分配期间如何逗号操作符的工作？ [英] How does comma operator work during assignment?

问题描述

  int类型的= 1;
INT B =（1,2,3）;
COUT＆LT;＆LT; A + B＆LT;＆LT; ENDL; //这个打印4
 

1. 是（1,2,3）某种在C结构++（一些原始类型的列表，也许？）


2. 为什么 B 分配值 3 ？该编译器简单地从列表中取最后价值？

解决方案

是的，这正是它：编译器会的最后一个值。这是逗号运算符，并评估其运算由左到右，并返回最右边之一。它还决心左到右。为什么有人会写code这样的，我不知道：）

所以 INT B =（1，2，3）等同于 INT B = 3 。这不是任何一种原始的列表，逗号操作符，主要用于在一个前pression的背景下，像<$ C $，以评估多个命令C> A + = 5，b + = 4，C + = 3，D + = 2，E + = 1，F 例如

int a = 1;
int b = (1,2,3);
cout << a+b << endl; // this prints 4

1. Is (1,2,3) some sort of structure in c++ (some primitive type of list, maybe?)
2. Why is b assigned the value 3? Does the compiler simply take the last value from the list?

解决方案

Yes, that's exactly it: the compiler takes the last value. That's the comma operator, and it evaluates its operands left-to-right and returns the rightmost one. It also resolves left-to-right. Why anyone would write code like that, I have no idea :)

So int b = (1, 2, 3) is equivalent to int b = 3. It's not a primitive list of any kind, and the comma operator , is mostly used to evaluate multiple commands in the context of one expression, like a += 5, b += 4, c += 3, d += 2, e += 1, f for example.

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