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Old 10 December 2020, 12:42   #1
Ernst Blofeld
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gcc and 64 bit, long long ints, not behaving for me?

I'm looking into using 64 bit long long ints in my code for one small part of a calculation that overflows a 32 bit int.

While trying them out I saw things that surprised me, so I started writing some test code. I've now got three functions to test what happens when the 64 bit numbers overflow. I think all three functions should behave the same, but they don't, only Test2 behaves as I expect and exits when the 64 bit number overflows and becomes negative.

Can anyone explain why?

Code:
typedef long long LONG64;

void Test1(void) {
	KPrintF("Test 1:");
	KPrintF("-------");
	
	LONG64 x = 1;
	int i = 0;
	for (;;) {
		i++;
		x = 2 * x;

		if (x > 0) {
			KPrintF("1 - Yes: %ld", i);
		} else {
			KPrintF("1 - No:  %ld", i);
			break;
		}
	}
}

void Test2(void) {
	KPrintF("Test 2:");
	KPrintF("-------");

	LONG64 x = 1;
	int i = 0;
	for (;;) {
		i++;
		LONG64 new_x = 2 * x;

		if (new_x > x) {
			KPrintF("2 - Yes: %ld", i);
		} else {
			KPrintF("2 - No:  %ld", i);
			break;
		}

		x = new_x;
	}
}

void Test3(void) {
	KPrintF("Test 3:");
	KPrintF("-------");

	LONG64 x = 1;
	int i = 0;
	for (;;) {
		i++;
		LONG64 new_x = 2 * x;

		if (new_x > 0) {
			KPrintF("3 - Yes: %ld", i);
		} else {
			KPrintF("3 - No:  %ld", i);
			break;
		}

		x = new_x;
	}
}
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Old 10 December 2020, 12:48   #2
Thomas Richter
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Quote:
Originally Posted by Ernst Blofeld View Post
Can anyone explain why?
Your code triggers undefined behaivour as nothing is guaranteed if a signed type "overflows". The compiler can, for example, assume that a positive number times 2 is never becoming negative.


Note that this is different from unsigned numbers which are guaranteed to implement modulo arithmetics.
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Old 10 December 2020, 12:58   #3
Ernst Blofeld
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Quote:
Originally Posted by Thomas Richter View Post
Your code triggers undefined behaivour as nothing is guaranteed if a signed type "overflows". The compiler can, for example, assume that a positive number times 2 is never becoming negative.


Note that this is different from unsigned numbers which are guaranteed to implement modulo arithmetics.
Thank you, exactly the answer I needed.
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