Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
342 views
in Technique[技术] by (71.8m points)

c - Are there sequence points in the expression a^=b^=a^=b, or is it undefined?

The allegedly "clever" (but actually inefficient) way of swapping two integer variables, instead of using temporary storage, often involves this line:

int a = 10;
int b = 42;

a ^= b ^= a ^= b; /*Here*/

printf("a=%d, b=%d
", a, b); 

But I'm wondering, compound assignment operators like ^= are not sequence points, are they? Does this mean it's actually undefined behavior?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)
a ^= b ^= a ^= b; /*Here*/

It is undefined behavior.

You are modifying an object (a) more than once between two sequence points.

(C99, 6.5p2) "Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression.

Simple assignments as well as compound assignments don't introduce a sequence point. Here there is a sequence point before the expression statement expression and after the expression statement.

Sequence points are listed in Annex C (informative) of the c99 and c11 Standard.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...