What does ~I mean in C++

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I am reading a code and unable to understand ~i in code
Here's is the code

What I have tried:

C++

<pre>class Solution {
public:
    int romanToInt(string S) {
        int ans = 0, num = 0;
        for (int i = S.size()-1; ~i; i--) {
            switch(S[i]) {
                case 'I': num = 1; break;
                case 'V': num = 5; break;
                case 'X': num = 10; break;
                case 'L': num = 50; break;
                case 'C': num = 100; break;
                case 'D': num = 500; break;
                case 'M': num = 1000; break;
            }
            if (4 * num < ans) ans -= num;
            else ans += num;
        }
        return ans;        
    }
};

Posted 28-Jun-22 6:04am

Hamza Jameel

Updated 28-Jun-22 11:40am

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jeron1 28-Jun-22 12:09pm

~ is the One's complement operator, (inverts all the bits).

https://www.tutorialspoint.com/cprogramming/c_bitwise_operators.htm

KarstenK 29-Jun-22 3:22am

at most: it is bad style becaus hardly to undestand and easily an issue for compiler bugs and changes. So avoid it better.

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merano99 · Answer 1 · 2022-06-28T11:40:00

Solution 2

Quote:
I am reading a code and unable to understand ~i in code

In this case it simply means i>=0.

Posted 28-Jun-22 11:40am

merano99

Comments

Greg Utas 28-Jun-22 18:38pm

I think it actually means i != 0, which will have the same effect. What amazes me is that some C and C++ programmers think it's brilliant to save a few keystrokes, so they write crap like this.

jeron1 28-Jun-22 20:08pm

Greg Utas wrote: "What amazes me is that some C and C++ programmers think it's brilliant to save a few keystrokes, so they write crap like this."

:thumbs up: Indeed.

merano99 28-Jun-22 20:25pm

Unfortunately, the result at i != 0 is wrong and thus does not have the same effect at all.
If you look at the generated code, there is no question of saving, on the contrary.

A look at the generated assembler code shows that the creative coding rather means more effort and is also more difficult to understand for humans.

for (int i = S.size() - 1; i >= 0; i--) {
 
00A17B45 83 E8 01             sub         eax,1  
00A17B48 89 45 D4             mov         dword ptr [ebp-2Ch],eax  
00A17B4B 83 7D D4 00          cmp         dword ptr [ebp-2Ch],0  
00A17B4F 0F 8C A2 00 00 00    jl          $LN13+2Bh (0A17BF7h)

for (int i = S.size() - 1; ~i; i--) {

00A07B45 83 E8 01             sub         eax,1  
00A07B48 89 45 D4             mov         dword ptr [ebp-2Ch],eax  
00A07B4B 8B 45 D4             mov         eax,dword ptr [ebp-2Ch]  
00A07B4E F7 D0                not         eax  
00A07B50 85 C0                test        eax,eax  
00A07B52 0F 84 A2 00 00 00    je          $LN13+2Bh (0A07BFAh)

(not optimized code, as you can see)

Greg Utas 28-Jun-22 23:19pm

You're right. It's ~(-1), not ~0, that will be interpreted as false.

OriginalGriff · Answer 2 · 2022-06-28T08:06:00

Solution 1

Invert each bit in the number: C++ Bitwise Operators[^]

A very simple google would have found that for you in seconds: in future you will save yourself a considerable amount of time if you do basic research first ...

Posted 28-Jun-22 8:06am

OriginalGriff