How to get the binary of a double in C?

Question

For a char or int, I can do this:

void print2(char c)//print char c in binary
{
    for(int i=7;i>=0;i--)
        if ((1<<i)&c) printf("1");
        else printf("0");
}

But for double I can't use operator &.

And I can't use reinterpret_cast(It's C++).

Then, How to do that?

memcpy? I never knew it. Sounds interesting. @DanielFischer Thanks. — Buddha, May 09 '13 at 10:56
Oh wait, I think I've misinterpreted "And I can't use reinterpret_cast(It's C++).". You're working in C, then using a union is simpler (less code) than `memcpy`. If you can `assert(sizeof (double) == sizeof(uint64_t))`, I'd suggest a `union pun { double d; uint64_t u; };`, otherwise using an `unsigned char[sizeof (double)]` in the union would be an option. — Daniel Fischer, May 09 '13 at 11:02
@DanielFischer Plus: you would need to add an `assert (sizeof(double) == sizeof(unsigned long long))` somehere. IMHO the memcpy to char[sizeof(double)] is more flexible wrt portability. (especially if stdint.h is not available) — wildplasser, May 09 '13 at 11:14

score 2 · Answer 1 · answered May 09 '13 at 11:17

2

A solution with union rather than memcpy or pointer:

void print2(double x)//print double x in binary
{
    union {
        double x;
        char c[sizeof(double)];
    } u;

    assert(sizeof(char) == 1);

    u.x = x;

    for (unsigned ofs = 0; ofs < sizeof(double); ofs++) {
        for(int i = 7; i >= 0; i--) {
            printf(((1 << i) & u.c[ofs]) ? "1" : "0");
        }
        printf(" ");
    }
}

answered May 09 '13 at 11:17

Axel Kemper

10,544
2
31
54

4

LOL@ `assert(sizeof(char) == 1);` Priceless! – wildplasser May 09 '13 at 11:18
Priceless, indeed. What was the point of that? – This isn't my real name May 09 '13 at 14:59

stralep · Answer 2 · 2013-05-09T10:56:26.907

1

Create char pointer and make it point to this double. Print first sizeof(double) chars using your function.

edited May 09 '13 at 10:56

answered May 09 '13 at 10:51

stralep

858
8
16

Eric Postpischil · Answer 3 · 2013-05-09T11:44:15.907

Presuming a double has 64 bits, this reinterprets the bytes of a double x as an unsigned 64-bit integer:

(union { double d; uint64_t u; }) {x} .u;

This is legal C, with the bytes of the double being implementation-dependent. It defines a compound literal that is a union, initializes the union with the double, and accesses the uint64_t member. The C standard says that when a member other than the last-stored member is accessed, the bytes are reinterpreted as the new type.

score 1 · Accepted Answer · answered May 09 '13 at 11:44

Naive implementation with memcpy(), as per @Daniel Fischer's comment:

#include <stdio.h>
#include <string.h>
#include <limits.h>

void double_to_bits(double val);

int main(void)
{
unsigned idx;

double vals[] = { -1.0, 0.0, 1.0, 2.0 };

for (idx = 0; idx < 4; idx++ ) {
        printf("\nvals[%u]= %+lf-->>", idx, vals[idx] );
        double_to_bits(vals[idx] );
        }
printf("\n");

return 0;
}

void double_to_bits(double val)
{
unsigned idx;
unsigned char arr[sizeof val];

memcpy (arr, &val, sizeof val);

for (idx=CHAR_BIT * sizeof val; idx-- ; ) {
        putc(
        ( arr[idx/CHAR_BIT] & (1u << (idx%CHAR_BIT) ) )
        ? '1'
        : '0'
        , stdout
        );
        }
}

, and the same with a pointer (omitting the char array and the memcpy)

void double_to_bits2(double val)
{
unsigned idx;
unsigned char *ptr = (unsigned char*) &val;

for (idx=CHAR_BIT * sizeof val; idx-- ; ) {
        putc(
        ( ptr[idx/CHAR_BIT] & (1u << (idx%CHAR_BIT) ) )
        ? '1'
        : '0'
        , stdout
        );
        }
}

How to get the binary of a double in C?

4 Answers4