why Q.15 multiplication isn't working

Question

Refering Notation for fixed point representation i have tried following, but i am totally confused.

int q = 1<<15;
printf("(sys3.b0 * q) = %hx \n",((0.2929 * q)));

output is

((0.2929 * q)) = cc60

I expect to see 257d because from calculator, 0.2929 * 32768 in hex = 257d

What is the problem in using this Q.15 format, or is this just because of not properly printing

---

EDIT

Guys, actually i've never thought of it.

My actual code tries to print two values. I thought 2nd value printing is the problem. After comments of @Soren i copied his code and is working. Then i removed 1st value from printf, my code also worked. So i repost the printf statement

printf("(sys3.b0 * q) = %X \t((0.2929 * q)) = %X\n",(sys3.b0 * q),(unsigned)((0.2929 * q)));
printf("(sys3.b0 * q) = %X \n",((unsigned)(0.2929 * q)));

where the second line is as suggested by Seron.

The value of sys3.b0 = 0.0000000001295

Now the output is

(sys3.b0 * q) = 4DB173CF        ((0.2929 * q)) = 3ED1CC60
(sys3.b0 * q) = 257D

Answer 1

What you have is undefined behavior the result of:

0.2929 * q

is a double but you are telling printf it is a unsigned short. If you cast the result to unsigned short you will get the result you expect (see it live):

printf("(sys3.b0 * q) = %hx \n",(unsigned short)(0.2929 * q));
                                ^^^^^^^^^^^^^^^^

Having warning turned on or up should have indicated that this was a problem, both gcc and clang give similar errros:

warning: format '%hx' expects argument of type 'int', but argument 2 has type 'double' [-Wformat=]

Although apparently Visual Studio does not.

Answer 2

The format string expects a type different from the one you're actually giving it. My compiler says:

foo.c:5:33: warning: format specifies type 'unsigned short' but the argument has type 'double' [-Wformat]

So you need a cast to convert 0.2929 * q from a double to an unsigned int. Like this:

int main() {
    int q = 1<<15;
    printf("(sys3.b0 * q) = %X \n",((unsigned)(0.2929 * q)));
}

EDIT: When you feed printf an argument of the wrong type, it'll read only as many bytes as the type it thought it was getting. In your case:

printf("(sys3.b0 * q) = %X \t((0.2929 * q)) = %X\n",
       (sys3.b0 * q),
       (unsigned)((0.2929 * q)))

The format string says to expect two unsigned int, that is, two times 4 bytes*. But you gave it a double and an unsigned int, which are respectively 8 and 4 bytes*. printf just reads 4 bytes for each %X, so effectively it reads the 8-result of sys3.b0 * q, thinking those 8 bytes are two integers, which it then prints.