Find value of struct-variable in assembly code

Question

I'm preparing for my exam of computer architecture this year and I'm solving some exams from previous years, but I have this one exercise which I can't solve. I'm asked to find the value of g.z when the following C and assembly-code is given:

You can see in the assembly-code that the value $0x41300000 is written to z, but what could this be in decimal? I tried to find some pattern between the values of g and g + 4 (x and y), but I can't seem to find anything.

Any ideas?

Maybe you should have a look at the format for ieee754 binary single-precision floating-point numbers? — EOF, May 10 '16 at 17:12
@Olaf: I'm sorry, but I don't have a textual source of this information. It's a bit too much to retype. — , May 10 '16 at 17:14
`g.z = 11;`. I cheated - http://www.h-schmidt.net/FloatConverter/IEEE754.html — Ian Abbott, May 10 '16 at 17:15
So it is a bit too much to help you either. Seriously: you expect us to answer a question **you** got, but don't want to spent even the least effort. — too honest for this site, May 10 '16 at 17:15
@IanAbbott I should have cheated, I originally posted an answer of 14! Not cheating gets you nowhere in mathematics. — Alexei Barnes, May 10 '16 at 17:44

score 5 · Accepted Answer · edited May 23 '17 at 12:08

5

This looks to me like typical IEEE754 (thanks for the correction) floating point number representation, on modern hardware a floating point number is represented in memory via a sign bit, a series of bits for an exponent, and the remaining bits in a mantissa (which in this case is dictated by the IEEE).

The value of z in this case is the hex 0x4130 0000, which comes out to 11.

Here is how you work that out for both an example of 1.0 (0x3F80 0000) and for your value below that.

0x3    F    8    0    0    0    0    0
 b0011 1111 1000 0000 0000 0000 0000 0000

 sign     = b0                               = 0         = +
 exponent = b0111 1111                       = 127 - 127 = 0
 mantissa = b000 0000 0000 0000 0000 0000    = 0   + 1   = 1
          = 1 * 2^0 = 1 * 1 = 1



0x4    1    3    0    0    0    0    0
 b0100 0001 0011 0000 0000 0000 0000 0000

 sign     = b0                               = 0          = +
 exponent = b1000 0010                       = 130  - 127 = 3
 mantissa = b011 0000 0000 0000 0000 0000    = .375 + 1   = 1.375
          = 1.375 * 2^3 = 1.375 * 8 = 11

source: How are floating point numbers are stored in memory?

edited May 23 '17 at 12:08

Community

1
1

answered May 10 '16 at 17:29

Alexei Barnes

268
1
9

@PieterVerschaffelt Sorry, I made a mistake calculating the mantissa, somehow I shifted the bits in my head. The correct answer for z is 11, not 14. – Alexei Barnes May 10 '16 at 17:42
That is not correct. C does not enforce a specific floating point representation. – too honest for this site May 10 '16 at 19:16
1

You can see the IEEE754 bit patterns of a float in C only because you're running C on a platform that uses the standard IEEE754 representation for floating point. (i.e. nearly every modern processor with hardware floating point.) – Peter Cordes May 10 '16 at 20:57
@PeterCordes thanks, you are correct. Not sure why I was lazy and put C as if C standardized that, but I've reworded the answer to be accurate. – Alexei Barnes May 11 '16 at 20:26

Find value of struct-variable in assembly code

1 Answers1