3 ms·
> The one-liner and two-liner compile to the same number of instructions, so I don't see how either "avoids inconsistent state". It's about performance, or th
by moonchild 6y ago
> The one-liner and two-liner compile to the same number of instructions, so I don't see how either "avoids inconsistent state".
It's about performance, or thread safety, or anything like that; it's about having a coherent mental model of the code. A statement should, if possible, represent a single, complete operation. Invariants should not be violated by a statement, with respect to its environment. (This more true for 'push' than 'pop'.) One way of solving that is to bundle the 'push' and 'pop' operations up into functions; someone else in this thread did that. But why bother with the mental overhead of a function call when you could just represent the operation directly? To be sure, there are cases where the abstraction is warranted, but a two~three-line stack operation isn't abstraction, it's just indirection.
> For someone who doesn't have the operator precedence rules memorized, it isn't clear whether the above code means [snipped] or [snipped]
> The two-liner [...] requires less knowledge of operator precedence rules
It's not operator precedence—that's a separate issue; despite having implemented c operator precedence, I don't know all of them by heart—but simply behaviour of pre- and post-increment/decrement operations. It's even mnemonic—when the increment symbol goes before the thing being incremented, the increment happens first; else after—but even if not, it's a fairly basic language feature.
Even beyond that, though, it's an idiom. Code is not written in a vacuum. Patterns of pre- and post-increment fall into common use over time and become part of an established lexicon which is not specified anywhere. Natural language works the same way. Nothing wrong with that.
- cellularmitosis 6y ago> It's not operator precedence—that's a separate issue > It's even mnemonic—when the increment symbol goes before the thing being incremented, the increment happens first; else after—but even if not, it's a fairly basic language feature. I think you missed the issue. This is 100% about operator precedence, and has nothing to do with the decrement operator being in front of or behind the variable. This expression: *stack-- means either this: (*stack)-- or this: *(stack--) depending on the operator precedence rules. If this is the layout of memory: ~~~~~~ stack-1: | 52 | stack: | 23 | stack+1: | 19 | ~~~~~~ (* stack)-- evaluates to 22, while *(stack--) evaluates to 52. https://godbolt.org/z/P7Ghfc https://godbolt.org/z/P7Ghfc
- moonchild 6y ago> operator precedence Right, yes. I got confused by your example, because the example is definitely about pre- vs post-increment. My point about idioms still stands, though. > (* stack)-- evaluates to 22, while * (stack--) evaluates to 52. Actually, (* stack)-- evaluates to 23, but changes *stack to 22 :)
- cellularmitosis 6y ago> Actually, (* stack)-- evaluates to 23, but changes *stack to 22 :) Thanks, I just realized I had that wrong :) https://pastebin.com/n7sHzW3p https://pastebin.com/n7sHzW3p