C++ Convert 4 Hex Values to Float

Question

I am trying to convert 4 hexadecimal values to a float.

The hex values are for example 3F A0 00 00. In binary representation they would correspond to 00111111 10100000 00000000 00000000. If these 4 binary values are interpreted as one 32-bit float (accordingly to IEEE754) the decimal value of the float should be 1,25.

How to automatically make the conversion from hex values to decimal float in C++ (I am using Qt as Framework)?

Do you have the values in an array? You can memcpy them into the float, but don't forget the endianess: https://ideone.com/AKUkSE — mch, Jan 27 '21 at 13:31
yeah @CppProgrammer23 is right. Your question is a bit unclear. Do you want to reinterpret 4byte of memory to be a float? Are your hex values stings? — Robert Andreas Fritsch, Jan 27 '21 at 13:32

StPiere · Accepted Answer · 2021-01-27T13:47:47.883

2

#include <iostream>
#include <cstring>
using namespace std;

int main()
{
    unsigned char bytes[] = {0x3F, 0xA0, 0x00, 0x00};
    
    uint32_t x = bytes[0];
    for (int i = 1; i < std::size(bytes); ++i) x = (x << 8) | bytes[i];

    static_assert(sizeof(float) == sizeof(uint32_t), "Float and uint32_t size dont match. Check another int type");
    
    float f{};
    memcpy(&f, &x, sizeof(x));
    // or since C++20 if available: float f = std::bit_cast<float>(x)
    
    cout << "f = " << f << endl;
}

Live

edited Jan 27 '21 at 13:47

answered Jan 27 '21 at 13:33

StPiere

4,113
15
24

Note that this program interprets that bytes in big endian order. – eerorika Jan 27 '21 at 13:35
1

OP didn't say what order in which they want to interpret the bytes. Edit: Oh, they said that they want the value to be 1.25. Then they seem to want big endian indeed. – eerorika Jan 27 '21 at 13:38
I think he did implicitelly .. his binary representation of those float value correspond to the bytes in BE order. – StPiere Jan 27 '21 at 13:40
Slightly nicer C++20 alternative: `float f = std::bit_cast(x);` – eerorika Jan 27 '21 at 13:43
Also, to avoid magic numbers, I recommend as loop end condition: `i < std::size(bytes)` – eerorika Jan 27 '21 at 13:46
Wow thank you so much for this fast answer! – Lipa Jan 27 '21 at 13:55

Ted Lyngmo · Answer 2 · 2021-01-30T12:42:08.540

You have two possible values stored in that array (5.74855e-41 and 1.25) if your implementation uses 24 digit IEEE 754 floats - so you definitely need to deal with endianess.

Example:

#include <algorithm>
#include <bit>
#include <cstring>
#include <iostream>
#include <iterator>
#include <limits>

int main() {
    static_assert(std::numeric_limits<float>::is_iec559 &&
                  std::numeric_limits<float>::digits == 24,
                  "Only 24 digit IEEE 754 floats supported");

    unsigned char src[] = {0x3F, 0xA0, 0x00, 0x00};
    float dest;

    if constexpr (std::endian::native == std::endian::little) {
        // swap the byte order
        std::reverse(std::begin(src), std::end(src));
    }

    std::memcpy(&dest, src, sizeof(float));
    
    std::cout << dest << '\n';    // prints 1.25
}

Demo

score -1 · Answer 3 · edited Nov 03 '21 at 07:58

-1

float Uint32ToFloat(uint32_t data)
{
    float returnData = 0.0;

    static_assert(sizeof(float) == sizeof(uint32_t), "Float and uint32_t size dont match. Check another int type");
    memcpy(&returnData, &data, sizeof(data));
    return returnData;
}

edited Nov 03 '21 at 07:58

kiatng

2,847
32
39

answered Nov 03 '21 at 00:44

Louise

1

I cannot see how this adds anything to the accepted answer – klutt Nov 03 '21 at 08:53
data is uint32 hex in memory. when you return a float, it handled by machine . similar as casting. – Louise Nov 11 '21 at 16:55
Yes, but what does it add to the accepted answer? StPiere has already explained that. – klutt Nov 12 '21 at 01:51

C++ Convert 4 Hex Values to Float

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