Is this undefined behavior in C/C++ (Part 2)
This question already has an answer here:
It's undefined because of 2 reasons:
The value of i is twice used without an intervening sequence point (the comma in argument lists is not the comma operator and does not introduce a sequence point).
You're calling a variadic function without a prototype in scope.
The number of arguments passed to printf() are not compatible with the format string.
the default output stream is usually line buffered. Without a 'n' there is no guarantee the output will be effectively output.
所有参数在调用函数时都会被计算，即使它们没有被使用，所以，由于函数参数的计算顺序是未定义的，所以您再次使用UB。
I think it's well defined. The printf matches the first % placeholder to the first argument, which in this instance is a preincremented variable.
链接地址: http://www.djcxy.com/p/73240.html上一篇: 单个语句中的多个增量运算符