this is probably an essentially trivial thing, but it somewhat escapes me, thus far..
char * a3[2];
a3[0] = "abc";
a3[1] = "def";
char ** p;
p = a3;
this works:
printf("%p - \"%s\"\n", p, *(++p));
this does not:
printf("%p - \"%s\"\n", a3, *(++a3));
the error i'm getting at compilation is:
lvalue required as increment operand
what am i doing wrong, why and what is the solution for 'a3'?
a3 is a constant pointer, you can not increment it. "p" however is a generic pointer to the start of a3 which can be incremented.
You can't assign to a3, nor can you increment it. The array name is a constant, it can't be changed.
a3 is not a pointer, period. You're probably missing a * in your question too (char *a3[2] ?)
Try
char *a3Ptr = a3;
printf("%p - \"%s\"\n", a3, *(++a3Ptr));
In C, a char array[] is different from char*, even if you can use a char* to reference the first location of an array of char.
aren't both "p" and "a3" pointers to pointers?
Yes but a3 is constant. You can't modify it.
a3 is a name of an array. This about it as a constant pointer.
You cannot change it. You could use a3 + 1 instead of ++a3.
Another problem is with the use of "%s" for the *(++a3) argument. Since a3 is an array of char, *a3 is a character and the appropriate format specifier should be %c.
You cannot increment or point any char array to something else after creating. You need to modify or access using index. like a[1]
