Cannot display in hexadecimal
unsigned int i = 0x02081;
cout << std::hex << i;
This displays 2081 when compiled 开发者_运维问答with VS2010 but I think it should display 0x02081. Am I right, and if so, how can this be fixed?
By default the base is not printed:
cout << std::hex << std::showbase << i;
The easiest solution, of course, is:
cout << "0x" << std::hex << i;
Leading zeroes can vary in amount because they don't matter. You can choose any amount you like.
I think it should display 0x02081. Am I right
No, you're not. It will display the value in hex, which is 2081
. The 0x
isn't part of the number, per se, it's just a notational convenience. The leading zero is also not a part of the number.
If you want the exact output you said you expected, you can do this:
cout << std::hex << std::showbase << std::setw(5) << std::setfill('0') << i;
It should display the value of i
as a hexadecimal number - and does so. The prefix 0x
is just something some programming languages use to indicate that a literal should be considered hexadecimal.
cout << "0x" << std::hex << i;
As Let_Me_Be points out: "0x" << std::hex
could be replaced with std::hex << std::showbase
if you want automatic printing of 0x/0/nothing for hex/octal/decimal.
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