How do I assign -0x80000000 to long long variable in VS2019?

Question

I am using Visual Studio 2019 Community edition. I have the following C++ code snippet:

long long l = 0x80000000;
assert(l > 0); //true
l = -0x7fffffff;
assert(l < 0); // true
l = -0x80000000;
assert(l < 0); // FALSE

I expect -0x80000000 to end up sign-extended to 64-bit value of 0xFFFFFFFF80000000 but it doesn't. Instead it carries the positive 0x80000000 value. What do I miss?

Try `l = -0x80000000LL;` – Retired Ninja Oct 26 '20 at 08:19 — Retired Ninja, Oct 26 '20 at 08:19
That does the trick. Thank you! – Kok How Teh Oct 26 '20 at 08:23 — Kok How Teh, Oct 26 '20 at 08:23

score 2 · Accepted Answer · answered Oct 26 '20 at 08:23

2

There are rules for integral literals without suffixes.

In short, 0x80000000's type is unsigned int because it's an hexadecimal number bigger than int's max value. In order to make it long long, you must add the LL suffix, i.e. 0x80000000LL.

answered Oct 26 '20 at 08:23

Joel Filho

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Correct for Visual Studio. Other implementations could have other maximum values for `int`, but 32 bits is by far the most common size today. – MSalters Oct 26 '20 at 09:31

How do I assign -0x80000000 to long long variable in VS2019?

1 Answers1