Hello I am beginner in c++ , can someone explain to me this
char a[]="Hello";
char b[]=a; // is not legal
whereas,
char a[]="Hello";
char* b=a; // is legal
If a array cannot be copied or assigned to another array , why is it so that it is possible to be passed as a parameter , where a copy of the value passed is always made in the method
void copy(char[] a){....}
char[] a="Hello";
copy(a);
It isn't copying the array; it's turning it to a pointer. If you modify it, you'll see for yourself:
void f(int x[]) { x[0]=7; }
...
int tst[] = {1,2,3};
f(tst); // tst[0] now equals 7
If you need to copy an array, use std::copy:
int a1[] = {1,2,3};
int a2[3];
std::copy(std::begin(a1), std::end(a1), std::begin(a2));
If you find yourself doing that, you might want to use an std::array.
std::array and the arrays used in the example?The array is silently (implicitly) converted to a pointer in the function declaration, and the pointer is copied. Of course the copied pointer points to the same location as the original, so you can modify the data in the original array via the copied pointer in your function.
pass by referencestd::vectorwhich has the behaviour that arrays do in most high-level languages.char *a. Calling this pass by reference conflicts with actually passing by reference, via the parameter being a reference (i.e. it has the&).