在C ++中将1 ＜<2和1 ＜＜ 3相加时的输出很奇怪 [英] Weird output when summing 1&lt;&lt;2 and 1&lt;&lt;3 in C++

问题描述

所以我只是尝试在C ++中进行一些操作.这是我尝试过的:

So I was just trying some bit manipulation in C++. Here is what I tried:

int a = 1<<2;
cout<<a;

这将输出显示为4.

int a = 1<<3;
cout<<a;

这将输出显示为8

但是当我这样做时:

int a = 1<<2 + 1<<3;
cout<<a;

它给出的输出为64.为什么这样?

It gives the output as 64. Why so?

我也尝试过:

int a = 1<<2;
int b = 1<<3;
cout<<a + b;

这将输出如预期的那样作为12.

Which gives the output as 12 as expected.

推荐答案

这是因为加法运算符的优先级高于移位.换句话说，您的第二个示例等效于1 << (2 + 1) << 3

This is because addition has a higher operator precedence than bitshift. In other words, your second example is equivalent to 1 << (2 + 1) << 3

此外，由于位移是左关联的，因此它与(1 << (2 + 1)) << 3相同.简化为8 << 3，即64.

Furthermore, since bitshifting is left-associative, it's the same as (1 << (2 + 1)) << 3. This simplifies to 8 << 3, which is 64.

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