Converting float to 32-bit hexadecimal C++

Question

Could anyone please tell me if there is any code to convert floating point number to hexadecimal format?

For Ex: float num = 70.482 and hexadecimal function should return 428CF6C9.

If there are any codes already done before on this, kindly link me.

Cheers.

That's what the online converter returned the value as. babbage.cs.qc.edu/IEEE-754/Decimal.html — Neophile
– Neophile, Commented Aug 30, 2011 at 15:39
Did you look at the JavaScript for that page? You would need to translate to c++, but at least you can look at their algorithm. — crashmstr
– crashmstr, Commented Aug 30, 2011 at 15:43

Kerrek SB · Accepted Answer · 2011-08-30 18:30:56Z

17

You can just trivially write that yourself:

float x;
const unsigned char * pf = reinterpret_cast<const unsigned char*>(&x);

for (size_t i = 0; i != sizeof(float); ++i)
{
  // ith byte is pf[i]
  // e.g. printf("0x02X ", pf[i]);
}

In fact, you can do that to obtain the binary representation of any (standard-layout*) variable.

_{*) thanks, @R. Martinho Fernandes!}

If you decide to try this on a long double (or rather, an 80-bit extended precision float), beware that that has only 10 bytes, but is padded to 12 or 16 on x86/x64, respectively.

edited Aug 30, 2011 at 18:30

answered Aug 30, 2011 at 15:38

Kerrek SB

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11 Comments

Praetorian Over a year ago

+1 And you don't need to worry about long doubles if you're using MSVC :-)

R. Martinho Fernandes Over a year ago

Not any variable though. In C++11 the value is unspecified if either the source or the destination pointer types is not a standard-layout type. I'm pretty sure there's some similar restriction in C++03.

phoxis Over a year ago

will the byte ordering of the architecture effect the order of bytes which are printed ?

R. Martinho Fernandes Over a year ago

Well, if A is, for example, a type with a virtual member function and given A a; auto b = reinterpret_cast<const char*>(&a); the value of b is unspecified. It bet in pretty much all implementations you will get to see the vptrs and all that, though. Oh, and the guarantee that you can roundtrip is still valid: auto x = reinterpret_cast<const A*>(b); will give you the original pointer, even though the intermediate value is unspecified.

Semjon Mössinger Over a year ago

By looping from 0 to 3 you flip the result from left to right! Instead of "428CF6C9" you will get "C9F68C42".

|

Michael · Accepted Answer · 2011-08-31 15:45:08Z

4

something like this is the floattohex conversion i use.. (it also performs a bit swap, if you dont need this take it out)

CString Class::FloatToHex(float* yourfloat)
{
unsigned char ch[4];
Cstring output;

memcpy(ch,yourfloat,sizeof(float));
output.Format("%X%X%X%X",ch[3],ch[2],ch[1],ch[0]);

return output;
}

answered Aug 31, 2011 at 15:45

Michael

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1 Comment

Kerrek SB Over a year ago

Isn't this making the dangerous and unchecked assumption that sizeof(float) does not exceed 4, and would otherwise trash your memory?

phoxis · Accepted Answer · 2011-08-30 15:59:07Z

2

Consider:

union float_bits {
    unsigned int i;
    float f;
} bits;

Assign the floating point number into bits.f, and then interpret the whole number as an unsigned integer and read it with bits.i. By doing this the floating point representation bytes will stay intact, and no implicit type conversion will be done as it would be when assigning a float to an int type variable. In this case we assume that the size of integer is same as the size of float.

Or you can do:

float f;
char *c = (char *) &f;

And then access the individual bytes of f through c[index].

edited Aug 30, 2011 at 15:59

answered Aug 30, 2011 at 15:39

phoxis

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4 Comments

pezcode Over a year ago

that is assuming sizeof(int) == sizeof(float)

Kerrek SB Over a year ago

That's also undefined behaviour.

phoxis Over a year ago

@R. Martinho Fernandes: right, i went the c way. corrected now.

Voo Over a year ago

@Kerrek SB While the union example is technically undefined, it's an often used idiom and quite certainly will be supported by any compiler and every other solution is ugly and unhandy. I'd still mention it (and use it myself, bad me) but a warning is certainly in place I agree..

Retired Ninja · Accepted Answer · 2011-08-30 15:56:27Z

1

float f = 70.482f;
int i = *(reinterpret_cast<int*>(&f));
printf("%08x\n", i);

answered Aug 30, 2011 at 15:56

Retired Ninja

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2 Comments

Mooing Duck Over a year ago

@Kerrek: What is the subtlety that allows the casting to const unsigned char that you did, and not int? (I believe you, I just don't see it)

Kerrek SB Over a year ago

The difference is that casting to char is allowed, but casting to other types isn't :-)

Grim Fandango · Accepted Answer · 2017-07-13 19:25:22Z

1

you could use a union:

union FloatToChar {
    float f;
    char  c[sizeof(float)];
};

FloatToChar x;
x.f = 10.42f;
for (size_t i=0; i<sizeof(float); i++)
    printf( "%X", x.c[i] );

you could do this vice-versa, too.

edited Jul 13, 2017 at 19:25

answered Nov 14, 2011 at 12:58

Grim Fandango

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2 Comments

Toby Speight Over a year ago

This seems to assume that sizeof (float) is 4 - can you justify that?

Grim Fandango Over a year ago

@TobySpeight, I changed the code to cater for all cases of sizeof(float)

Mikarnage · Accepted Answer · 2011-08-30 15:40:31Z

-1

Simply use sprintf() : http://www.cplusplus.com/reference/clibrary/cstdio/sprintf/

For example,

char out[8];
float myFloat = 70.482;
sprintf(out,"%X",myFloat);

answered Aug 30, 2011 at 15:40

Mikarnage

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2 Comments

Neophile Over a year ago

Yea, we tried that. Did not work the way we wanted it to. Did not return the correct value.

Kerrek SB Over a year ago

Careful: float gets promoted to double when passed through variadic arguments!

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