Question

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Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)

In Java the evaluation order is specified to be left-to-right. Is this the case for C and C++ as well, or is it implementation dependent? I do remember that the evaluation order is unspecified for function arguments, but what about sub-expressions?

Answer 1

It is unspecified which of the arguments to + is evaluated first – but that doesn’t even matter, because in C and C++, modifying the same object twice without an intervening sequence point is completely undefined behaviour.

Here you’re modifying x three times without an intervening sequence point, so you’re well into here be dragonnes territory 😉

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The relevant part of the C99 standard is “6.5 Expressions”:

2 Between the previous and next
sequence point an object shall have
its stored value modified at most once
by the evaluation of an expression.
Furthermore, the prior value shall be
read only to determine the value to be
stored.

and

3 The grouping of operators and
operands is indicated by the
syntax. Except as specified later
(for the function-call (), &&, ||, ?:,
and comma operators), the order of
evaluation of subexpressions and the
order in which side effects take place
are both unspecified.

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It’s possible to write legal code that demonstrates the unspecified order of evaluation – for example:

#include <stdio.h>

int foo(void)
{
    puts("foo");
    return 1;
}

int bar(void)
{
    puts("bar");
    return 2;
}

int main()
{
    int x;

    x = foo() + bar();
    putchar('\n');

    return x;
}

(It is unspecified whether you get output of foobar or barfoo).