int num = 8;

printf("%d", num++ - ++num + --num);
In the C programming language, the expression inside the `printf` function involves multiple uses of the `num` variable along with the increment (`++`) and decrement (`--`) operators both in postfix and prefix form. It is important to note that the use of such operators multiple times on the same variable in a single expression can lead to undefined behavior according to the C standard.\n\nHowever, if we were to assume a specific order of evaluation (which you should not do in production code), we could try to break down what happens in the expression, step by step:\n\n1. The postfix increment (`num++`) evaluates to the current value of `num` before incrementing it. So it takes the value 8 but sets `num` to 9 after this part of the expression is evaluated.\n\n2. The prefix increment (`++num`) increments `num` first before evaluating to its new value. Since `num` was set to 9, it is now incremented to 10, and then this part of the expression evaluates to 10.\n\n3. The prefix decrement (`--num`) decrements `num` first before evaluating to its new value. Since `num` was 10, it is now decremented to 9, and then this part of the expression evaluates to 9.\n\nIf the expression was evaluated strictly from left to right (not considering operator precedence or associativity, and ignoring potential undefined behavior), we would have:\n\n```c\nnum++ // num is 8, then num becomes 9\n- ++num // num is incremented to 10, then evaluates to 10, we have 8 - 10 = -2\n+ --num // num is decremented to 9, then evaluates to 9, we have -2 + 9 = 7\n```\n\nSo it would print \"7\". But remember, you can't rely on this in a real C program because of undefined behavior; different compilers can evaluate such expressions differently, leading to different outputs or possibly even runtime errors. For well-defined behavior, you should separate these operations into distinct statements.