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What is the difference between char *a and char a[]?
There is a lot of difference!
char a[] = \"string\";
char *a = \"string\";
The declaration char a[] asks for space for 7 characters and see that
its known by the name \"a\". In contrast, the declaration char *a, asks
for a place that holds a pointer, to be known by the name \"a\". This
pointer \"a\" can point anywhere. In this case its pointing to an
anonymous array of 7 characters, which does have any name in
particular. Its just present in memory with a pointer keeping track of
its location.
char a[] = \"string\";
+----+----+----+----+----+----+------+
a: | s | t | r | i | n | g | \'\\0\' |
+----+----+----+----+----+----+------+
a[0] a[1] a[2] a[3] a[4] a[5] a[6]
char *a = \"string\";
+-----+ +---+---+---+---+---+---+------+
| a: | *======> | s | t | r | i | n | g | \'\\0\' |
+-----+ +---+---+---+---+---+---+------+
Pointer Anonymous array
It is curcial to know that a[3] generates different code depending on
whether a is an array or a pointer. When the compiler sees the
expression a[3] and if a is an array, it starts at the location \"a\",
goes three elements past it, and returns the character there. When it
sees the expression a[3] and if a is a pointer, it starts at the
location \"a\", gets the pointer value there, adds 3 to the pointer
value, and gets the character pointed to by that value.
If a is an array, a[3] is three places past a. If a is a pointer, then
a[3] is three places past the memory location pointed to by a. In the
example above, both a[3] and a[3] return the same character, but the
way they do it is different!
Doing something like this would be illegal.
char *p = \"hello, world!\";
p[0] = \'H\';
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