Getting memory address of char returns strange result

Asked Jul 31 '16 at 18:49

Active Jul 31 '16 at 18:49

Viewed 34 times

I have this code, that gets the memory address of the integer myAge and then prints it out:

int myAge = 13;
cout << "myAge is stored at: " << &myAge << endl;

The result of that is a normal memory address, as expected. However when I try to apply the same method to a variable with type char, it gives a strange result:

char myGrade = 'A';
cout << "myGrade is stored at: " << &myGrade << endl;

Output:

A╠╠╠╠╠╠╠╠

Why is this happening?

asked Jul 31 '16 at 18:49

carefulnow1

5

Cast to `void*`, `char*` has a specialization. Looking up the duplicate ... – πάντα ῥεῖ Jul 31 '16 at 18:50
After some testing, this seems to work also: `cout << "myGrade is stored at: " << (int *)&myGrade << endl;` – carefulnow1 Jul 31 '16 at 18:52
1

@odixon But that doesn't make any sense. Just use `void*` for an arbitrary pointer. – Jonathon Reinhart Jul 31 '16 at 18:52
3

Here it is: http://stackoverflow.com/questions/4860788/why-is-address-of-char-data-not-displayed – πάντα ῥεῖ Jul 31 '16 at 18:53

Getting memory address of char returns strange result

0 Answers0