i = ++i + ++i; in C++ [duplicate]

Answer 1

Simple... you compiler is evaluating BOTH increments before performing the sum, without caching the intermediate results. This means that when you add i twice, it now has the value of 7.

If you do

int j=++i; 
int k=++i;

i = j+k;

you'll see 13 as expected.

Answer 2

Because the prefix increment has precedence:

int i = 5;
i = i+1; // First ++i, i is now 6
i = i+1; // Second ++i, i is now 7
i = i + i // i = 7 + 7
cout << i // i = 14

Answer 3

The order of side effects is undefined in C++. Additionally, modifying a variable twice in a single expression has no defined behavior (See the C++ standard, §5.0.4, physical page 87 / logical page 73).

Solution: Don't use side effects in complex expression, don't use more than one in simple ones. And it does not hurt to enable all the warnings the compiler can give you: Adding -Wall(gcc) or /Wall /W4(Visual C++) to the command line yields a fitting warning:

test-so-side-effects.c: In function 'main':
test-so-side-effects.c:5: warning: operation on 'i' may be undefined
test-so-side-effects.c:5: warning: operation on 'i' may be undefined

Obviously, the code compiles to:

i = i + 1;
i = i + 1;
i = i + i;