3 ms·

>good analogy in the real world: If you have 0 apples, you do not have any apples. unless I owe some apples - in which case, by your "real-world" test, I do ha

>good analogy in the real world: If you have 0 apples, you do not have any apples. unless I owe some apples - in which case, by your "real-world" test, I do have some?

logicallee 9y ago
I only received a downvote. To clarify, I mean that most people know what it means to "have -1 apples" (it would mean you owe one apple.) So why should negative values be truthful, if it's a simple analogy with the real world?
- userbinator 9y ago
  So why should negative values be truthful, if it's a simple analogy with the real world? Do you owe anything if you have 0 apples?
  - logicallee 9y ago
    >Do you owe anything if you have 0 apples? This phrasing shows the error even more starkly. No, you don't owe anything if you have 0 apples. If you use this as a test, however, then you also either "don't owe" anything if you have -5 apples, or you do owe something if you have 5 apples. You simply can't use this as the test. To put this into code directly based on the logic you say it's an analogy for: numapples = 0; // Do you owe anything if you have 0 apples? if (!numapples) { printf("I don't owe any apples.\n"); } else { printf("I owe some apples.\n"); } This simply doesn't work. (Gives wrong result at 5 for example.) The correct tests are, respectively: numapples = 5; // have no apples if (numapples == 0) { printf("I don't have or owe any apples.\n"); } if (numapples < 0) { printf("I owe some apples.\n"); } if (numapples > 0) { printf("I have some apples.\n"); } There is another, subtler, reason for this. What is special about the value "0" here is that it's the default value - you're born without any apples, or something. But by making someone write the test like this: (currentapples - startingapples) == 0 you can test whether there is any offset from a different starting value. The "0 == false" version of the test completely breaks that: it applies when you're testing an offset at 0, but breaks anywhere else. In other words as an offset it's equivalent to: ( currentapples == zeroapples ) && (zeroapples == 0) don't get me wrong, I use it in code. But I'm not under the misapprehension that there is a real-world analogy. There isn't one. It's simply the definition of C/C++. Whenever I see == 0 in an if test, I feel free to remove those 4 characters, don't get me wrong. Because it's just the definition of C. Not because I think there's any analogy with anything. That said, which version of this code is cleaner and less likely to allow a programmer to make a logical mistake: 1. if ((currentapples - startingapples) != 0) { 2. if (currentapples - startingapples) { to me, the intent of 1 is super-clear, as for the intent of 2, who knows, but of course after half of a second I can parse what its effect is, by running through the definition of C in my head. That's not great for reading and debugging programs. :)

>good analogy in the real world: If you have 0 apples, you do not have any apples. unless I owe some apples - in which case, by your "real-world" test, I do have some?

logicallee 9y ago

I only received a downvote. To clarify, I mean that most people know what it means to "have -1 apples" (it would mean you owe one apple.) So why should negative values be truthful, if it's a simple analogy with the real world?

userbinator 9y ago
So why should negative values be truthful, if it's a simple analogy with the real world? Do you owe anything if you have 0 apples?
- logicallee 9y ago
  >Do you owe anything if you have 0 apples? This phrasing shows the error even more starkly. No, you don't owe anything if you have 0 apples. If you use this as a test, however, then you also either "don't owe" anything if you have -5 apples, or you do owe something if you have 5 apples. You simply can't use this as the test. To put this into code directly based on the logic you say it's an analogy for: numapples = 0; // Do you owe anything if you have 0 apples? if (!numapples) { printf("I don't owe any apples.\n"); } else { printf("I owe some apples.\n"); } This simply doesn't work. (Gives wrong result at 5 for example.) The correct tests are, respectively: numapples = 5; // have no apples if (numapples == 0) { printf("I don't have or owe any apples.\n"); } if (numapples < 0) { printf("I owe some apples.\n"); } if (numapples > 0) { printf("I have some apples.\n"); } There is another, subtler, reason for this. What is special about the value "0" here is that it's the default value - you're born without any apples, or something. But by making someone write the test like this: (currentapples - startingapples) == 0 you can test whether there is any offset from a different starting value. The "0 == false" version of the test completely breaks that: it applies when you're testing an offset at 0, but breaks anywhere else. In other words as an offset it's equivalent to: ( currentapples == zeroapples ) && (zeroapples == 0) don't get me wrong, I use it in code. But I'm not under the misapprehension that there is a real-world analogy. There isn't one. It's simply the definition of C/C++. Whenever I see == 0 in an if test, I feel free to remove those 4 characters, don't get me wrong. Because it's just the definition of C. Not because I think there's any analogy with anything. That said, which version of this code is cleaner and less likely to allow a programmer to make a logical mistake: 1. if ((currentapples - startingapples) != 0) { 2. if (currentapples - startingapples) { to me, the intent of 1 is super-clear, as for the intent of 2, who knows, but of course after half of a second I can parse what its effect is, by running through the definition of C in my head. That's not great for reading and debugging programs. :)