This is normal: when you start the clock, there is no visual indicator until 1 second has passed, and the time ticks down to the next second. So there is that short period where you aren't sure if your press registered or not.

Gates made his move without starting the clock and then they sat around chatting for a while before Magnus started the clock.

It's wrong then, those times were for the whole game. If you watch the video of the match, this is clear.

If you actually watch the video, they were playing a timed game. The World Champion (funny how the slashdot title omits that minor detail??) had 30 seconds, Bill had 2 minutes.

At least he knows how to play minesweeper. (A reddit thread recently revealed that a lot of people had been playing it by randomly clicking, and either ignored the numbers, or thought they were points).

‘A reader lives a thousand lives before he dies, said Jojen. The man who never reads lives only one.’

all the boring stories lately?

Sun’s Hemispheres Out Of Sync During Magnetic Field

Written by: Tara Dodrill Trending August 8, 2013 3 Comments

It may not even work on a flat address space, if "buf"'s allocated block is right at the end of the addressable space.

I think you must be mis-remembering the details slightly. The comma operator is a sequence-point, so "tmp" must be assigned the value of "a", and f() and g() must both be called with a value that is the value of "a" converted to the type of "tmp". The two functions can be called in either order though (or in parallel) but there is no issue there.

Of course, the compiler can do anything it likes so long as the program's output is equivalent to what I just described. So, for example, it might not allocate a memory location to "tmp", it could just push the value of "a" onto a register and then call f and g with it. Or if f or g do nothing and have no side-effects, the assembly code might not show calls to f and g. But there is no way you could know these things by running the program, which is the whole point.

"Overflows of unsigned values" is NOT undefined. You can assign out-of-range values to unsigned types, and also perform arithmetic operations which exceed the bounds of the type; and the value is adjusted using modular arithmetic.

Some would be facetious and say that "unsigned types cannot overflow", meaning that they always have well-defined behaviour on operations that would generate an out-of-range value, but that's just an issue of pedantry with English.

>The dereference is undefined, and therefore

Stop right here. Once undefined behaviour occurs, "all bets are off" as they say; the remaining code may have any behaviour whatsoever. C works like this on purpose , and it's something I agree with. It means the compiler doesn't have to insert screeds of extra checks , both at compile-time and run-time.

There are plenty of other languages you can use if you want a different language definition :)

>a == b is not a use of the argument that has been invalidated

Yes it is. Evaluating the expression "a" causes undefined behaviour if "a" is
indeterminate. "a" is considered to no longer have a value, any attempt to
refer to its value causes UB. (It has the same status as a variable that has
been defined but not initialized, i.e. "int a;"

The only thing that can be done with "a" thereafter is to assign a new value to it
(or take its address, or do "sizeof a" .. can't think of any other exceptions)

The behaviour is also undefined if realloc returns NULL. Also, sizeof(char) is 1 by definition.

>If my C code contains *foo=2, the compiler can't just leave that out

Well, it could if the program produces no further output before exiting, or if "foo" is unassigned.