之间的区别和|| ，或&和&& [英] Difference between | and || , or & and &&

问题描述

这些是用Dev-cpp C ++ 5.4.2编写的C ++中的两个简单示例：

These are two simple samples in C++ written on Dev-cpp C++ 5.4.2:

float a, b, c;

if (a | b  & a | c)
   printf("x = %.2f\tF = %.0f\n", x, F);
else
   printf("x = %.2f\tF = %.2f\n", x, F);

和以下代码：

float a, b, c;

if (a || b  && a || c)
   printf("x = %.2f\tF = %.0f\n", x, F);
else
   printf("x = %.2f\tF = %.2f\n", x, F);

有人可以告诉我 || 之间的区别> | 和& > & 。第二个代码有效，但第一个无效。
并且编译器给出错误消息：

Can somebody tell my difference between || > | and & > &&. The second code works , but first does not. And compiler gives an error message :

[错误]类型为'float'和'float'的无效操作数为二进制'

[Error] invalid operands of types 'float' and 'float' to binary 'operator&'.

推荐答案

运算符 | ，& 和〜并行作用于各个位。它们只能用于整数类型。 a | b 对 a 的每个位与相应的 b 的位进行独立的或运算生成结果的那一部分。

The operators |, &, and ~ act on individual bits in parallel. They can be used only on integer types. a | b does an independent OR operation of each bit of a with the corresponding bit of b to generate that bit of the result.

运算符 || ，&& ，和！对每个整个操作数进行操作，作为单个 true / false 值。可以使用任何隐式转换为 bool 的数据类型。许多数据类型，包括 float 都隐式转换为bool，它暗含了！= 0 操作。

The operators ||, &&, and ! act on each entire operand as a single true/false value. Any data type can be used that implicitly converts to bool. Many data types, including float implicitly convert to bool with an implied !=0 operation.

|| 和& 也短路。这意味着只要第一个操作数就可以确定结果的值，则第二个操作数就不会被评估。示例：

|| and && also "short circuit". That means whenever the value of the result can be determined by just the first operand, the second is not evaluated. Example:

ptr&& （* ptr == 7）如果 ptr 为零，则结果为假，而不会因引用零而导致段错误。

ptr && (*ptr==7) If ptr is zero, the result is false without any risk of seg faulting by dereferencing zero.

您可以将其与（int）ptr& （* ptr）。忽略这样一个事实，即使（int）ptr 为零，整个结果也将为零，所以人类可能会认为您不希望这样做。在这种情况下，不需要第二个操作数。但是该程序仍然可能会计算两者。

You could contrast that with (int)ptr & (*ptr). Ignoring the fact that this would be a bizarre operation to even want, if (int)ptr were zero, the entire result would be zero, so a human might think you don't need the second operand in that case. But the program will likely compute both anyway.

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