类型转换为无符号用C [英] typecasting to unsigned in C

问题描述

int a = -534;
unsigned int b = (unsigned int)a;
printf("%d, %d", a, b);

打印 -534，-534

为什么不强制转换发生？

Why is the typecast not taking place?

我希望它是 -534，534

如果我修改code到

int a = -534;
unsigned int b = (unsigned int)a;
if(a < b)
  printf("%d, %d", a, b);

它不是打印毕竟什么... A 小于 B ??

推荐答案

首先，你不需要投：的 A 隐式转换为<值code> unsigned int类型与分配给 b 。所以，你的语句等价于：

First, you don't need the cast: the value of a is implicitly converted to unsigned int with the assignment to b. So your statement is equivalent to:

unsigned int b = a;

现在，一个重要的特性无符号整型的C和C ++的是，它们的值总是在范围[0，的最大的] ，其中的最大的为 unsigned int类型是 UINT_MAX （它定义在 limits.h中）。如果您分配一个值，这不是在这个范围内，它被转换到该范围内。因此，如果该值是负的，您可以添加 UINT_MAX + 1 反复使其在区间[0， UINT_MAX ]。对于上面的code，这是因为如果我们写： unsigned int类型B =（UINT_MAX + A）+ 1 。这不等于 -a （534）。

Now, an important property of unsigned integral types in C and C++ is that their values are always in the range [0, max], where max for unsigned int is UINT_MAX (it's defined in limits.h). If you assign a value that's not in that range, it is converted to that range. So, if the value is negative, you add UINT_MAX+1 repeatedly to make it in the range [0, UINT_MAX]. For your code above, it is as if we wrote: unsigned int b = (UINT_MAX + a) + 1. This is not equal to -a (534).

请注意，上面的是真实的底层重新presentation是否补，一补数，或登录幅度（或任何其它特殊编码）。人们可以看到的东西，如：

Note that the above is true whether the underlying representation is in two's complement, ones' complement, or sign-magnitude (or any other exotic encoding). One can see that with something like:

signed char c = -1;
unsigned int u = c;
printf("%u\n", u);
assert(u == UINT_MAX);

在一个典型的补机具有4个字节的 INT ， C 是 0xFF的和 U 是为0xffffffff 。编译器必须确保当值 1 分配给 U ，它被转换成等于值 UINT_MAX 。

On a typical two's complement machine with a 4-byte int, c is 0xff, and u is 0xffffffff. The compiler has to make sure that when value -1 is assigned to u, it is converted to a value equal to UINT_MAX.

现在回到你的code时，的printf 格式字符串是错误的 B 。您应该使用％U 。当你这样做，你会发现它打印的值UINT_MAX - 534 + 1 而不是 534

Now going back to your code, the printf format string is wrong for b. You should use %u. When you do, you will find that it prints the value of UINT_MAX - 534 + 1 instead of 534.

当在比较操作使用＆LT; ，因为 B 是无符号整型， A 也转换为 unsigned int类型。这与给出B = A ;早些时候，意味着 A＆LT; b 是假的： A 为 unsigned int类型等于乙。

When used in the comparison operator <, since b is unsigned int, a is also converted to unsigned int. This, given with b = a; earlier, means that a < b is false: a as an unsigned int is equal to b.

让我们假设你有一个一补数机，以及你做的：

Let's say you have a ones' complement machine, and you do:

signed char c = -1;
unsigned char uc = c;

假设一个字符（符号或无符号）是机器上的8位。然后 C 和 UC 将存储下列值，位模式：

Let's say a char (signed or unsigned) is 8-bits on that machine. Then c and uc will store the following values and bit-patterns:

+----+------+-----------+
| c  |  -1  | 11111110  |
+----+------+-----------+
| uc | 255  | 11111111  |
+----+------+-----------+

注意的ç和 UC 的位模式是不一样的。编译器必须确保 C 值为 1 和 UC 值为 UCHAR_MAX ，这是255在此计算机上。

Note that the bit patterns of c and uc are not the same. The compiler must make sure that c has the value -1, and uc has the value UCHAR_MAX, which is 255 on this machine.

有上<更详细href=\"http://stackoverflow.com/questions/1863153/why-unsigned-int-0xffffffff-is-equal-to-int-1/1863219#1863219\">my回答一个问题在这里SO 。

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