问题描述
关于优先级和评估顺序的概念有很多疑问，但我找不到与我的特殊情况有关的问题.
There are lots of questions on concepts of precedence and order of evaluation but I failed to find one that refers to my special case.
考虑以下语句:
if(f(0) && g(0)) {};
保证会首先评估f(0)吗?请注意，运算符是&&.
Is it guaranteed that f(0) will be evaluated first? Notice that the operator is &&.
我的困惑源于我在"The C ++ Programming Language，(Stroustrup，4ed，2013)"中阅读的内容.
My confusion stems from what I've read in "The C++ Programming Language, (Stroustrup, 4ed, 2013)".
这本书的10.3.2节说:
In section 10.3.2 of the book, it says:
int x = f(2)+ g(3); //不确定是先调用f()还是g()
int x = f(2)+g(3); // undefined whether f() or g() is called first
这似乎适用于所有运营商，包括和&&运算符，但在以下段落中表示:
This seems to apply to all operators including && operator, but in a following paragraph it says:
在11.1.1节中还另外提到了这一点:
There is also another mention of this in section 11.1.1:
while(p&！whitespace(p))++ p;
while (p && !whitespace(p)) ++p;
在这里，如果p是nullptr，则不会取消引用.
Here, p is not dereferenced if it is the nullptr.
最后一个引号表示&&和||首先评估他们的第一个参数，因此似乎加强了我的假设，即第二个引号中提到的运算符是第一个引号的例外，但我也不能从最后一个例子中得出明确的结论，因为该表达式仅包含一个子表达式，而不是我的示例，其中包含两个.
This last quote implies that && and || evaluate their 1st argument first, so it seems to reinforce my assumption that operators mentioned in 2nd quote are exceptions to 1st quote, but I cannot draw a definitive conclusion from this last example either, as the expression contains only one subexpression as opposed to my example, which contains two.
推荐答案
&&，||和,的特殊排序行为在C和C ++中已得到很好的确立.您引用的第一句话应为通常未指定表达式中子表达式的求值顺序"或除少数特定例外情况之外，未指定表达式中子表达式的求值顺序".
The special sequencing behavior of &&, ||, and , is well-established in C and C++. The first sentence you quoted should say "The order of evaluation of subexpressions within an expression is generally unspecified" or "With a few specific exceptions, the order of evaluation of subexpressions within an expression is unspecified".
您问过有关C ++的问题，但此问题 http://c-faq.com/expr/seqpointops.html"rel =" nofollow noreferrer> C常见问题解答列表是相关的.
You asked about C++, but this question in the C FAQ list is pertinent.
附录:我刚刚意识到，在这些规则中，未指定"比未定义"更好.编写类似f() + g()的东西不会给您未定义的行为.您只是无法知道先调用f还是g.
Addendum: I just realized that "unspecified" is a better word in these rules than "undefined". Writing something like f() + g() doesn't give you undefined behavior. You just have no way of knowing whether f or g might be called first.
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