Convert Binary to Float

Question

I'm trying to assign a float from a raw binary string, but I don't get what I would expect.

int main()
{
    char recBuffer[] = {0x44, 0x42, 0x96, 0x12, 0x34, 0x56, 0x78};
    float tempFloat;
    int position = 3;

    printf( "recBuffer=0x%x%x%x%x\n", recBuffer[3], recBuffer[4], recBuffer[5], recBuffer[6]);
    memcpy( &tempFloat, recBuffer + position, 4 );
    printf( "tempFloat=%f (0x%x)\n", tempFloat, tempFloat );
    return 0;
}

My output looks like:

recBuffer=0x12345678
tempFloat=*************************************** (0x40000000)

The above procedure works for integers:

int main()
{
    char recBuffer[] = {0x44, 0x42, 0x96, 0x12, 0x34, 0x56, 0x78};
    int tempFloat;
    int position = 3;

    printf( "recBuffer=0x%x%x%x%x\n", recBuffer[3], recBuffer[4], recBuffer[5], recBuffer[6]);
    memcpy( &tempFloat, recBuffer + position, 4 );
    printf( "tempFloat=%d (0x%x)\n", tempFloat, tempFloat );
    return 0;
}

with an output:

recBuffer=0x12345678
tempFloat=2018915346 (0x78563412)

(I'm aware of Endianness.)

I've tried to assign the float directly, but I still get something weird as an output (what do all the *'s mean?).

int main()
{
    char recBuffer[] = {0x44, 0x42, 0x96, 0x12, 0x34, 0x56, 0x78};
    float* tempFloat;
    int position = 3;

    printf( "recBuffer=0x%x%x%x%x\n", recBuffer[3], recBuffer[4], recBuffer[5], recBuffer[6]);
    tempFloat = (float*)(recBuffer + position);
    printf( "tempFloat=%f (0x%x)\n", *tempFloat, *tempFloat );
    return 0;
}

with output:

recBuffer=0x12345678
tempFloat=*************************************** (0x40000000)

Any binary sequence should give me an output, as 0x78563412 = 1.73782443614495040019632524267E34. I'm not sure why 0x40000000 = *. It should be 2.0E0. What am I doing wrong?! Any help would be appreciated!

(unfortunately I'm working on an old QNX machine, with no debugger. Just simple printf()'s to help me along).

The %x format won't print the float because the call to printf() automatically converts the float to a double (and anything shorter than int to int). What happens when you use %e instead of %f? — Jonathan Leffler
– Jonathan Leffler, Commented May 17, 2013 at 19:46
I get: recBuffer=0x12345678 tempFloat=*************************************** (1.737824e+034) Amazing. It's never where you'd expect. Thank you so much. — Taylor Moore
– Taylor Moore, Commented May 17, 2013 at 20:04

masoud · Accepted Answer · 2013-05-17 19:48:02Z

1

printf( "tempFloat=%d (0x%x)\n", tempFloat, tempFloat);
                          ^                     |
                          |                     |
                          +---------------------+

The %x specifier is useful for integer numbers but you're passing a float value to the printf. So the output is not a meaningful value.

answered May 17, 2013 at 19:48

masoud

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1 Comment

Taylor Moore Over a year ago

That did it! recBuffer=0x12345678 tempFloat=*************************************** (1.737824e+034) I still don't understand why %f would return ****, but I'm happy memcpy() does what I'd expect. Thank you for your help!

BLUEPIXY · Accepted Answer · 2013-05-17 20:06:16Z

0

int main(){
    char recBuffer[] = {0x44, 0x42, 0x96, 0x12, 0x34, 0x56, 0x78};
    float tempFloat;
    int position = 3;

    printf( "recBuffer=0x%x%x%x%x\n", recBuffer[3], recBuffer[4], recBuffer[5], recBuffer[6]);
    memcpy( &tempFloat, recBuffer + position, 4 );
    printf( "tempFloat=%f (0x%x)\n", tempFloat, *(unsigned*)&tempFloat );
    return 0;
}
/*
recBuffer=0x12345678
tempFloat=17378244361449504000000000000000000.000000 (0x78563412)
*/

answered May 17, 2013 at 20:06

BLUEPIXY

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1 Comment

Taylor Moore Over a year ago

%f still gives me *******, but using your advice: printf( "tempFloat=%e (0x%x)\n", tempFloat, *(unsigned*)&tempFloat ); allows me to get the raw hex value!: recBuffer=0x12345678 tempFloat=1.737824e+034 (0x78563412). That helps me alot, thanks!

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