Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
920 views
in Technique[技术] by (71.8m points)

c - Converting float to an int (float2int) using only bitwise manipulation

I am wondering if someone could set me in the right direction with a problem I am working on. I am trying to do what the following C function does using only ARM assembly and bit manipulation:

int float2int(float x) {
return (int) x;
}

I have already coded the reverse of this (int2float) without many issues. Im just unsure of where to start with this new problem.

For example:

3 (int) = 0x40400000 (float) 
0011 = 0 10000000 10000000000000000000000

Where 0 is the Sign Bit, 10000000 is the exponent, and 10000000000000000000000 is the mantissa/fraction.

Can someone simply point me in the right direction with this problem? Even a C pseudocode representation would be helpful. I know I need to extract the sign bit, extract the exponent and reverse the bias (127) and also extract the fraction but I just have no idea where to begin.

There is also the issue of if the float cannot be represented as an integer (because it overflows or is a NaN).

Any help would be appreciated!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Start with a number whose mantissa is your number (00000011 in your example) and whose exponent is 01111111 (127, which is how 0 is stored in excess-of-127) Count how many bits are from the LSb to the last set bit (not included). For each bit counted, add 1 to the exponent.

In your example: there are only one bit from the LSb to the last (most significant) bit set, so the exponent is added by 1 resulting in 128 (10000000).

Shift left your mantissa (your original number) so the left-most set bit is lost. Take into account that the shift must be performed using a variable capable of holding at least 23 bits. So in your example, the original mantissa is 00000000000000000000011 . You must shift left it until the left-most '1' is lost, resulting in 10000000000000000000000 About the sign, if the original number is in 2-complement, simply take the MSb, and that will be your sign. In your example, 0 (positive)

So your result will be: Sign : 0 Exponent: 10000000 Mantissa: 10000000000000000000000

Another example: convert the short int number -234 into a float. -234 using 2-complement is stored as 1111111100010110 (16 bits)

From here it's easy to get the sign: 1 (the MSb)

We must work with the absolute magnitude, so we complement the number to get the positive (magnitude) version. We can do it by xoring it with 1111111111111111, then adding 1. This gives us 0000000011101010 (234)

Initial mantissa (using 23 bits): 00000000000000011101010 Initial exponent: 01111111 (127) Count how many bits are from the LSb to the left-most set bit, withouth including it. There are 7 bits. We add this to our exponent: 127+7=134 = 10000110 The mantissa is shifted left until the left-most set bit is gone. This gives us: 11010100000000000000000

Our number will be: 1 10000110 11010100000000000000000


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...