++x + ++x

No prob!

Vig.

Viggy had it right.

Operator precedence would evaluate the ++x first, so x = 5 after this is done. Then x + x = 10

Quote:

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Originally Posted by INeedHelp2

Operator precedence would evaluate the ++x first, so x = 5 after this is done. Then x + x = 10

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It is only one possibility.
There are plenty others:

• compile-time error.
• crash
• It is very probable that it may yield any value in range [7, 10]

And, saying simply that "side effects appear in any order" is not exact.
Side effects can appear in any order and may even be interlaced!

For example, if there were a primitive operator "swap bytes" (little endian <-> big endian), doing something like:

Code:

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swapbytes(swapbytes(x))

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Would probably make some bytes appear twice, and others not at all, and in any order...

Indeed any program that make two side effects on the same lvalue between two sequence points, has undefined behaviour.
It may play the 9th symphony of Beethoven...