char *y="WHATIZIT";
Here y is a charecter pointer meaning it contains the address of a character. But "WHATIZIT" has 9 characters (including the implicit NULL character at end) so how it works? The C/C++ compiler does a trick here. Whenever it encounters a string literal like "WHATIZIT", it stores it at a location (usually read only as literals are not meant to be modified like integer constants 2,3..) and returns the starting address of it. The starting address of a sequence of characters and that of the first character of the sequence is one and the same. And so we can assign it to a character pointer and this technique enables us to use strings in C even without a native string data type. But the thing here is we are not explicitly allocating a memory for the characters and the compiler is thus treating it as a constant- side effect is we cannot modify the string. i.e., the following is not allowed:
y[0] = 'a';//We are not allowed to change the contents of the memory location of constants.
Constant data might be in RO segment of memory and accessing them for Write can cause a segmentation fault from OS.
But we can do
y = "Hello"; //y is made to point to another string literal
char x[]="WHATIZIT";
Here, x is an array of characters and the size of the array is 9 (including NULL at end). In C/C++ an array is same as a pointer but is constant. i.e., it cannot be made to point to any other location like we can do for a pointer. But the values of the array can be changed as long as we do not overflow the bound. i.e., the following are valid:
x[0] = 't';
x[1] = 'l';
But the following are not
x[10] = 'p'; //overflow which can cause runtime error
x = "hello"; //x is a constant pointer
In short *x here works like a pointer to a constant and x[] works like a constant pointer.
