incrementing an array as formal arg

We are not allowed to pre and post increment any array like

Code:

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int array[10];
array++;//Illegal because base address of an array cannot be changed.

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But in a function(if it is formal argument why it is legal?

Code:

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void function(char *array)
{
array++;//Legal
}

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Because an array is not a pointer and a pointer is not an array. And a modifiable copy of a location is not the same as a nonmodifiable location.

Compare

Code:

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3++; // illegal

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With

Code:

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func(3);

...

void func ( int a ) {
  a++; // legal - changes the value stored in a, not the value of 3
}

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Function arguments are variables containing copies of the original data.

I think my question is not clear to you.So I am repeating

Code:

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function(char *a[])
{
char *b[3];
b++;//Error
a++;//Legal Why????
}

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Or more precisely how the Lvalue is comming in formal argument.

Code:

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char *b[3];

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b is an array of 3 elements of character pointers. If you use 'b' you really mean
the address of the first element (i.e. &b[0]). When you attempt to use b++,
you are telling the compiler to change the address of the constant address of the
array which is called 'b' in this case. Just like you said in your opening post. Thus
this is illegal since you are trying to change a constant address. 'b' itself also is
only an rvalue I believe. So you can't assign/change it in anyway.

Code:

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function(char *a[])
...
a++;

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Is "legal" because "arrays" get represented as pointers when passed as parameters.
i.e. 'a' in this case becomes a pointer to the first element. It is legal to apply
pointer arithmetic on non-constant pointers - thus this statement is legal.

Not sure if I explained it correct though, do correct me if it's wrong :)

Quote:

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Originally Posted by vaibhav

I think my question is not clear to you.So I am repeating

Code:

---

function(char *a[])
{
char *b[3];
b++;//Error
a++;//Legal Why????
}

---

---

I think my answer is not clear to you. So I am repeating.

Here

Code:

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function(char *a[])

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a is not an array. It is a pointer to a pointer to char.

Here

Code:

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char *b[3];

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b is an array (of pointers to char).

You can't increment an array, you can increment a pointer.