cout中的逻辑运算（C ++） [英] Logical operations in cout (C++)

问题描述

考虑：

int i = 56, j = 0;
    int n = i&&j;
    cout << i&&j;
    cout << endl << n;

输出将是：

56
0

我想是因为运算符优先级或逻辑短路，但我似乎无法弄清楚是哪个原因或原因

I imagine its either because of operator precedence or logical short circuit, but I can't seem to figure out which or the reason

推荐答案

表达式 cout<< i& j 等效于（cout<< i）&& j 。这两个操作数都将求值并转换为 bool 。该语句整体上没有任何作用，但是对子表达式 cout<<的求值是无效的。 i 具有通常的副作用，即向标准输出中写入内容。

The expression cout << i&&j is equivalent to (cout << i) && j. Both operands are evaluated and converted to bool. The statement as a whole has no effect, but the evaluation of the subexpression cout << i has the usual side effects, of course, namely writing something to the standard output.

&& ; 运算符确实是短路的，并且仅在 cout<<<时才评估 j 。 i 的评估结果为 true 。此条件等效于 cout.good（），通常是这种情况（除非您以某种方式设法关闭了标准输出）。

The && operator is indeed short-circuited and j is only evaluated if cout << i evaluates as true. This condition is equivalent to cout.good(), which is usually the case (unless you somehow managed to close your standard output).

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