无法分配-2147483648到long long类型的变量 [英] Can't assign -2147483648 to variable of type long long

问题描述

我编译下面，由于某种原因，我不能指定-2147483648到这是8个字​​节长，并签署了变量code。

I'm compiling the code below and for some reason I can't assign -2147483648 to the variable which is 8 bytes long and signed.

long long x = -2147483648;

当我跨过这条线，x的值是2147483648和观察在微软的Visual Studio窗口显示x的类型为__int64。一个的sizeof（X）也将返回8。

When I step over this line, the value of x is 2147483648 and the 'Watch' window in MS Visual Studio shows that the type of x is __int64. A sizeof(x) also returns 8.

据limit.h的限制符号long long是：

According to limit.h the limits for a signed long long are:

#define LLONG_MAX     9223372036854775807i64       /* maximum signed long long int value */
#define LLONG_MIN   (-9223372036854775807i64 - 1)  /* minimum signed long long int value */

和

/* minimum signed 64 bit value */
#define _I64_MIN    (-9223372036854775807i64 - 1)
/* maximum signed 64 bit value */
#define _I64_MAX      9223372036854775807i64

我只是不明白这一点！

I just don't get it!!!

有人请阐明这一些轻？

推荐答案

如果没有LL，编译器似乎演绎 2147483648 是一个32位的无符号长。的然后的适用的 - 运营商。其结果是 0 - 2147483648 。由于这是小于0，作为一个无符号长T ， 4294967296 添加，这是 2147483648 一次。那么这个值被分配给久长X 。

Without the LL, the compiler appears to deduce 2147483648 is a 32-bit unsigned long. Then it applies the - operator. The result is 0 - 2147483648. Since this is less than 0 and being an unsigned long t, 4294967296 is added, which is 2147483648 again. This value is then assigned to long long x.

建议：

long long x = -2147483648LL;
// or
long long x = -2147483647  - 1;

这篇关于无法分配-2147483648到long long类型的变量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！