My brain has gone a lot fuzzy just recently and I can't for the 开发者_开发百科life of me remember why the following C code:
char a[3][3] = { "123", "456", "789" };
char **b = a;
Generates the following warning:
warning: initialization from incompatible pointer type
Could someone please explain this for me.
Thank you.
char (*b)[3] = a;
This declares b
as a pointer to char arrays of size 3. Note that this is not the same as char *b[3]
, which declares b
as an array of 3 char pointers.
Also note that char *b = a
is wrong and still emits the same warning as char **b = a
.
Try this,
char a[3][3] = { {'1','2','3'}, {'4','5','6'}, {'7','8','9' }};
char *b = &a[0][0];
Since, a is character array of arrays you need to initialize them as a character.
the problem is that ** is not statically allocated.
you could accomplish this simple version with the following:
char a[3][3] = {"123", "456", "789"};
char *b[3] = {a[0], a[1], a[2]};
That is right.
a
is a pointer.
char *b
defines a pointer to char.
char **b
defines a pointer to pointer to char.
a
is still a pointer to a char:
char* b = a;
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