Can anyone explain me the output which is given by the program. Why the value of j is 22 why not 21? [duplicate]

#include <stdio.h>

int main(void)
{
   int i = 0;
   i = i++ + ++i;
   printf("%d\n", i); // 3

   i = 1;
   i = (i++);
   printf("%d\n", i); // 2 Should be 1, no ?

   volatile int u = 0;
   u = u++ + ++u;
   printf("%d\n", u); // 1

   u = 1;
   u = (u++);
   printf("%d\n", u); // 2 Should also be one, no ?

   register int v = 0;
   v = v++ + ++v;
   printf("%d\n", v); // 3 (Should be the same as u ?)

   int w = 0;
   printf("%d %d\n", ++w, w); // shouldn't this print 1 1

   int x[2] = { 5, 8 }, y = 0;
   x[y] = y ++;
   printf("%d %d\n", x[0], x[1]); // shouldn't this print 0 8? or 5 0?
}

Solution 1:

C has the concept of undefined behavior, i.e. some language constructs are syntactically valid but you can't predict the behavior when the code is run.

As far as I know, the standard doesn't explicitly say why the concept of undefined behavior exists. In my mind, it's simply because the language designers wanted there to be some leeway in the semantics, instead of i.e. requiring that all implementations handle integer overflow in the exact same way, which would very likely impose serious performance costs, they just left the behavior undefined so that if you write code that causes integer overflow, anything can happen.

So, with that in mind, why are these "issues"? The language clearly says that certain things lead to undefined behavior. There is no problem, there is no "should" involved. If the undefined behavior changes when one of the involved variables is declared volatile, that doesn't prove or change anything. It is undefined; you cannot reason about the behavior.

Your most interesting-looking example, the one with

u = (u++);

is a text-book example of undefined behavior (see Wikipedia's entry on sequence points).

Solution 2:

Most of the answers here quoted from C standard emphasizing that the behavior of these constructs are undefined. To understand why the behavior of these constructs are undefined, let's understand these terms first in the light of C11 standard:

Sequenced: (5.1.2.3)

Given any two evaluations A and B, if A is sequenced before B, then the execution of A shall precede the execution of B.

Unsequenced:

If A is not sequenced before or after B, then A and B are unsequenced.

Evaluations can be one of two things:

• value computations, which work out the result of an expression; and
• side effects, which are modifications of objects.

Sequence Point:

The presence of a sequence point between the evaluation of expressions A and B implies that every value computation and side effect associated with A is sequenced before every value computation and side effect associated with B.

Now coming to the question, for the expressions like

int i = 1;
i = i++;

standard says that:

6.5 Expressions:

If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined. [...]

Therefore, the above expression invokes UB because two side effects on the same object i is unsequenced relative to each other. That means it is not sequenced whether the side effect by assignment to i will be done before or after the side effect by ++.
Depending on whether assignment occurs before or after the increment, different results will be produced and that's the one of the case of undefined behavior.

Lets rename the i at left of assignment be il and at the right of assignment (in the expression i++) be ir, then the expression be like

il = ir++     // Note that suffix l and r are used for the sake of clarity.
              // Both il and ir represents the same object.  

An important point regarding Postfix ++ operator is that:

just because the ++ comes after the variable does not mean that the increment happens late. The increment can happen as early as the compiler likes as long as the compiler ensures that the original value is used.

It means the expression il = ir++ could be evaluated either as

temp = ir;      // i = 1
ir = ir + 1;    // i = 2   side effect by ++ before assignment
il = temp;      // i = 1   result is 1  

or

temp = ir;      // i = 1
il = temp;      // i = 1   side effect by assignment before ++
ir = ir + 1;    // i = 2   result is 2  

resulting in two different results 1 and 2 which depends on the sequence of side effects by assignment and ++ and hence invokes UB.

Solution 3:

I think the relevant parts of the C99 standard are 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 6.5.16 Assignment operators, §4:

The order of evaluation of the operands is unspecified. If an attempt is made to modify the result of an assignment operator or to access it after the next sequence point, the behavior is undefined.