c - struct look up table using string pointers issue -

i have written following code:

unsigned char *cptr[] = {     "first",     "second",     "third" };  typedef struct {     int a;     char *lut;  /* or char lut[20]*/ }grp_t;  const grp_t grp_st[]= {     { 0, cptr[0] },     { 1, cptr[1] },     { 2, cptr[2] } }; 

when try compile code got following error:

error: initializer element not constant

error: (near initialization 'grp_st[0].lut[0]')

but when replace *cptr[] cptr[][16] as

unsigned char cptr[][16] = {     "first",     "second",     "third" }; 

i able compile code.

please can 1 explain me concept missing here.

it limitation of c.

6.7.8.4 initialization:

all expressions in initializer object has static storage duration shall constant expressions or string literals.

and 6.6.9 constant expressions:

an address constant null pointer, pointer lvalue designating object of static storage duration, or pointer function designator; shall created explicitly using unary & operator or integer constant cast pointer type, or implicitly use of expression of array or function type. array-subscript [] , member-access . , -> operators, address & , indirection * unary operators, , pointer casts may used in creation of address constant, value of object shall not accessed use of these operators.

if can change code make grp_st not of static storage duration compiles:

#include <stdio.h> const char *cptr[] = {     "first",     "second",     "third" };  typedef struct {     int a;     const char *lut; } grp_t;   int main(void) {     grp_t grp_st[]= {     { 0, cptr[0] },     { 1, cptr[1] },     { 2, cptr[2] }     };     return 0; } 

in second case managed compile code because didn't access array element @ given index, pointer (in case of char cptr[][16], cptr[0] gives address of first array of 16 characters , cptr[0] decays type char * in expression) - therefore didn't violate 6.6.9 rule.