char 的指针数组与 int 的指针数组

Question

Why the first declaration is valid whereas the other is not?

    char* string[2] = { "Hello", "Bellow" };

    int* b[2] = { {1,2,3}, {2,3,4} };

Answer 1

The reason is that the compiler is not able to imply the type from {1,2,3} , which is desired to be int[3] . You can use array literal to specify it manually as int[3] or int[] :

int *b[2] = { (int[]){1,2,3}, (int[]){2,3,4} };

However, you need to be careful because the lifetime of the literals is bound only to the block where they are defined.

Answer 2

Unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T " will be converted, or "decay", to an expression of type "pointer to T " and the value of the expression will be the address of the first element of the array.

In the declaration

char* string[2] = { "Hello", "Bellow" };

the string literals are not being used to initialize a character array, but an array of pointers, so both strings "decay" to pointers to their first element, so you get

char *[2] = { char *, char * };

In the other declaration

int* b[2] = { {1,2,3}, {2,3,4} };

{1,2,3} and {2,3,4} are not array expressions - they're initializer lists, and they don't "decay" to pointers. As tstanisl shows, you can use compound literals like

int *b[2] = { (int[]){1,2,3}, (int[]){2,3,4} };

and each of the compound literal expressions will decay to a pointer.