Expression evaluation in C++ involving unary operators [duplicate]

问题

This question already has answers here:

Undefined behavior and sequence points (5 answers)

Why are these constructs using pre and post-increment undefined behavior? (14 answers)

Closed 5 years ago.

Why does not C/C++ evaluates expression in order of left to right in these cases: Initially x=1

Evaluating x + ++x gives 4.

If normal evaluation is carried out (precedence of ++ is higher than +) then the result should be 1 + 2 = 3

Similarly:

x + ++x + x gives 6
x + x + ++x gives 4

Why are results different?

More Cases:

  x + x++ +x gives 5

What rule is followed by C/C++ instead?

回答1:

Specifically the results of these expressions are not defined, this is because of Cs requirement for multiple accesses (excluding cases where all accesses are reads) to always have a sequence point between them (such as a ; or ,). The results you are getting there are effectively random and would depend on your compiler or theoretically could even change between runs of your program, please see here for information on sequence points:

http://en.wikipedia.org/wiki/Sequence_point

And here if you want to learn about undefined behavior (what your misuse of the variable causes):

http://en.wikipedia.org/wiki/Undefined_behavior#Examples_in_C_and_C.2B.2B

来源：https://stackoverflow.com/questions/21878364/expression-evaluation-in-c-involving-unary-operators