Question

Consider the code:

void foo(char a[]){
   a++;            //  works fine, gets compiled
   //... 
}

Now, consider this:

void foo(){
   char a[50];
   a++;            //  Compiler error
   //... 
}

I heard an array is equivalent to a constant pointer and can’t be incremented as it is not a lvalue…

Then why does first code gets compiled, is it so because array arguments to functions are passed as a pointer, i.e. T[] is converted to T* for passing..
So, foo(a) passes a as a pointer.

But is it not back converted to T[] again because is declared as:

void foo(char a[]);

Answer 1

When you pass an array as an argument to a function, it decays to a pointer.
So the thing you increment inside the function body is a pointer, not an array.