Array of Pointer for char Vs Array of pointer for int

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} };

Because in the first one, the (string literal) data resolves to the pointer type. In the second, the data is not pointers, and the definition should be `int b[2][3]` — Weather Vane, Dec 28 '20 at 12:23
There is a simpler version that shows the behaviour you talking about. Just a pointer, the array is not needed to show it. `const char *foo="blabla"; int *thisDoesNotWork={1,2,3,4};` — 12431234123412341234123, Dec 28 '20 at 12:29
You asking why the language designers made the decisions they did? — Weather Vane, Dec 28 '20 at 12:31
Related, but not a duplicate: https://stackoverflow.com/questions/30533439/string-literals-vs-array-of-char-when-initializing-a-pointer — 12431234123412341234123, Dec 28 '20 at 12:33
`string` should be `const char *string[2];` and not `char *string[2];`, since the string literals are `const`. (Not related to your question) — 12431234123412341234123, Dec 28 '20 at 12:34
https://stackoverflow.com/questions/14295147/initializing-char-pointer-as-string-vs-other-type-pointers-as-arrays — 12431234123412341234123, Dec 28 '20 at 12:35

tstanisl · Answer 1 · 2020-12-28T13:22:10.020

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.

edited Dec 28 '20 at 13:22

answered Dec 28 '20 at 12:33

tstanisl

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score 1 · Accepted Answer · answered Dec 28 '20 at 14:09

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.

Array of Pointer for char Vs Array of pointer for int

2 Answers2