c++ 中的增量和三元运算符优先级

Question

我想打印以分号分隔的矢量元素。下面的代码片段产生了一个奇怪的结果：结果从向量的第二个元素开始。任何人都可以写一个解释吗？在这种情况下，迭代器何时递增？

#include <iostream>
int main() {
    vector<int> v(10);
    for ( int i = 0; i < 10; ++i ) {
        v[i] = i;
    }

    for ( auto it = v.begin(); it != v.end(); )
        std::cout << *it << ( (++it != v.end() ) ? ";" : "" );
}

Answer 1

在C++17之前，*it和++it的携带顺序是未指定的。参见，例如，https://en.cppreference.com/w/cpp/language/eval_order

这可以在一个更简单的示例中看到

#include <iostream>

int main() {
    int i = 3;
    std::cout << i << ++i;
}

我的编译器（带有 -Wall 选项的 Apple LLVM）报告

warning: unsequenced modification and access to 'i' [-Wunsequenced]
    std::cout << i << ++i;
                 ~    ^

Answer 2

Generally:

Order of evaluation of the operands of almost all C++ operators (including the order of evaluation of function arguments in a function-call expression and the order of evaluation of the subexpressions within any expression) is unspecified.

自 C++17 起有一个特殊的移位规则：

In a shift operator expression E1<<E2 and E1>>E2, every value computation and side-effect of E1 is sequenced before every value computation and side effect of E2.

c++ 中的增量和三元运算符优先级

Increment and ternary operator precedence in c++

c++

iterator

operator-precedence