CODEWITHSUNDEEP
c++pointers
Freddie
Freddie

Reputation: 1707

C++ Printing value of pointer gives strange result

When I compile and run this C++ code, I am not getting the output I expected.

#include <iostream>
using namespace std;

int main()
{
    int * i = new int;
    long * l = new long;
    char * c = new char[100];
    float * f = new float[100];

    cout << "i " << i << endl;
    cout << "l " << l << endl;
    cout << "c " << c << endl;
    cout << "f " << f << endl;


    delete i;
    delete l;
    delete []c;
    delete []f;

    cin.get();
    return 0;
}

On a unix machine I get

i 0x967f008
l 0x967f018
c
f 0x967f090

On a windows machine the value for c prints as over a line of random characters.

Please can someone explain why it's not printing the pointer for the char array correctly.

Thanks

Upvotes: 10

Views: 4355

Answers (4)

Armen Tsirunyan
Armen Tsirunyan

Reputation: 132974

operator << for std::ostream and std::wostream is defined in special way for char pointers(char*, const char*, wchar_t* and const wchar_t* to print out a null-terminated string. This enables you to write 

const char* str = "Hello, World";
std::cout << str;

and see a nice string on your stdout.

To get the pointer value, cast to void *

std::cout << static_cast<void*>(c)

Upvotes: 20

FelixCQ
FelixCQ

Reputation: 2028

The char * is actually a C string. So I'm guessing it is trying to print it as a string.

One way to force the pointer address to be printed is to use printf:

printf("%p\n", c);

Upvotes: 1

Felice Pollano
Felice Pollano

Reputation: 33242

Since there is no initialization in the char's pointed by c, you dump some random memory values till it found one with value 0. It will sometimes crash, when zero is not find until your readable memory finish. Writing code like this is forbidden and wrong in Unix too, and in any other operating system.

Upvotes: 0

Mat
Mat

Reputation: 206669

The operator<< is overloaded for char*. It will consider that you are trying to print a C-style string, and print all the chars until it finds a 0x00. Since you're not initializing the allocated memory, it will print out random garbage.

Upvotes: 4

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