sequence-points

问题 Can someone tell me how prefix / postfix operators really work? I've been looking online a lot but haven't found anything. From what I can tell prefex first increments, then does the operation and then assigns. Postfix would do the operation first, then assign and then increment. But I'm having a bit of trouble with my code: int x, y; x = 1; y = x + x++; // (After operation y = 2)(x=2) However when I do: y = x++ + x; // (After operation y = 3)(x=2) I'm not sure why these operations would be

问题 Can someone tell me how prefix / postfix operators really work? I've been looking online a lot but haven't found anything. From what I can tell prefex first increments, then does the operation and then assigns. Postfix would do the operation first, then assign and then increment. But I'm having a bit of trouble with my code: int x, y; x = 1; y = x + x++; // (After operation y = 2)(x=2) However when I do: y = x++ + x; // (After operation y = 3)(x=2) I'm not sure why these operations would be

问题 Can someone tell me how prefix / postfix operators really work? I've been looking online a lot but haven't found anything. From what I can tell prefex first increments, then does the operation and then assigns. Postfix would do the operation first, then assign and then increment. But I'm having a bit of trouble with my code: int x, y; x = 1; y = x + x++; // (After operation y = 2)(x=2) However when I do: y = x++ + x; // (After operation y = 3)(x=2) I'm not sure why these operations would be

问题 Wikipedia says that: In computer science, an operation, function or expression is said to have a side effect if it modifies some state variable value(s) outside its local environment, that is to say has an observable effect besides returning a value (the main effect) to the invoker of the operation. But how can we access a variable outside its local environment, can anyone explain this situation, side effect, main effect and sequence point comprehensibly? 回答1: A function is (should be) a

问题 In this recent question, some code was shown to have undefined behavior : a[++i] = foo(a[i-1], a[i]); because even though the actual call of foo() is a sequence point , the assignment is unsequenced , so you don't know whether the function is called after the side-effect of ++i took place or before that. Thinking further about this, the sequence point at a function call only guarantees that side effects from evaluating the function arguments are carried out once the function is entered, e.g.

问题 In this recent question, some code was shown to have undefined behavior : a[++i] = foo(a[i-1], a[i]); because even though the actual call of foo() is a sequence point , the assignment is unsequenced , so you don't know whether the function is called after the side-effect of ++i took place or before that. Thinking further about this, the sequence point at a function call only guarantees that side effects from evaluating the function arguments are carried out once the function is entered, e.g.

问题 In this recent question, some code was shown to have undefined behavior : a[++i] = foo(a[i-1], a[i]); because even though the actual call of foo() is a sequence point , the assignment is unsequenced , so you don't know whether the function is called after the side-effect of ++i took place or before that. Thinking further about this, the sequence point at a function call only guarantees that side effects from evaluating the function arguments are carried out once the function is entered, e.g.

来源： https://stackoverflow.com/questions/63670099/why-this-loop-runs-10-times-instead-of-5

来源： https://stackoverflow.com/questions/63670099/why-this-loop-runs-10-times-instead-of-5

来源： https://stackoverflow.com/questions/63505019/why-does-this-code-print-1-2-2-and-not-the-expected-3-3-1