What is the difference between char *a and char a[]?

There is a lot of difference!


char a[] = "rupali";

char *a = "rupali";

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[] = "rupali";


+----+----+----+----+----+----+------+

a: |r|u|p|a|l|i|'\0\'| 


+----+----+----+----+----+----+------+

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

char *a = "rupali";

+-----+           +---+---+---+---+---+---+------+


a: *======> |r|u|p|a|l|i|'\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';