For the bounty: How can this behavior can be disabled on a case-by-case basis without disabling or lowering the optimization level?
The following conditional expression was compiled on MinGW GCC 3.4.5, where a is a of type signed long, and m is of type unsigned long.
if (!a && m > 0x002 && m < 0x111)
The CFLAGS used were -g -O2. Here is the corresponding assembly GCC output (dumped with objdump)
120: 8b 5d d0 mov ebx,DWORD PTR [ebp-0x30]
123: 85 db test ebx,ebx
125: 0f 94 c0 sete al
128: 31 d2 xor edx,edx
12a: 83 7d d4 02 cmp DWORD PTR [ebp-0x2c],0x2
12e: 0f 97 c2 seta dl
131: 85 c2 test edx,eax
133: 0f 84 1e 01 00 00 je 257 <_MyFunction+0x227>
139: 81 7d d4 10 01 00 00 cmp DWORD PTR [ebp-0x2c],0x110
140: 0f 87 11 01 00 00 ja 257 <_MyFunction+0x227>
120-131 can easily be traced as first evaluating !a, followed by the evaluation of m > 0x002. The first jump conditional does not occur until 133. By this time, two expressions have been evaluated, regardless of the outcome of the first expression: !a. If a was equal to zero, the expression can (and should) be concluded immediately, which is not done here.
How does this relate to the the C standard, which requires Boolean operators to short-circuit as soon as the outcome can be determined?
As others have mentioned, this assembly output is a compiler optimization that doesn't affect program execution (as far as the compiler can tell). If you want to selectively disable this optimization, you need to tell the compiler that your variables should not be optimized across the sequence points in the code.
Sequence points are control expressions (the evaluations in if, switch, while, do and all three sections of for), logical ORs and ANDs, conditionals (?:), commas and the return statement.
To prevent compiler optimization across these points, you must declare your variable volatile. In your example, you can specify
volatile long a;
unsigned long m;
{...}
if (!a && m > 0x002 && m < 0x111) {...}
The reason that this works is that volatile is used to instruct the compiler that it can't predict the behavior of an equivalent machine with respect to the variable. Therefore, it must strictly obey the sequence points in your code.