c++ - Hexadecimal representations of 32-bit numbers

Question

I have a script that generates two random 32-bit floating point numbers a and b, and divides them to make an output c.

I would like to store all 3 of these floating point numbers in hexadecimal format, as a string that contains 8 characters.

I found a clever way to do this online here:

http://forums.devshed.com/c-programming-42/printing-a-float-as-a-hex-number-567826.html

And here's my implementation of their advice in C++

    fileout << hex << *(int*)&a[i] << endl;
    fileout << hex << *(int*)&b[i] << endl;
    fileout << hex << *(int*)&c[i] << endl;

This works for most of the cases. However, some of the cases, the strings are not 8 characters wide. Sometimes they are only one bit long. Here is the sample of the output:

                            af1fe786
                    ffbbff0b
                    fffbff0b
                    7fbcbf00  <-I like it, but has zeros at the end 
                    77fefe77
                    7ffcbf00
                    fdad974d
                    f2fc7fef
                    4a2fff56
                    67de7744
                    fdf7711b
                    a9662905
                    cd7adf0   <-- problem
                    5f79ffc0
                    0         <--- problem
                    6ebbc784
                    cffffb83
                    de3bcacf
                    e7b3de77
                    ec7f660b
                    3ab44ae4
                    aefdef82
                    fffa9fd6
                    fd1ff7d2
                    62f4      <--why not "62f40000"
                    ebbf0fa6
                    ddd78b8d
                    4d62ebb3
                    ff5bbceb
                    3dfc3f61
                    ff800000 <- zeros at end, but still 8 bytes?
                    df35b371
                    e0ff7bf1
                    3db6115d
                    fbbfbccc
                    ddf69e06
                    5d470843
                    a3bdae71
                    fe3fff66
                    0         <--problem
                    979e5ba1
                    febbe3b9
                    0         <-problem
                    fdf73a80
                    efcf77a7
                    4d9887fd
                    cafdfb07
                    bf7f3f35
                    4afebadd
                    bffdee35
                    efb79f7f
                    fb1028c   <--problem

I want 8-character representations. As for the case of zero, I want to convert it to "00000000".

But I am really confused about the ones that are only 4, 5, 6, 7 characters long. Why do some numbers get zero filled at the end and others get truncated? If an int is 32 bits, why does sometimes only one bit show up? Is this due infamous "subnormal" numbers?

Thanks.

Answer 1

If I understand correctly your requirement, you want to add zeros as a filling character to the left of your hexadecimal representations if the number of digits is less than 8.

In that case, you can simply use the std::setfill() the std::setw() stream manipulators:

#include <iomanip> // Necessary for the setw and setfill

int n = ...;
std::cout << std::hex << std::setw(8) << std::setfill('0') << n;

For n = 1024, for instance, the output will be:

00000400