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How do I modify bits of float/double without undefined behavior?

I want to set/clear bits of a float and double in C++ and print out the result. I'm positive long long bits = *(long long*)&doubleVar; invokes undefined behavior. I'm not sure if putting it in a union is correct either. How do I get/set bits of a float and double in C++ without causing undefined behavior?

long long bits = *(long long*)&doubleVar;

char* pChar = reinterpret_cast< char* >(&doubleVar);

reinterpret_cast

2 Answers
2

To avoid undefined behavior you need to use memcpy() A simple way to do this is using the proposed template std::bit_cast<to_type>(from_type from)

memcpy()

std::bit_cast<to_type>(from_type from)

You can get the template that wraps this operation (it's only proposed) in https://en.cppreference.com/w/cpp/numeric/bit_cast and is simple to use:

double pun_me=3.0;
std::uint64_t ui64=my_namespace::bit_cast<uint64_t>(pun_me);


It's a good idea to put it in a separate namespace so you don't get conflicts when/if it comes out in C++20. You can them modify whatever you wish and convert it back.

You can use unsigned char * for that, that's the simplest yet standard conformant way, if you need to modify some of the bits of an existing float/double. See:

unsigned char *

float

double

float a = 1.0f;
unsigned char *x = reinterpret_cast<unsigned char *>(&a);
x[3] ^= 0x80; // let's change sign of a (assuming IEEE-754, and little-endian, 8-bit chars)


Doing this with memcpy or bit_cast is more complex.

memcpy

bit_cast

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