Using 'char' to represent text

[calioranged]

*** Newbie question ***

1). char day[] = "Thursday";
std::cout << "Today is " << day << std::endl;

2). char Day[9] = "Thursday";
std::cout << "Today is " << Day << std::endl;

Both of the above methods produce the same output; "Today is Thursday".

I have also seen programmers use char pointers to represent text:

3). char* DAY = Day;

So why does this work? If this were a 'real' pointer, wouldn't the code be written like so:

4). char* DAY = &Day;

In example 3, 'DAY' returns the value of 'Day' rather than a pointer to 'DAY', so I figured why the need for the pointer(*) at all if this doesn't actually operate as a pointer:

5). char DAY = Day;

But this produces the error:

" error C2440: 'initializing': cannot convert from 'char [9]' to 'char' "

What exactly is going on here? Why do some people use pointers(*) to represent text and why do these not actually operate as pointers?

[salem c]

3). char* DAY = Day;
So why does this work?

Because array names almost always decay to "pointer to first element of array".

So
char* DAY = Day;
is the same as
char* DAY = &Day[0];

Try this.
char* DAY = &Day[5];
std::cout << "Today is " << DAY << std::endl;

[calioranged]

Ah okay. So:

char* DAY = Day;

or

char* DAY = &Day[0];

Simply stores all elements of the character array 'Day' into the char* 'DAY'. Whilst:

char* DAY = &Day[5];

Stores all elements after and including the fifth element of the character array 'Day' into the char* 'DAY'.

Apologies for my beginner questions, but why does the char* 'DAY' not actually store the memory address of 'Day', like a normal pointer would? For example:

int x = 1;
void* pointer = &x;
std::cout << "Today is " << pointer << std::endl;

Outputs "Today is 0x0093f84c", which is obviously a pointer to the memory address of the variable 'x'. Why does the same not happen in our example? Our example:

char Day[9] = "Thursday";
char* DAY = &Day[0];
std::cout << "Today is " << DAY << std::endl;

Outputs 'Today is Thursday". But if I use a 'void' pointer rather than a 'char' pointer to the character array 'Day':

char Day[9] = "Thursday";
void* pointer = &Day;
std::cout << "Today is " << pointer << std::endl;

Then the output is "Today is 0x00bdfbcc" , which is what I would have expected a pointer to do. So I guess my question is why does a char* output the actual value of 'Day', whilst the void* outputs the memory address of 'Day'.

Again sorry if this question seems really simple, I just can't really get my head around why these two examples do not operate in the same way.

[salem c]

> Simply stores all elements of the character array 'Day' into the char* 'DAY'. Whilst:
No, there are no additional copies of the elements of the array.

It's like a postcard with your address on it.
Writing your address on a 2nd postcard doesn't make a new house for you.

> Apologies for my beginner questions, but why does the char* 'DAY' not actually store the memory address of 'Day', like a normal pointer would?
It does store a memory address like a normal pointer.

I think you're confusing the 'special case' in std::cout where if you have a char* or const char * pointer, you get what the pointer points to (the string) rather than the 0xnnnnnnnn representation of a memory address you would otherwise get with any other type of pointer.

#include <iostream>

int main ()
{
    char today[] = "Thursday";
    char *day1 = today;
    char *day2 = &today[5];

    std::cout << "Today is " << today << std::endl;
    std::cout << "Today is " << day1 << std::endl;
    std::cout << "Today is " << day2 << std::endl;

    std::cout << "Today is " << reinterpret_cast<void*>(today) << std::endl;
    std::cout << "Today is " << reinterpret_cast<void*>(day1) << std::endl;
    std::cout << "Today is " << reinterpret_cast<void*>(day2) << std::endl;
}

[Grime]

Note that when you assign a pointer say A to another pointer say B then the pointer B isn't pointing to the address of the pointer A but rather it's pointing to what A was pointing to (correct me if I'm wrong).

[calioranged]

> I think you're confusing the 'special case' in std::cout where if you have a char* or const char * pointer, you get what the pointer points to (the string) rather than the 0xnnnnnnnn representation of a memory address you would otherwise get with any other type of pointer.

Yes you are right. So would it be correct to say that the char*/const char* is still pointing to the memory address, but 'std::cout' will effectively output 'what the pointer points to' rather than its memory address.

And that if any other type of pointer is used (as opposed to char*/const char*), this would not be the case, and the pointer would simply output the memory address in 'cout' rather than what the pointer points to?

[salem c]

Yes calioranged, that's correct.

[calioranged]

Thanks for your help salem c