why Hex data output differently?

Question

I am writing a very simple code to test the Hex data on c++

int main()
{
    unsigned char bytestosend[4] = {0xB5, 0x62, 0x06, 0x08};

    cout << &bytestosend << endl;
}

The data that comes on the terminal is : 0xbfa1ef58

how could this happen ? and when I remove the '&' it gives me strange symbols

Using the ampersand is giving you the memory location of your array, not the contents of it. If you remove the ampersand, you're printing a non-null-terminated string — Hybrid
– Hybrid, Commented May 27, 2013 at 13:51

Simon Broadhead · Accepted Answer · 2013-05-27 13:50:43Z

5

When you use the & it is outputting the address of the array and when you don't it is printing it as ASCII data.

answered May 27, 2013 at 13:50

Simon Broadhead

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7 Comments

David G Over a year ago

Doesn't it also need a null byte 0x0?

Simon Broadhead Over a year ago

Yes, it will most likely print a bunch of gibberish as it goes off the end of the array.

SirGuy Over a year ago

@0x499602D2 Only string literals initialized with "..." have a terminating null character put in for you.

David G Over a year ago

@GuyGreer Yes I know but since he's using a char array and he isn't initializing it with a string literal, isn't he suppose to explicitly add the null byte?

SirGuy Over a year ago

@0x499602D2 If he wants to be able to use it as a string, yes.

|

David Hammen · Accepted Answer · 2013-05-27 15:18:27Z

1

The '&' means 'address of' in this context. You are printing the location of bytestosend in memory rather than the contents of it.

Remove that '&' and you won't get what you want either. Now that stream insertion operator (operator >>) sees a char array, so it's going to try to print your array as if it was a string. Your array isn't null terminated, so that output might go on for a while.

If you want to print all four of the elements of bytestosend in hex use something like this:

std::cout.flags (std::ios::hex | std::ios::showbase);
std::cout << int(bytestosend[0]) << ", "
          << int(bytestosend[1]) << ", "
          << int(bytestosend[2]) << ", "
          << int(bytestosend[3]) << std::endl;

edited May 27, 2013 at 15:18

answered May 27, 2013 at 13:56

David Hammen

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2 Comments

David Hammen Over a year ago

Oops. Code fixed. std::ostream & operator<<(char) interprets the input as a char rather than an integer. You need to cast to non-char integral type to have the input interpreted as a number rather than a character.

David Hammen Over a year ago

Zeyad Serag - You do not need those brackets around int. I rolled back your edit. I prefer not to use things like (TypeName) x. This is a C-style cast. I always compile with warnings on use of C-style casts. TypeName(x) is a functional style cast. This is illegal syntax in C, but keep in mind that C++ is not C.

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