在C ++中，1和1i64有什么区别？ [英] In C++, what is the difference between 1 and 1i64?

问题描述

我正在将一些32位兼容代码转换为64位-我遇到了麻烦。我正在编译VS2008 x64项目，并且收到此警告：

I'm converting some 32-bit compatible code into 64-bit - and I've hit a snag. I'm compiling a VS2008 x64 project, and I receive this warning:

warning C4334: '<<' : result of 32-bit shift implicitly converted to 64 bits
(was 64-bit shift intended?)

这是原始代码行：

if ((j & (1 << k)) != 0) {

这就是我遵循微软的建议：

if ((j & (1i64 << k)) != 0) {

在32位和64位系统上都可以编译代码时，这样做安全吗？如果是这样，请解释为什么我必须在末尾添加 i64，以及为什么这不会影响32位编译。否则，将不胜感激。

Is this safe to do, when the code will be compiled on both 32-bit and 64-bit systems? If so, please explain why I must add "i64" to the end, and why this will not affect a 32-bit compilation. Otherwise, a work-around would be much appreciated.

除此之外，我的代码看起来更棘手。

Beyond this, I have what looks like an even trickier bit of code.

if (id[j] != id[j ^ (1u << k)]) {

我了解 u表示数字是无符号的，但是在不超过有符号最大值的值上指定该数字有什么意义呢？ ...我猜这与位移有关？

I understand that the "u" means that the digit is unsigned, but what's the point in specifying that on a value that doesn't exceed the signed maximum value... I'm guessing this has something to do with the bit shift?

推荐答案

1 根据C ++标准的类型为 int 。在64位Microsoft编译器上， int 的sizeof = 4字节，这意味着 int 是32位变量。 1i64 的类型为 __ int64 。

1 has type int according to C++ Standard. On 64-bit Microsoft compiler int has sizeof = 4 bytes, it means that int is 32-bit variable. 1i64 has type __int64.

使用移位运算符时，结果的类型与左操作数的类型相同。这意味着将 1 移位会得到32位结果。 Microsoft编译器假定它可能不是您所期望的（在64位平台上），并给您警告消息。

When you use shift operator the type of the result is the same as the type of the left operand. It means that shifting 1 you'll get 32-bit result. Microsoft compiler assumes that it could be not what you are expecting (on 64-bit platform) and gives you warning message.

当您使用 1i64时结果在两个平台上均为64位。 j 和 0 将隐式转换为64位。因此，整个表达式将使用64位变量进行计算，结果将为 bool 。

When you use 1i64 result will be 64-bit on both platforms. j and 0 will be implicitly casted to 64-bit. So the whole expression will be calculated in 64-bit variables and result will be bool.

因此使用 1i64 在两个（32/64）平台上都是安全的。

So using 1i64 is safe on both (32/64) platforms.

这篇关于在C ++中，1和1i64有什么区别？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！