这个逻辑表达式是如何计算的

Question

#include<stdio.h>
int main()
{
    int a=-10,b=3,c=0,d;
    d= a++||++b &&c++;
    printf("%d %d %d %d ",a,b,c,d);
}

How above expression is evaluates. I have read preceedence but still i am getting confused. Kindly give me the right solution of this in step by step.

Answer 1

In case of ||and && operators the order of evaluation is indeed defined as left-to-right.So, operator precedence rules tell you that the grouping should be

(a++) || ((++b) && (c++))

Now, order-of-evaluation rules tell you that first we evaluate a++ , then (if necessary) we evaluate ++b , then (if necessary) we evaluate c++ , then we evaluate && and finally we evaluate || .

Answer 2

EDIT: The question was originally tagged C++ so I wrote this answer for C++.

I have a feeling this is a homework question, so I'll try to give extra explanation. There are actually a lot of concepts going on here!

Here are the main concepts:

1. Pre- and post- increment and decrement. ++a increments a before the value is used in an expression, while a++ increments a after the value is used in the expression.
2. Operator precedence . Specifically, && has higher precedence than || , which means that the expression assigned to d should be should be read as d = (a++) || (++b && c++); d = (a++) || (++b && c++);
3. Short-circuit boolean evaluation . This says that, for operators && and || , if the value of the left-hand term is enough to determine the result of the expression, then the right-hand term is not evaluated.
4. Expression evaluation order . The link shows a whole list of evaluation rules for C++ (I'm assuming C++11). In particular, item 6 says (paraphrasing) that the left-hand arguments of || and && are always evaluated before right-hand arguments, provided those operators are not overridden.
5. Implicit casting to boolean. In C++, if an int is cast to a bool , a nonzero value becomes true , while 0 becomes false . When converting the other way, false becomes 0 and true becomes 1 .

Putting this all together, here is what happens:

• After the first line, a is 10, b is 3, c is 0 and d is unset.
• Next, the variable d needs to be assigned. To determine the value assigned to d , the left hand term of (a++) || (++b && c++) (a++) || (++b && c++) is evaluated, which is a++ . The value USED in the expression is 10, however the value of a after this expression is -9 due to the post-increment.
• When cast to a boolean, the value 10 becomes true . This means that the value of the ||expression must be true . Because of short-circuit evaluation, this means that ++b && c++ is not evaluated at all, so the increments do not happen.
• The value to be assigned to d is true , but cast to an int . The result is 1, so we have d = 1 .
• Finally, the values are: a = -9, b = 3, c = 0, d = 1.

So the program prints out -9 3 0 1 .