如何编译器知道，在一个函数调用中的逗号不是逗号操作符？ [英] How does the compiler know that the comma in a function call is not a comma operator?

问题描述

考虑函数调用（调用 INT总和（INT，INT））

printf("%d", sum(a,b));

编译器如何决定，，在函数调用中使用 SUM（INT，INT）不是逗号操作符？

How does the compiler decide that the , used in the function call sum(int, int) is not a comma operator?

注意：我不想在函数调用实际上用逗号。我只是想知道编译器如何知道这是不是一个逗号操作符。

NOTE: I didn't want to actually use the comma operator in the function call. I just wanted to know how the compiler knows that it is not a comma operator.

推荐答案

看语法的C语言。它的上市，完全在标准的附录A 。它的工作方式是，你可以在C程序通过每个令牌退一步，在语法中的下一项匹配起来。在每一个步骤中，只有选项的数量有限，所以任何给定的字符的相互pretation将取决于上下文中出现的。在语法每个规则里面，每一行给出了程序匹配的有效替代品。

Look at the grammar for the C language. It's listed, in full, in Appendix A of the standard. The way it works is that you can step through each token in a C program and match them up with the next item in the grammar. At each step you have only a limited number of options, so the interpretation of any given character will depend on the context in which it appears. Inside each rule in the grammar, each line gives a valid alternative for the program to match.

特别是，如果你看看参数列表，你会看到它包含一个明确的逗号。因此，每当编译器的C语言解析器中的参数列表的模式，逗号，它发现将被理解为参数分隔符，而不是逗号操作符。这同样适用于括号真（也可发生于前pressions）。

Specifically, if you look for parameter-list, you will see that it contains an explicit comma. Therefore, whenever the compiler's C parser is in "parameter-list" mode, commas that it finds will be understood as parameter separators, not as comma operators. The same is true for brackets (that can also occur in expressions).

这工作，因为参数列表规则是慎重使用分配-EX pression 规则，而不仅仅是纯前pression 规则。一个前pression 可以包含逗号，而一个分配-EX pression 不能。如果不是这样的语法是含糊不清，编译器会不知道该怎么办时，它遇到了一个参数列表里加上逗号。

This works because the parameter-list rule is careful to use assignment-expression rules, rather than just the plain expression rule. An expression can contain commas, whereas an assignment-expression cannot. If this were not the case the grammar would be ambiguous, and the compiler would not know what to do when it encountered a comma inside a parameter list.

不过，左括号，例如，是的不的一个函数的定义/调用的一部分，或者如果，，而或为语句，将PTED作为一名前pression的部分（除$ p $，因为没有其他选项​​，但只有一个前pression的开始是在这一点上一个有效的选择），然后在括号里面，前pression 语法规则将适用，并允许逗号运营

However, an opening bracket, for example, that is not part of a function definition/call, or an if, while, or for statement, will be interpreted as part of an expression (because there's no other option, but only if the start of an expression is a valid choice at that point), and then, inside the brackets, the expression syntax rules will apply, and that allows comma operators.

这篇关于如何编译器知道，在一个函数调用中的逗号不是逗号操作符？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！