Convert int to float: how it is done

Question

I'm new to C programming language, and I'd like to ask a question.

Integer i here is casting to float then f (somehow) successfully represents 5.0:

int i = 5;
float f = i;   //Something happened here...

However if we try this approach:

int i = 5;
float f = *(float *)&i;

f would NOT get 5.0 since it interprets the bits stored in i in "float's way". So what magic the complier actually does in the first case? It seems a quite effort-taking job... Can someone specify that? Thanks.

Answer 1

convert int bits to float

float IntBitsToFloat(long long int bits)
{
    int sign     = ((bits & 0x80000000) == 0) ? 1 : -1;
    int exponent = ((bits & 0x7f800000) >> 23);
    int mantissa =  (bits & 0x007fffff);

    mantissa |= 0x00800000;
   // Calculate the result:
   float f = (float)(sign * mantissa * Power(2, exponent-150));
   return f;
}

Answer 2

Well, I just compiled the code in question under VC++ and looked at the disassembly:

   int i = 5;
00A613BE  mov         dword ptr [i],5 

   float f = i;
00A613C5  fild        dword ptr [i] 
00A613C8  fstp        dword ptr [f] 

First assembly statement moves 5 into the memory represented by i. Second statement, fild, converts the memory represented by i into a floating point number and pushes it onto the FPU stack. The third statement, fstp, takes the memory on the FPU stack and moves it to f.