这与运算符优先级无关。
您正在使用
std::cout << ((i != 0) ? "Not zero " : "zero ") << ++i << std::endl;
// Equivalent to:
operator<<(operator<<(operator<<(std::cout, ((i != 0) ? "Not zero " : "zero ")), ++i), std::endl);
这里唯一的规则是在调用函数之前必须对参数进行完全评估。对于参数的求值顺序没有限制,即使它们的求值与调用交错(甚至部分求值)也是如此。
解读一:
1) ((i != 0) ? "Not zero " : "zero "))
2) ++i
3) operator<<(std::cout, (1));
4) operator<<((3), (2));
5) operator<<((4), std::endl);
注意:限制是:
A) (1) happens before (3)
B) (2) happens before (4)
C) (3) happens before (4)
D) (4) happens before (5)
鉴于这些限制,我们可以按如下方式重新排序。
解读2:
1) ++i
2) ((i != 0) ? "Not zero " : "zero "))
3) operator<<(std::cout, (2));
4) operator<<((3), (1));
5) operator<<((4), std::endl);
解读3:
1) ((i != 0) ? "Not zero " : "zero "))
2) operator<<(std::cout, (1));
3) ++i
4) operator<<((2), (3));
5) operator<<((4), std::endl);