char * and char[]

Why is this right ?

#include<iostream>
using namesp开发者_开发问答ace std;
int main()
{
    char *s="raman";
    char *t="rawan";
    s=t;
    cout<<s;

return 0;
}

But this is wrong?

#include<iostream>
using namespace std;
int main()
{
    char s[]="raman";
    char t[]="rawan";
    s=t;
    cout<<s;

return 0;
}

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In the first example, s=t does a pointer assignment. In the second, s=t tries to assign a pointer value (resulting from the implicit conversion, or "decay", of the array expression t) to an array object. C++ doesn't permit array assignments.

C and C++ happen to be very similar in this area; section 6 of the comp.lang.c FAQ covers the relationship between arrays and pointers very well.

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The first example assigns an pointer to another which is valid.

The second example assigns an array to another array which in not allowed in C & C++ both.

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This excellent C++ FAQ entry and this answer should be a good read for you.

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In addition to what the other guys said:

Contrary to popular belief, arrays are actually not pointers. They just share a lot of similarities when working with them and have a couple of implicit conversions to pointers which is why it's easy to work with them as if they are pointers.

Arrays are a standalone feature of (C and) C++. It doesn't behave exactly like a pointer would.

For example, it's possible to allocate array objects on the stack, which is not possible when you allocate objects using new (which returns a pointer) and pointers.

And the example that you showed is another one: You can't use arrays as if they are pointers. But you can use pointers to point to a continuous piece of memory (array).

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Array name is a const pointer. meaning, when you declare an array, the name is a pointer, which cannot be altered.

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char *s means:
address and value both are not constant.

const char *s:

value is constant, not address.

const char * const s;

address and value both are constant.

char *s[] 

is an array. Array base address is always contant. You can not change the base address of it, that is not allowed in c.