警告：左移位数＆GT;类型=宽度 [英] warning: left shift count >= width of type

问题描述

我很新的处理比特和编译时已经就死在以下警告：

I'm very new to dealing with bits and have got stuck on the following warning when compiling:

7: warning: left shift count >= width of type

我的第7行看起来像这样

My line 7 looks like this

unsigned long int x = 1 << 32;

这将使意义，如果长在我的系统大小为32位。然而，的sizeof（长）返回8 CHAR_BIT被定义为8暗示长应该是8×8 = 64位。

This would make sense if the size of long on my system was 32 bits. However, sizeof(long) returns 8 and CHAR_BIT is defined as 8 suggesting that long should be 8x8 = 64 bits long.

我缺少的是在这里吗？是的sizeof和CHAR_BIT不准确或有我误解了一些基本的东西？​​

What am I missing here? Are sizeof and CHAR_BIT inaccurate or have I misunderstood something fundamental?

推荐答案

长可能是64位的，但 1 仍然是一个 INT 。你需要让 1 A 使用→长整型 后缀：

long may be a 64-bit type, but 1 is still an int. You need to make 1 a long int using the L suffix:

unsigned long x = 1UL << 32;

（你也应该让无符号使用 U 后缀我已经证明，以避免问题的左移一个符号整数。有没有问题，当一个长宽64位，您可以通过32位的移位，但如果你转向63位），这将是一个问题

(You should also make it unsigned using the U suffix as I've shown, to avoid the issues of left shifting a signed integer. There's no problem when a long is 64 bits wide and you shift by 32 bits, but it would be a problem if you shifted 63 bits)

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