（无符号）-INT_MIN [英] (unsigned)-INT_MIN
问题描述
大家好
int i = INT_MIN;
unsigned int u = -i;
是你保证有INT_MIN的绝对值吗？
为什么它可能没有：-i有类型（int），-INT_MIN可能超过
INT_MAX。
演员表也可能不起作用，因为在（unsigned）-i中;演员是太晚了
迟到，而且 - （无符号）我然后我不能被表示为无符号。
如果长于int，我可以做 - （长）我，但如果我们正在谈论
关于可用的最长整数类型怎么办？
TIA
viza
7月21日上午11:53，viza< tom.v ... @ gm-il.com.obviouschange.invalid>
写道：
大家好
int i = INT_MIN;
unsigned int u = -i;
你保证绝对值是INT_MIN吗？
为什么它不会： -i具有type（int），-INT_MIN可能大于
INT_MAX。
那又怎样？相应的整数类型具有相同的值范围;如果
int有N个值，unsigned int有N个值。
保证。
< snip>
7月21日，1：53 * pm，viza< tom.v ... @ gm-il.com.obviouschange.invalid>
写道：
大家好
int i = INT_MIN;
unsigned int u = -i;
你保证有INT_MIN的绝对值吗？
在我的系统上，u变为0，因为INT_MIN是-2147483648而INT_MAX是
2147483647.因此，-INT_MIN回绕到0。
为什么不这样：-i有类型（int），-INT_MIN可能超过
INT_MAX。在大多数系统上，
-INT_MIN大于INT_MAX。
如果长比int更长，我能做 - （长）我，但如果我们正在谈论
关于可用的最长整数类型呢？
我们在看什么？它是关于在unsigned int中获取
signed int的绝对值吗？
7月21日下午12:21，vipps。 .. @ gmail.com写道：
7月21日上午11:53，viza< tom.v ... @ gm-il.com.obviouschange .invalid>
写道：大家好
int i = INT_MIN;
unsigned int u = -i ;
您是否保证绝对值为INT_MIN？
为什么它可能不会：-i具有type（int），-INT_MIN可能大于
INT_MAX。
那又怎样？相应的整数类型具有相同的值范围;如果
int有N个值，unsigned int有N个值。
保证。
< snip>
对不起，我误解了你。你对
INT_MIN< -INT_MAX。
那么，在这种情况下，计算-INT_MIN会调用未定义的行为;
（溢出）
试试这个
if（i == INT_MIN）
u + = INT_MAX +（无符号） - （i + INT_MAX）;
u = -i;
Hi all
int i= INT_MIN;
unsigned int u= -i;
Is u guaranteed to have the absolute value of INT_MIN?
Why it might not: -i has type (int), and -INT_MIN might be more than
INT_MAX.
A cast might not work either, because in (unsigned)-i; the cast is too
late, and in -(unsigned)i then i cannot be represented as unsigned.
If long is longer than int, I can do -(long)i, but what if we are talking
about the longest integer type available?
TIA
viza
On Jul 21, 11:53 am, viza <tom.v...@gm-il.com.obviouschange.invalid>
wrote:Hi all
int i= INT_MIN;
unsigned int u= -i;
Is u guaranteed to have the absolute value of INT_MIN?
Why it might not: -i has type (int), and -INT_MIN might be more than
INT_MAX.So what? corresponding integer types have the same range of values; If
int has N values, unsigned int has N values.
It is guaranteed.
<snip>
On Jul 21, 1:53*pm, viza <tom.v...@gm-il.com.obviouschange.invalid>
wrote:Hi all
int i= INT_MIN;
unsigned int u= -i;
Is u guaranteed to have the absolute value of INT_MIN?On my system, u becomes 0 as INT_MIN is -2147483648 and INT_MAX is
2147483647. So, -INT_MIN wraps around to 0.
Why it might not: -i has type (int), and -INT_MIN might be more than
INT_MAX.-INT_MIN is greater than INT_MAX on most of the systems.
If long is longer than int, I can do -(long)i, but what if we are talking
about the longest integer type available?What are we looking at? Is it about getting the absolute value of a
signed int in an unsigned int?
On Jul 21, 12:21 pm, vipps...@gmail.com wrote:On Jul 21, 11:53 am, viza <tom.v...@gm-il.com.obviouschange.invalid>
wrote:Hi all
int i= INT_MIN;
unsigned int u= -i;
Is u guaranteed to have the absolute value of INT_MIN?
Why it might not: -i has type (int), and -INT_MIN might be more than
INT_MAX.
So what? corresponding integer types have the same range of values; If
int has N values, unsigned int has N values.
It is guaranteed.
<snip>
I''m sorry, I misunderstood you. You''re interested in the case that
INT_MIN < -INT_MAX.
Well, in that case, computing -INT_MIN invokes undefined behavior;
(overflow)
Try this
if(i == INT_MIN)
u += INT_MAX + (unsigned)-(i + INT_MAX);
u = -i;
这篇关于（无符号）-INT_MIN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！
