C++ NOT(~) vs NEGATION(!)
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NOT(~) vs NEGATION(!)
提问by tez
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
int i=-5;
while(~(i))
{
cout<<i;
++i;
}
}
The output is -5,-4,-3,-2. Shouldn't it print values till -1?Why is it only till -2. And please explain me the difference between 'not' and 'negation' operators.When ever I write a program they were the source for bugs.
输出为 -5,-4,-3,-2。它不应该打印值到 -1 吗？为什么只打印到 -2。请向我解释“非”和“否定”运算符之间的区别。当我编写程序时，它们都是错误的来源。
while(i)
I know that the loop condition will be true for positive and negative i's except 0.
我知道循环条件对于正负 i 都是真的，除了 0。
while(!i) vs while(~i)
For what values of 'i' the above two loops get executed?
对于 'i' 的什么值，上述两个循环被执行？
回答by Ry-
When igets to -1, the value of ~iis ~-1, or 0, so the whileloop stops executing. The !operator works because it does something completely different; it results in 1for 0values and 0for all other values. ~is a bitwise negation.
当i到达-1，的值~i是~-1，或者0，所以while循环停止执行。该!运营商工作，因为它会产生完全不同的; 它导致1了0价值观和0所有其他值。~是按位否定。
A little more in detail:
再详细一点：
~takes each bit in a number and toggles it. So, for example, 100102would become 011012-1is all ones in binary when a two's complement signed integer.~0b…11111111is0.
~取数字中的每一位并切换它。因此，例如，10010 2将变为 01101 2-1当二进制补码有符号整数时，所有二进制都是 1。~0b…11111111是0。
However:
然而：
!0is1,!anythingElseis0-1is not0!-1is still0
!0是1，!anythingElse是0-1不是0!-1还是0
And if you actually want to loop including i == -1, just use while (i)instead of while (~i).
如果你真的想循环包括i == -1，只需使用while (i)代替while (~i)。
回答by dasblinkenlight
You are correct about i == -1being the exit condition: your loop is equivalent to
你是正确i == -1的退出条件：你的循环相当于
int i=-5;
while(i != -1)
{
cout<<i;
++i;
}
// i == -1 immediately after the loop
When written this way, it should be clear why -1is not printed the value is first printed, and only then incremented, that's why -2is the last value that you print.
以这种方式编写时，应该清楚为什么-1不打印该值首先打印，然后才递增，这就是-2您打印的最后一个值的原因。
The !operator, on the other hand, will produce 1only when it is given a zero. That's why the loop would print -1when the !operator is used in the loop condition.
该!运营商，而另一方面，会产生1只有当它被赋予一个零。这就是为什么在循环条件中使用运算符-1时循环会打印的原因!。
回答by aasa
'~' is the operator that : ~x = -x-1 and when i = -1, then ~i = 0. if you wonder the value of ~i, you can just print them out:
'~' 是操作符：~x = -x-1，当 i = -1 时，~i = 0。如果你想知道 ~i 的值，你可以打印出来：
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
int i=-5;
for (int i = -5; i <= 3; i++)
{
cout<<i<<" "<<(~i)<<endl;
}
}
and then you will find: -5 4 -4 3 -3 2 -2 1 -1 0 0 -1 1 -2 2 -3 3 -4
然后你会发现： -5 4 -4 3 -3 2 -2 1 -1 0 0 -1 1 -2 2 -3 3 -4
回答by Steve Taylor
!is true/false logic flipping
!是真/假逻辑翻转
!means any nonzero becomes 0, and 0 becomes 1
!意味着any nonzero becomes 0，和0 becomes 1
eg1. !0b1010 -> 0b0000
例如1。 !0b1010 -> 0b0000
eg2. !0b0000 -> 0b0001
例如2。 !0b0000 -> 0b0001
eg3. !0b1111 -> 0b0000
例如3。 !0b1111 -> 0b0000
generalised, out = in?0:1
概括， out = in?0:1
while...
尽管...
~is bit flipping
~有点翻转
~means flip each and every bit
~方法 flip each and every bit
eg1. ~0b1010 -> 0b0101
例如1。 ~0b1010 -> 0b0101
eg2. ~0b0000 -> 0b1111
例如2。 ~0b0000 -> 0b1111
eg3. ~0b1111 -> 0b0000
例如3。 ~0b1111 -> 0b0000
generalised, out = in^0b1111
概括， out = in^0b1111
