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If 'b' is actually used then the compiler has to put 'a' in memory. Theoretically. He says that b is not changed, and does not "escape" (the scope of the code i

by gens 6y ago

If 'b' is actually used then the compiler has to put 'a' in memory. Theoretically. He says that b is not changed, and does not "escape" (the scope of the code in question). In that case b is useless, and thus it doesn't really even exist. A compiler can "collapse" a lot of code into a fraction of it. That's a part of the "optimizing" part of "optimizing compiler". A compiler can put a value in a register. A compiler can do whatever it wants as long as the results are always the same as if it didn't do anything other then just translate to machine code. Pointer aliasing in C can be bad for the compiler as it can't be sure that it can optimize away something. That's why Fortran is faster, and why the restrict keyword was added to the C standard. And yea, i agree with him completely that a context should be made clear. Personally i'd like less voodoo in programming topics, but, on the other hand, this is still a young science and a lot of terminology is all around the place. ..Actually, scratch that, i found what i was interested in and don't really care about everything else.

klodolph 6y ago
> If 'b' is actually used then the compiler has to put 'a' in memory. Theoretically, no—the “as if” rule applies. In practice, also no, because real compilers perform escape analysis and there are aggressive optimization passes for eliminating dead loads and dead stores. Just like the compiler can optimize out b and replace it with a reference to a, the compiler can optimize out the value stored in a but keep the reference to it stored in b. > In that case b is useless, and thus it doesn't really even exist. I don’t think this notion of “exist” has legs.
- gens 6y ago
  That is what i said, that it can be optimized away. "exist" is the wrong word, "is useless" would be better. It is like.. you have a spanner in a box labeled "a" and you have a box "b" that has a piece of paper that says "look in a". You ask for a spanner and you get told to look in box "a". The box labeled "b" is then completely useless. One day some guy, weirdly named "compiler", decides to trow away box "b", and nobody cares. But we have to assume that a compiler will compile the program just as we wrote it. Otherwise we are not talking about the language, but about the compiler. This is gone off topic, if there even was one.
  - klodolph 6y ago
    A variable is useless because it is optimized away? I don’t agree with that terminology. The variable is useful / exists in the source code, and that is enough. Variables / objects only exist conceptually in the original program anyways, we just infer their existence in the compiled program by correspondence with the original code, or educated guesses about the original code.

by gens 6y ago

klodolph 6y ago

> If 'b' is actually used then the compiler has to put 'a' in memory. Theoretically, no—the “as if” rule applies. In practice, also no, because real compilers perform escape analysis and there are aggressive optimization passes for eliminating dead loads and dead stores. Just like the compiler can optimize out b and replace it with a reference to a, the compiler can optimize out the value stored in a but keep the reference to it stored in b. > In that case b is useless, and thus it doesn't really even exist. I don’t think this notion of “exist” has legs.

gens 6y ago
That is what i said, that it can be optimized away. "exist" is the wrong word, "is useless" would be better. It is like.. you have a spanner in a box labeled "a" and you have a box "b" that has a piece of paper that says "look in a". You ask for a spanner and you get told to look in box "a". The box labeled "b" is then completely useless. One day some guy, weirdly named "compiler", decides to trow away box "b", and nobody cares. But we have to assume that a compiler will compile the program just as we wrote it. Otherwise we are not talking about the language, but about the compiler. This is gone off topic, if there even was one.
- klodolph 6y ago
  A variable is useless because it is optimized away? I don’t agree with that terminology. The variable is useful / exists in the source code, and that is enough. Variables / objects only exist conceptually in the original program anyways, we just infer their existence in the compiled program by correspondence with the original code, or educated guesses about the original code.