Trying to understand pointers. Why does this give unexpected results?

[rob]

This is my test program

#include everything relevant here
int main {
int a = 0;
float b = 1.0;
char c = 'c';

int *ptrA;
float *ptrB;
char *ptrC;

ptrA = &a;
ptrB = &b;
ptrC = &c;

cout << "value of a: " << a << "; address of a: " << ptrA << endl;
cout << "vaue of b: " << b << "; address of b: " << ptrB << endl;
cout << "value of c: " << c << "; address of c: " << ptrC << endl;

return 0;
}


This program, compiled w/ gcc 9.3 on Pop_OS 20.04, does not print an
address for c. It just prints "c"; this does show addresses for a and
b.

Why can't I show the address for the char variable c?

Thx,
Rob

[Ben Bacarisse]

rob <r...@drrob.com> writes:

> This is my test program
>
> #include everything relevant here
> int main {

And you need ()s there and at least a "using namespace std;" unless you
are using a museum version of C++.

> int a = 0;
> float b = 1.0;
> char c = 'c';
>
> int *ptrA;
> float *ptrB;
> char *ptrC;
>
> ptrA = &a;
> ptrB = &b;
> ptrC = &c;
>
> cout << "value of a: " << a << "; address of a: " << ptrA << endl;
> cout << "vaue of b: " << b << "; address of b: " << ptrB << endl;
> cout << "value of c: " << c << "; address of c: " << ptrC << endl;

What would you like this code to do:

char *string = "Hello world\n";
cout << string;

? Can you see the problem now?

> return 0;
> }
>
>
> This program, compiled w/ gcc 9.3 on Pop_OS 20.04, does not print an
> address for c. It just prints "c"; this does show addresses for a and
> b.
>
> Why can't I show the address for the char variable c?

You can if you do this:

cout << "value of c: " << c << "; address of c: " << (void *)ptrC << endl;

--
Ben.

[rob]

Thanks, that helps me.

[Real Troll]

You could also do something like this:

> #include <iostream>
>
> using namespace std;
>
> int main()
> {

>   int a = 0;
>   float b = 1.0;
>   char c[] = "c";
>
>   int *ptrA;
>   float *ptrB;
>   char *ptrC;
>
>   ptrA = &a;
>   ptrB = &b;

>   ptrC = c;

>
>   cout << "value of a: " << a << "; address of a: " << ptrA << endl;
>   cout << "vaue of b: " << b << "; address of b: " << ptrB << endl;

>   cout << "value of c: " << c << "; address of c: " << &ptrC << endl;
>
>   return 0;
> }

[Juha Nieminen]

rob <r...@drrob.com> wrote:
> int *ptrA;
> float *ptrB;
> char *ptrC;
>
> ptrA = &a;
> ptrB = &b;
> ptrC = &c;

I'm genuinely wondering why you are writing it like that, instead of the
simpler:

int *ptrA = &a;
float *ptrB = &b;
char *ptrC = &c;

> cout << "value of c: " << c << "; address of c: " << ptrC << endl;

A char* pointer is overloaded to print the string pointed to by that pointer,
so that will severely malfunction. You can cast it to void* instead:

std::cout << static_cast<void*>(ptrC) << std::endl;

[Fred. Zwarts]

Op 04.jul..2021 om 00:04 schreef Real Troll:

Doesn't that print the address of ptrC, instead of the address of c?

>>
>>   return 0;
>> }
>
>
>