Cast char* to double - as bytes

Question

I have a byte array that represents double:

char number[8];

I need to cast this to actual double (which has 8 bytes as well). Based on advice I tried this, but it failed:

std::cout<<(*((*double)number))<<" is my number.\n";

Why did it fail and what should I do? I can, of course, extract the data using some << magic, but I don't want to do this - it would consume memory and make code too robust.

Type-punning like this is rather dangerous (though valid in the specific case of `char` IIRC, at least for now and as long as you only use one type). In general, a `union` is safer, and more convenient too. With pointers, strict aliasing is always out to get you. — , Mar 17 '13 at 18:25
@delnan, But reading from a member of a union that wasn't set last is UB, where as this is implementation-defined (at least the casting is IIRC). — chris, Mar 17 '13 at 18:27
@chris You won't get anything better than implementation-defined, the standard doesn't even specify how many bits there are in a `char`. However, implementation-defined is much better than undefined behavior (which is what you get with stuff like `*(int*)&a_double` IIRC). More importantly, even if both are equally broken by the standard, many compilers perform type-based alias analysis which can break code doing type punning with pointers, but doesn't break type punning via unions IIUC. — , Mar 17 '13 at 18:30
I hate it when my code is too robust. Use `memcpy` to copy the bytes into a double, the compiler will make it just as fast. — Jonathan Wakely, Mar 17 '13 at 22:12

Drew Dormann · Accepted Answer · 2013-03-17T18:37:57.147

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Why did it fail?

You have a typo here.

std::cout<<(*((*double)number))<<" is my number.\n";

It should be:

std::cout<<(*((double*)number))<<" is my number.\n";

and what should I do?

You could reduce the number of parenthesis used.

std::cout<< *(double*)number <<" is my number.\n";

You should use C++ casts instead of C casts, so it's clear what you're doing.

std::cout<< *reinterpret_cast<double*>(number) <<" is my number.\n";

edited Mar 17 '13 at 18:37

answered Mar 17 '13 at 18:24

Drew Dormann

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My typos will kill me one day. Thank you :) – Tomáš Zato Mar 17 '13 at 18:24
+1 I would recommend to use `reinterpret_cast(number)` instead of `(double*) number` though. – LihO Mar 17 '13 at 18:26
Thanks for stealing my c++ answer. – poitroae Mar 17 '13 at 18:30
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@poitroae: This was the first answer, that actually explained why the original code didn't work. – LihO Mar 17 '13 at 18:32
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Don't cry @poitroae. Do not forget it took you 5 edits to make it correct :P. Rather, I'm wondering why hint about parenthesizes reduction dissapeared. – Tomáš Zato Mar 17 '13 at 18:33
@TomášZato Restored. I worried that I was going on a tangent. :) – Drew Dormann Mar 17 '13 at 18:38

poitroae · Answer 2 · 2013-03-17T18:29:27.613

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If you use c++, then use reinterpret_cast. C++ have much more expressive, as you see.

// cool c++
double value = *reinterpret_cast<double*>(number);

// c style
double value = (*((double*)number));

edited Mar 17 '13 at 18:29

answered Mar 17 '13 at 18:22

poitroae

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Can I change type of the original variable? – Tomáš Zato Mar 17 '13 at 18:23
@OmriBarel: Will it work as `double value = reinterpret_cast(*number);` ? – Tomáš Zato Mar 17 '13 at 18:24
@TomášZato No, it should be `*reinterpret_cast(a)` – poitroae Mar 17 '13 at 18:26
@poitroae Your answer works now. I've tried with `{ 0xF3, 0x8E, 0x53, 0x74, 0x24, 0x97, 0xBF, 0x3F }` which is another way of saying 0.1234 on a little-endian machine. – Omri Barel Mar 17 '13 at 18:29
`reinterpret_cast` is not *cool c++*, it's just expressing the intent of the cast explicitly. – LihO Mar 17 '13 at 18:30
@LihO I think they're cool. *This is c++*. The old-style casts are there only for compat. – poitroae Mar 17 '13 at 18:31
@OmriBarel Okay, great! :D – poitroae Mar 17 '13 at 18:35

score 1 · Answer 3 · answered Mar 17 '13 at 18:24

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char number[8];
double d;
// number is assumed to be filled with a buffer representing a double.
memcpy(&d, &number, sizeof(double));
std::cout << d;

Not sure if the sizeof is needed. The damage was already dealt when the assumption that a double is 8 bytes was made. I don't know what it says in the standard about doubles.

answered Mar 17 '13 at 18:24

user123

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The standard says that a double most be large enough to hold any `float` value, although I'm not sure what the exact wording used is. – Cubic Mar 17 '13 at 18:28

Cast char* to double - as bytes

3 Answers3