将字节重新解释为C中的浮点数（IEEE 754单精度二进制）

Question

I want to reinterpret 4 bytes as IEEE 754 single-precision binary in C.

To obtain the bytes that represent float , I used:

num = *(uint32_t*)&MyFloatNumber;
aux[0] = num  & 0xFF;
aux[1] = (num >> 8) & 0xFF;
aux[2] = (num >> 16)  & 0xFF;
aux[3] = (num >> 24)  & 0xFF;

• num is a uint32_t.
• aux[] is int[4].

To reinterpret the bytes as a float , I'm trying:

Buff = *(float*)&aux;

In that second case nothing apears on "Buff"

• Buff is a float .

What am I doing wrong in the second case?

Answer 1

2 problems:

1. Buff = *(float*)&aux; attempts to use the address of an array of 4 int as a pointer to a float. aux[] is perhaps 16 bytes long and a IEEE 754 single-precision binary float is expected to be 4 bytes.

2. Both casts: (uint32_t*) and (float*) invoke undefined behavior as well as alignment and anti-aliasing issues. Better to use a union .

    int main(void) { union { float f; unsigned char uc[sizeof(float)]; } x; // float to bytes xf = 1.23f; printf("%x %x %x %x\\n", x.uc[0], x.uc[1], x.uc[2], x.uc[3]); // bytes to float x.uc[0] = 0xA4; x.uc[1] = 0x70; x.uc[2] = 0x9D; x.uc[3] = 0x3F; printf("%.8e\\n", xf); } 

Output

a4 70 9d 3f
1.23000002e+00