cout and sizeof are not returning what I expect in C++

Question

Here is sth I find but I can't understand one address ,when I use "cout<

#include<iostream>
using namespace std;
int main()
{
    char  a[2]={'a','b'};
    char  b[3]="ab";
    cout<<&a<<endl;
    cout<<&b<<endl;
    cout<<sizeof(a)<<endl<<cout<<sizeof(b);//the result of this I am puzzled
    return 0;
}

The result is :

0x28ff2e
0x28ff10
2
0x4453c43

Since you seem to be just starting out with C++ programming, I'll mention a style thing: it's easier to read your code if you put spaces around operators, like this `cout << sizeof(a) << endl << cout << sizeof(b);` :) — Almo, Mar 14 '14 at 15:37

score 5 · Accepted Answer · edited May 23 '17 at 11:49

5

0x28ff2e is an address of a
0x28ff10 is an address of b
2 is the size of a
0x4453c43 is an address of the result of converting cout to void* followed by sizeof(b) (See Does std::cout have a return value?)

Maybe you did want this instead:

cout << sizeof(a) << endl;
cout << sizeof(b) << endl;

Or this:

cout << sizeof(a) << endl << sizeof(b) << endl;

edited May 23 '17 at 11:49

Community

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answered Mar 14 '14 at 15:39

mirelon

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Or rather, `0x4453c43` is the output from streaming `cout` followed by `sizeof(b)`. – Mike Seymour Mar 14 '14 at 15:47
`sizeof(b)` is 2 and there 3 at the end of the output. This is weird as though the `<<` operator is left-associative – mirelon Mar 14 '14 at 15:58
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`sizeof(b) == sizeof(char[3]) == 3`. – Mike Seymour Mar 14 '14 at 16:02

score 3 · Answer 2 · edited May 23 '17 at 10:25

3

When you do this line:

cout<<sizeof(a)<<endl<<cout<<sizeof(b)

You shouldn't use cout second time. When you do, you printf address of it:

0x4453c4

or rather Does std::cout have a return value?

and then you print size of b, is the 3 on the end of this 0x4453c43

Rather you should just use this:

cout<< sizeof(a) << endl << sizeof(b) << endl;

edited May 23 '17 at 10:25

Community

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answered Mar 14 '14 at 15:40

Ardel

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It's not (necessarily) the address of `cout`. That's `&cout`. Instead, it's `cout` converted to a `void *` via the conversion operator. – chris Mar 14 '14 at 15:41

score 1 · Answer 3 · answered Mar 14 '14 at 15:38

1

You're printing the address of cout ;)

cout<<sizeof(a)<<endl<<cout<<sizeof(b)

answered Mar 14 '14 at 15:38

mcopik

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The address of `cout` is `&cout`, not `cout`. The reason this works is that `cout` can be converted to a `void *`. – chris Mar 14 '14 at 15:39
Yeah, my mistake. I didn't notice this. – mcopik Mar 14 '14 at 15:41
@chris Implementations typically like to return `this` cast to `void*`, so afterall, it *is* the address of `cout` :) – jrok Mar 14 '14 at 15:42
@jrok, Yeah, I added "not necessarily" to my other comment since implementations like to do that for `std::ostream`. – chris Mar 14 '14 at 15:43

score 1 · Answer 4 · answered Mar 14 '14 at 15:43

The problem is that you're streaming the cout object to itself, so it prints whatever the cout can be converted to that operator<< if overloaded for - which happens to be void* (before C++11, which you evidently aren't using).

Either break cout<<sizeof(a)<<endl<<cout<<sizeof(b); into two lines with a semicolon after endl, or remove the second cout. You should put in a final endl or '\n' too... on some systems you won't be able to read the output otherwise (as the shell prompt will return to the left-of-screen then overwrite it).

score 1 · Answer 5 · answered Mar 14 '14 at 15:45

You probably don't want this bit:

cout<<sizeof(a)<<endl<<cout<<sizeof(b);
                     ^^^^^^

Older implementations of streams had an operator void*(), to allow constructs like if (cout) without allowing implicit conversions to bool and other numeric types; you are seeing the result of that, concatenated with the final value. A C++11 implementation should have an explicit operator bool() insead, and so should cause a compile error here.

Removing that gives something like:

0x28ff2e
0x28ff10
2
3

as you would expect.

score 0 · Answer 6 · answered Mar 14 '14 at 15:39

0

"the result of this I am puzzled"

cout< < sizeof(a)< <endl << cout <<sizeof(b);
                            ^^^ The "puzzling" address comes from this

Remove it and see the result

cout output to standard output is implementation defined

answered Mar 14 '14 at 15:39

P0W

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score 0 · Answer 7 · answered Mar 14 '14 at 15:40

You have a stray cout there. Note that you're effectively printing cout into cout here:

cout << ... << cout << sizeof(b);

cout needs some form of conversion to boolean. Before C++11, this was provided by a conversion to void*; this happens in your case, as the only way cout can be converted to something streamable. Note that there's an extra 3 after 6 hexa digits of the pointer - that's the 3 produced by sizeof(b)

cout and sizeof are not returning what I expect in C++

7 Answers7