I do not understand the following:
int main() {
char ch[13] = "hello world";
function(ch);
return 0;
}
void function( char *ch) {
cout << ch;
}
This outputs "hello world". But if I derefference the pointer ch the program outputs the first letter i.e. "h". I cannout figure out why.
cout << *ch;
As someone stated in the comments section the result of derefferencing the pointer to an array is a plain char.
Let me explain why:
Your ch pointer indicates the adress of the start of the char array, so when you call cout<<ch it will show on the screen all you have in the memory starting from the ch adress and goes secquentially till a first NULL value appears and stops.
And when you call cout<<*ch it will take the value that you have stored on the start adress of the array which is h.
Derefferencing means that you take the value from a specific adress.
Hope it helped! :)
When you pass an array into a function, either directly or with an explicit pointer to that array, it has decayed functionality, in that you lose the ability to call sizeof() on that item, because it essentially becomes a pointer.
So it's perfectly reasonable to dereference it and call the appropriate overload of the stream << operator.
More info: https://stackoverflow.com/a/1461449/1938163
Also take a look at the following example:
#include <iostream>
using namespace std;
int fun(char *arr) {
return sizeof(arr);
}
int fun2(char arr[3]) {
return sizeof(arr); // It's treating the array name as a pointer to the first element here too
}
int fun3(char (&arr)[6]) {
return sizeof(arr);
}
int main() {
char arr[] = {'a','b','c', 'd', 'e', 'f'};
cout << fun(arr); // Returns 4, it's giving you the size of the pointer
cout << endl << fun2(arr); // Returns 4, see comment
cout << endl << fun3(arr); // Returns 6, that's right!
return 0;
}
or try it out live: http://ideone.com/U3qRTo
char*results in a plaincharvalue.