使用C中的两个字符串初始化const char * const *

Question

Does anyone know the correct way of initializing a const char * const * with two literal strings ( "abcdefg" and "hijklmnop" )? I read it was difficult/not possible to convert from a char ** , but I may be wrong.

Any help much appreciated.

Answer 1

Instead of pointers-to-pointers, use an array of pointers and initialize them like this:

const char * const strs[] = {"abcdefg", "hijklmnop"};

So instead of a constant pointer to string constants you now have a constant array to string constants. C does not allow initializing pointers with braces (unless there is only a single value in them), but it does allow for arrays to be initialized this way.

Answer 2

In C99 and up, you can use a compound literal to create an array and use that as an initializer for your pointer.

const char * const *p = (const char *[]){"abcdefg", "hijklmnop"};

This is equivalent to using a named array like in Kninnug's answer, plus a pointer with an initializer pointing to the array by name:

const char * const strs[] = {"abcdefg", "hijklmnop"};
const char * const *p = strs;

Except for the fact that the array has a name ( strs ) in one and is anonymous in the other.

And also let me add that none of these solutions involves converting a char ** to a const char * const * . The way they are constructed, each char * is converted to a const char * individually by virtue of being used as an initializer for an array element that is individually a const char * . There aren't any char ** 's in the code.