奇怪的布尔转换（？） [英] Weird boolean conversion (?)
问题描述
解释pleasy为什么第二个表达式返回false
Explain pleasy why second expression returns false
cout << (4==4) << endl; //1
cout << (4==4==4) << endl; // 0
推荐答案
== 4 == 4）基本上是（（4 == 4）== 4）这是 == 4）这是（1 == 4） 1 / code> 2 这是打印为 0 。
(4==4==4) is basically ((4==4)==4) which is (true == 4) which is (1==4) 1 which is false 2 which is getting printed as 0.
== 具有关联性左对齐 - 右，但这并不重要（在 情况下），因为即使它具有从右到左的关联性，结果将是相同的。
Note that == has associativity left-to-right, but that doesn't matter (in this case) because even if it had associativity right-to-left, the result would have been the same.
1。由于整体推广。
2。请注意，有人可能会认为（true == 4）中的 4 可以被视为 true （毕竟 4 不是零，因此 true ）。这个想法可能会得出结论（true == 4）是（true == true）这是 true 。但这不是它的工作原理。它是bool，它被提升为int，而不是int到bool。
1. Due to integral promotion.
2. Note that one might tempted to think 4 in (true==4) could be treated as true (after all 4 is non-zero, hence true). This thinking might conclude (true==4) is (true==true) which is true. But that is not how it works. It is the bool which gets promoted to int, instead of int to bool.
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