Displaying char array in gcc does not work

Question

I wrote a piece of code and tested with gcc compiler

#include <iostream>

int main()
{
    char arr[ 1000 ];
    for( int index( 0 ); index < 1000; ++index )
    {
        std::cout << arr[ index ] << std::endl;
    }
    return 0;
}

I was hoping it to print the garbage values but to my surprise, it did not print anything. When I simply changed the datatype of arr from char to int, it displayed the garbage values as expected. Could somebody please explain this to me?

James Kanze · Accepted Answer · 2012-11-21 16:23:39Z

3

The overloads for << for character types do not treat them as integral types, but as characters. If the garbage value corresponds to a printable character (e.g. 97, which corresponds to 'a'), you will see it. If it doesn't (e.g. 0), you won't. And if the garbage values correspond to some escape sequence which causes your terminal to use a black foreground on a black background, you won't see anything else, period.

If you want to see the actual numerical values of a char (or any character type), just convert the variable to int before outputting it:

std::cout << static_cast<int>( arr[index] ) << std::endl;

answered Nov 21, 2012 at 16:23

James Kanze

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1 Comment

Saurabh Sinha Over a year ago

I got it. So basically what is happening over here is that the random value are not printable ascii character codes that is why I am not seeing anything. Thank you so much for pointing it out.

emartel · Accepted Answer · 2012-11-21 16:16:17Z

2

What you're trying to do has an undefined behavior. Some compilers will clear out the memory for you, others will leave it as it was before the creation of your buffer.

Overall, this is a useless test.

answered Nov 21, 2012 at 16:16

emartel

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1 Comment

bames53 Over a year ago

It's not really 'undefined behavior' in the sense that the C++ standard uses that term. All values of char are required to be valid and so it should be perfectly legal and well behaved to read indeterminate char values.

user1842591 · Accepted Answer · 2012-11-21 16:20:36Z

2

Some platforms may choose, for example for security purposes, to fill the uninitialized char array with zeroes, even though it's not static and wasn't explicitly initialized. Therefore, that is why no garbage is showing up - your char array was just automatically initialized.

answered Nov 21, 2012 at 16:20

user1842591

Comments

john · Accepted Answer · 2012-11-21 16:14:15Z

1

On your platform garbage characters don't print. On another platform it might be different.

As an experiment try this

std::cout << '|' << arr[ index ] << '|' << std::endl;

See if anything appears between the || characters.

answered Nov 21, 2012 at 16:14

john

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2 Comments

juanchopanza Over a year ago

More likely the "garbage values" are 0, no?

James Kanze Over a year ago

@juanchopanza More likely, the garbage values are whatever happens to be left there from previous functions. The system will initialize any memory it gives the process to zero, for security reasons, but crt0 has probably called a few functions before calling main, and these will probably have left non-zero values here and there.

Joseph Mansfield · Accepted Answer · 2012-11-21 16:17:37Z

0

You're getting undefined behaviour because you're attempting to use values from an uninitialised array. You can't expect anything in particular to happen. Maybe every character happens to be a non-printing character. Maybe it just decided that it didn't want to print anything because it doesn't like your little games. Anything goes.

answered Nov 21, 2012 at 16:17

Joseph Mansfield

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1 Comment

James Kanze Over a year ago

Actually, char is an exception, I think. (At least unsigned char is, and I think C++ extended this guarantee to char.) The standard does not allow char to have any trapping representation, and does allow using it to read "raw memory", including uninitialized raw memory.

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