CODEWITHSUNDEEP
c++cfloating-pointunions
Cole Tobin
Cole Tobin

Reputation: 9426

Runtime error in a program supposed to convert a float to a byte array

I was given a floating point variable and wanted to know what its byte representation is. So I went to IDEOne and wrote a simple program to do so. However, to my surprise, it causes a runtime error:

#include <stdio.h>
#include <assert.h>

int main()
{
    // These are their sizes here. So just to prove it.
    assert(sizeof(char) == 1);
    assert(sizeof(short) == 2);
    assert(sizeof(float) == 4);

    // Little endian
    union {
        short s;
        char c[2];
    } endian;
    endian.s = 0x00FF; // would be stored as FF 00 on little
    assert((char)endian.c[0] == (char)0xFF);
    assert((char)endian.c[1] == (char)0x00);

    union {
        float f;
        char c[4];
    } var;
    var.f = 0.0003401360590942204;
    printf("%x %x %x %x", var.c[3], var.c[2], var.c[1], var.c[0]); // little endian
}

On IDEOne, it outputs:

39 ffffffb2 54 4a

along with a runtime error. Why is there a runtime error and why is the b2 actually ffffffb2? My guess with the b2 is sign extension.

Upvotes: 0

Views: 334

Answers (3)

Kerrek SB
Kerrek SB

Reputation: 477454

Your approach is all kinds of wrong. Here's how you print a general object's binary representation:

template <typename T>
void hexdump(T const & x)
{
    unsigned char const * p = reinterpret_cast<unsigned char const *>(&x);
    for (std::size_t i = 0; i != sizeof(T); ++i)
    {
        std::printf("%02X", p[i]);
    }
}

The upshot is that you can always interpret any object as a character array and thus reveal its representation.

Upvotes: 2

hivert
hivert

Reputation: 10667

Replace char by unsigned char in the struct and add a return 0; at the end fixes all the problems: http://ideone.com/ienG2b.

Upvotes: 5

user529758
user529758

Reputation:

char is a signed type. If it's 8 bits long and you put anything greater than 127 in it, it will overflow. Signed integer overflow is undefined behavior, so is printing a signed value using a conversion specifier that expects an unsigned one (%x expects unsigned int, but char is promoted [implicitly converted] to signed int when passed to the variadic printf() function).

Bottom line - change char c[4] to unsigned char c[4] and it will work fine.

Upvotes: 6

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