hello";
        p = "Merry";
        CCARRY arr1={p};//OK using the type of char * to initialize to type of const char *
        arr1[0]=p;//OK,assignment
        CCARRY arr2[5]={arr1};//1**compile error:cannot convert 'const char**' to 'const char*' in initialization
        CCARRY arr3[5]={*arr1};//3**ok,right
        arr3[5]=*arr1;//2**compile error:incompatible types in assignment of 'const char*' to 'const char* [10]

I have some question in ** .I think the type of element in arr2 is CCARRY that is const char* [10].But it is const char *,why?
And ,if the type of element in arr2 is const char* as compiler said in 1**,why in 2** compiler think the type of element in arr2 is const char* [10]?

PS:I believe the type of element in arr2 is const char* [10].
My test environment is vs2008 and GCC4.4.0

[weaknessforcats]

If you define this array:

const char* CCARRY[10];

then CCARRY is an array of 10 char* that are const.

The name of the array, CCARRY is the address of element 0. So CCARRY[0] is a const char* making CCARRY the address of a const char*. That is, CCARRY is a const char**.

In order to have CCARRY be a pointer to a array of 10 const char* you would need to code:

const char* CCARRY[][10];

[evenstar]

If const char* CCARRY[10].then CCARRY is an array of 10 char* that are const.And CCARRY is an array name

But,in my code it is typedef const char* CCARRY[10].And CCARRY is a type name.

[weaknessforcats]

Yes but remembeer that all typedef does is let you name your own types.

So:

CCARRY arr1={p};

just saves you from writing:

const char* arr1[10];

CCARRY may have meaning in your program that makes using that name convenient.