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Undefined behavior and sequence points reloaded

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问题 Consider this topic a sequel of the following topic: Previous installment Undefined behavior and sequence points Let\'s revisit this funny and convoluted expression (the italicized phrases are taken from the above topic *smile* ): i += ++i; We say this invokes undefined-behavior. I presume that when say this, we implicitly assume that type of i is one of the built-in types. What if the type of i is a user-defined type? Say its type is Index which is defined later in this post (see below).

Why are these constructs using pre and post-increment undefined behavior?

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问题 #include <stdio.h> int main(void) { int i = 0; i = i++ + ++i; printf(\"%d\\n\", i); // 3 i = 1; i = (i++); printf(\"%d\\n\", i); // 2 Should be 1, no ? volatile int u = 0; u = u++ + ++u; printf(\"%d\\n\", u); // 1 u = 1; u = (u++); printf(\"%d\\n\", u); // 2 Should also be one, no ? register int v = 0; v = v++ + ++v; printf(\"%d\\n\", v); // 3 (Should be the same as u ?) int w = 0; printf(\"%d %d\\n\", ++w, w); // shouldn\'t this print 1 1 int x[2] = { 5, 8 }, y = 0; x[y] = y ++; printf(\"%d

sequence points in c

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A sequence point in imperative programming defines any point in a computer program's execution at which it is guaranteed that all side effects of previous evaluations will have been performed, and no side effects from subsequent evaluations have yet been performed. What does this mean? Can somebody please explain it in simple words? When a sequence point occurs, it basically means that you are guaranteed that all previous operations are complete. Changing a variable twice without an intervening sequence point is one example of undefined behaviour. For example, i = i++; is undefined because

Undefined behavior and sequence points

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问题 What are \"sequence points\"? What is the relation between undefined behaviour and sequence points? I often use funny and convoluted expressions like a[++i] = i; , to make myself feel better. Why should I stop using them? If you\'ve read this, be sure to visit the follow-up question Undefined behavior and sequence points reloaded . (Note: This is meant to be an entry to Stack Overflow\'s C++ FAQ. If you want to critique the idea of providing an FAQ in this form, then the posting on meta that

Undefined behavior and sequence points reloaded

阅读更多关于 Undefined behavior and sequence points reloaded

Consider this topic a sequel of the following topic: Previous installment Undefined behavior and sequence points Let's revisit this funny and convoluted expression (the italicized phrases are taken from the above topic *smile* ): i += ++i; We say this invokes undefined-behavior. I presume that when say this, we implicitly assume that type of i is one of the built-in types. What if the type of i is a user-defined type? Say its type is Index which is defined later in this post (see below). Would it still invoke undefined-behavior? If yes, why? Is it not equivalent to writing i.operator+=(i