使用std :: cout << std :: hex填充x的倍数的前导零

Question

There are many questions regarding how to pad a fixed number of leading zeroes when using C++ streams with variables we want represented in hexadecimal format:

std::cout << std::hex << setfill('0') << setw(8) << myByteSizedVar;

My question regards how to do this for not a fixed width, but a multiple of some fixed amount - likely 8 for the obvious reason that when comparing outputs we might want:

0x87b003a
0xab07

To match up for width to be compared more easily (okay the top is larger - but try a bitwise comparison in your head? Easily confused.)

0x87b003a
0x000ab07

Nice, two bytes lined up nice and neatly. Except we only see 7 bits - which is not immediately obvious (especially if it were 15/16, 31/32, etc.), possibly causing us to mistake the top for a negative number (assuming signed).

All well and good, we can set the width to 8 as above.

However, when making the comparison next to say a 32-bit word:

0x000000000000000000000000087b003a
0x238bfa700af26fa930b00b130b3b022b

It may be more unneccessary, depending on the use, or even misleading if dealing with hardware where the top number actually has no context in which to be a 32-bit word.

What I would like, is to automagically set the width of the number to be a multiple of 8, like:

std::cout << std::hex << std::setfill('0') << setWidthMultiple(8) << myVar;

For results like:

0x00000000
0x388b7f62
0x0000000f388b7f62

How is this possible with standard libraries, or a minimal amount of code? Without something like Boost.Format .

Answer 1

How about this:

template <typename Int>
struct Padme
{
    static_assert(std::is_unsigned<Int>::value, "Unsigned ints only");

    Int n_;
    explicit Padme(Int n) : n_(n) {}

    template <typename Char, typename CTraits>
    friend
    std::basic_ostream<Char, CTraits> & operator<<(
        std::basic_ostream<Char, CTraits> & os, Padme p)
    {
        return os << std::setw(ComputeWidth(p.n_)) << p.n_;
    }

    static std::size_t ComputeWidth(Int n)
    {
        if (n <         0x10000) { return  4; }
        if (n <     0x100000000) { return  8; }
        if (n < 0x1000000000000) { return 12; }
        return 16;
    }
};

template <typename Int>
Padme<Int> pad(Int n) { return Padme<Int>(n); }

Usage:

std::cout << pad(129u) << "\n";

With some more work you could provide versions with different digit group sizes, different number bases etc. And for signed types you could stick std::make_unsigned into the pad function template.

Answer 2

There's no immediate support for it, because it involves more than one input value (if I understand what you're asking for correctly). The only solution which comes to mind is to use std::max to find the largest value, and then derive the number of digits you need from that, say by using a table:

struct DigitCount
{
    unsigned long long maxValue;
    int digits;
};
static const DigitCount digitCount[] =
{
    {        255UL, 2 },
    {      65535UL, 4 },
    {   16777215UL, 6 },
    { 4294967295UL, 8 },
};

and looking up the width in it.