Search code examples
cfloating-pointdoubleieee-754

How does double 5.2 is represented in a binary file?

For example the given program

FILE *f = fopen("test.txt","w");
double d = 5.2;
fwrite(&d, sizeof(typeof(d)), 1, f);
fclose(f);

when U use xxd to view the content I get

//binary
0000000: 11001101 11001100 11001100 11001100 11001100 11001100  ......
0000006: 00010100 01000000                                      .@
//hex
0000000: cdcc cccc cccc 1440                      .......@

but I found How to represent FLOAT number in memory in C that it should look like

0 10000001 01001100110011001100110

so I am not sure if its the same binary in memory and file or I am completely wrong. How can i determine the float value from given example?

Solution

  • The bit pattern you show, 0 10000001 01001100110011001100110, is for an IEEE-754 basic 32-bit binary floating-point representation, often used for float. You request the representation of a double, which often uses IEEE-754 basic 64-bit binary.

    The bytes you show, cdcc cccc cccc 1440, are a representation of approximately 5.2, in a little-endian order. In the high-value byte, the first bit, in position 0x80, is zero. It is the sign bit, and zero means positive. The next seven bits, 0x40, and the four bytes from the next byte, the 1 of 0x14, are the exponent. Together, they are 0x401. The exponent is biased by 0x3ff. So, with an exponent encoded as 0x401, the actual exponent is 2, meaning 22.

    The remaining bits encode the trailing bits of the significand. They are 0x4cccccccccccd. For normal numbers, the significand is put after a “radix point” (the general equivalent of a decimal point) with 1 before the point: 1.4cccccccccccd16. In decimal, that is approximately 1.3 (exactly 1.3000000000000000444089209850062616169452667236328125).

    Together, the value is + 22 • 1.3000000000000000444089209850062616169452667236328125 = 5.20000000000000017763568394002504646778106689453125.