将单精度浮点数乘以2

Question

This is a homework question. I already found a lot of code online, including some code in StackOverflow. But I just want the concept not the code. I want to implement it myself. So the function I want to implement is:

• float_twice - Return bit-level equivalent of expression 2*f for floating point argument f .
• Both the argument and result are passed as unsigned int 's, but they are to be interpreted as the bit-level representation of single-precision floating point values.

I want to know how to do this. I know floating point representation. And read wiki page on how to multiply two floats, but didn't understand it. I just want to know the concept/algorithm for it.

Edit:

Thanks everyone. Based on your suggestions I wrote the following code:

unsigned float_twice(unsigned uf) {
    int s = (uf >> 31) << 31;
    int e = ((uf >> 23) & 0xFF) << 23;
    int f = uf & 0x7FFF;

    // if exponent is all 1's then its a special value NaN/infinity
    if (e == 0xFF000000){
        return uf;

    } else if (e > 0){  //if exponent is bigger than zero(not all zeros', not al 1's, 
                        // then its in normal form, add a number to the exponent
        return uf + (1 << 23);

    } else { // if not exponent not all 1's and not bigger than zero, then its all 
             // 0's, meaning denormalized form, and we have to add one to fraction

        return uf +1;
    } //end of if

} //end of function

Answer 1

You could do something like this (although some would claim that it breaks strict-aliasing rules):

unsigned int func(unsigned int n)
{
    float x = *(float*)&n;
    x *= 2;
    return *(unsigned int*)&x;
}

void test(float x)
{
    unsigned int n = *(unsigned int*)&x;
    printf("%08X\n",func(n));
}

In any case, you'll have to assert that the size of float is equal to the size of int on your platform.

---

If you just want to take an unsigned int operand and perform on it the equivalent operation of multiplying a float by 2, then you can simply add 1 to the exponent part of it (located in bits 20-30):

unsigned int func(unsigned int n)
{
    return n+(1<<20);
}