Optimization Level 2 produces error

I've written this little sample program to illustrate the problem:

Code:

1 #include <stdio.h> 2 #include <stdlib.h> 3 4 int main() 5 { 6 double a = 2.3492e8,b; 7 extern double test(double); 8 9 b = test(a); 10 11 printf("%e\n",b); 12 13 } 14 15 double test(double a) 16 { 17 long tmp; 18 double ret; 19 20 tmp = *(long*)&a; // copy ieee-754 representation of "a" to "tmp" 21 tmp = tmp >> 5; 22 tmp = tmp << 5; 23 ret = *(double*)&tmp; 24 25 return ret; 26 }

It is supposed to shift the bit-representation of "a" to the right and to the left again, which works when compiling without optimization or with level 1 optimization (-O switch for gcc). But with level 2 optimization (-O2) the program outputs zero. My problem is also that I can't really debug with optimization, somehow the instructions are mixed or seem to be skipped.
Does anybody have an idea why this is? I really need to use second level optimization.

My system is Debian squeeze x86-64 with gcc 4.3.3

Why are you using extern with the declaration?

Tried it and work fine for me

I had the same result as the OP, gcc 4.3.2

It's because of this line:

Code:

tmp = *(long*)&a;

if you change it to:

Code:

tmp = (long)a;

It works...however, this does not yield the same value :(.

Here is a very weird thing. I tried:

Code:

memcpy(&tmp,&a,sizeof(long));

instead of the tmp assignment. Compiled without -O2, this gives the correct result (2.349200e+08). Compiled with -O2, it gives 2.072910e-317 -- which is the exact same if you compile the "tmp = (long)a" version optimized (unoptimized, that one is 1.160659e-315).

Go figure.

Is long 32 bits or 64 bits? If it's 32, then it's not big enough to hold the bit pattern of a double. Now, consider this line:

Code:

ret = *(double*)&tmp;

This will actually read past the end of tmp, because tmp is shorter than a double. What you find beyond the end of tmp is undefined, and can easily change with compiler settings.

This piece of code is absolutely horrific. I'm glad it's failing.

Quote:

Originally Posted by BEN10

Why are you using extern with the declaration?

I'm a beginner to C. Is it correct to just cross out the extern?

to brewbuck:
long is 64 bits, could you please help me to improve this piece of code?

Nb. I am on 64-bit so they are the same length.

I didn't consider that if gcc -O2 didn't like the first zany assignment it won't like the second!

Anyway, this works -02:

Code:

tmp = (long)a; tmp = tmp >> 5; tmp = tmp << 5; ret = (double)tmp;

But that still depends on long and double being the same size.

Quote:

Originally Posted by 0xa4

I'm a beginner to C. Is it correct to just cross out the extern?

In your case, extern is redundant, so you can remove it.

I'm sorry, I didn't clearly explain what I want to achieve. I want to want to manipulate the bit representation of a double, not cast the actual value to a long integer. I will now look into this memcopy function, maybe this is what I need (though I still don't see why my code would work with optimization level 1 but not with level 2...)

Quote:

Originally Posted by 0xa4

I'm sorry, I didn't clearly explain what I want to achieve. I want to want to manipulate the bit representation of a double, not cast the actual value to a long integer. I will now look into this memcopy function, maybe this is what I need (though I still don't see why my code would work with optimization level 1 but not with level 2...)

The normal way of doing this is "type punning."

Code:

union DoubleBits { double d; unsigned long a; /* unsigned long must be same size as double */ };

First you store the double value in the d element, then access its bits in the a element. This is getting into undefined territory (it's technically not defined to read one union element after writing some different element), but then again, your original approach is undefined as well. In practice it works.

Quote:

Originally Posted by 0xa4

I'm sorry, I didn't clearly explain what I want to achieve. I want to want to manipulate the bit representation of a double, not cast the actual value to a long integer.

Well you might as well do this then:

Code:

double a = 2.3492e8; a = (int)a << 5;

Quote:

though I still don't see why my code would work with optimization level 1 but not with level 2...)

Probably because *(cast*)&x is implicitly redundant.

Quote:

Originally Posted by MK27

Code:

double a = 2.3492e8; a = (int)a << 5;

That does not have the intended effect. The intention is to alter the bits in the double value without regard to their numerical meaning. The code above just casts the value to an integer with truncation. The results aren't even remotely the same.

thanks a lot for the help, the union finally worked. However i couldn't get it to work with your syntax, it kept giving me 'x' undeclared, first use in this function errors, but I found another syntax with a typedef keyword that worked here.