The excellent source to clear up the confusion is Peter Van der Linden, Expert C Programming, Deep C secrets - that arrays and pointers are not the same is how they are addressed in memory.
With an array,
char new_str[];
the compiler has given the new_str a memory address that is known at both compilation and runtime, e.g. 0x1234, hence the indexing of the new_str is simple by using
[]. For example
new_str[4], at runtime, the code picks the address of where
new_str resides in, e.g. 0x1234 (that is the address in physical memory). by adding the index specifier
[4] to it, 0x1234 + 0x4, the value can then be retrieved.
Whereas, with a pointer, the compiler gives the symbol
char *newstr
an address e.g. 0x9876, but at runtime, that address used, is an indirect addressing scheme. Supposing that newstr was malloc'd
newstr = malloc(10);
, what is happening is that, everytime a reference in the code is made to use newstr, since the address of newstr is known by the compiler i.e. 0x9876, but what is newstr pointing to is variable. At runtime, the code fetches data from physical memory 0x9876 (i.e. newstr), but at that address is, another memory address (since we malloc'd it), e.g 0x8765 it is here, the code fetches the data from that memory address that malloc assigned to newstr, i.e. 0x8765.
The char new_str[] and char *newstr are used interchangeably, since an zeroth element index of the array decays into a pointer and that explains why you could newstr[5] or *(newstr + 5) Notice how the pointer expression is used even though we have declared char *newstr, hence
*(new_str + 1) = *newstr;
OR
*(new_str + 1) = newstr[1];
In summary, the real difference between the two is how they are accessed in memory.
Get the book and read it and live it and breathe it. Its a brilliant book! :)
