在不支持双精度的C编译器上将64位双精度浮点数据转换为Uint32

Question

I need to decode a timestamp encoded as an IEEE double (from iOS NSTimeInterval ) and stored in a 8 byte array, so that I can use ctime to print out the timestamp in a human readable format. This is trivial on most systems, but not on mine.

Example: on iOS side

uint8_t data[8];
double x = 3.14;
memcpy(data,&x,8);

I'm running on a MSP430F5438A and I need to use the COFF ABI to link with 3rd part libraries. TI's Code Composer does not support 64-bit floating point types (IEEE double) if you use the COFF ABI. It treats double the same as float (ie single precision).

These don't work.

uint8_t data[8];
double x; 
memcpy(&x,data,8);

or

x = *(double*)data;

or

union {
   uint8_t data[8];
   double d;
} x;
memcpy(x.data,data,8);

I just get gibberish, because double is only 4 bytes using Code Composer. I need a way to directly convert the uint8_t[8] data (which is a legitimate IEEE double precision value) into an integer value.

Answer 1

This will convert an integral double to an exact uint32_t (as long as there is no overflow, up to 2^32-1). It will not work if the double is Nan or Inf, but that is unlikely.

static unsigned long ConvertDoubleToULong(void* d)
{
 unsigned long long x;
 unsigned long long sign ;
 long exponent;
 unsigned long long mantissa;

 memcpy(&x,d,8);

 // IEEE binary64 format (unsupported)
 sign     = (x >> 63) & 1; // 1 bit
 exponent = ((x >> 52) & 0x7FF); // 11 bits
 mantissa = (x >> 0) & 0x000FFFFFFFFFFFFFULL; // 52 bits
 exponent -= 1023;

 mantissa |=           0x0010000000000000ULL; // add implicit 1

 int rshift = 52 - exponent;
 if (rshift > 52) {
    x = 0;
 } else if (rshift >=0) {
    x = mantissa >> rshift;
 } else {
    x = 0x7FFFFFFF;
 }
 if (sign == 0) {
    return x;
 } else {
    return -x;
 }
}