Not understanding basic C problem

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I am expecting the answer should come as 1.But in GCC compiler and Turbo c compiler is showing the result as 0.In some DEV c compiler showing the result as 1.Which one is correct and why ?

#include<stdio.h>
int main()
{
   int z = 0;
   z = 0 || z++;
   printf ("%d", z);
   return 0;
 }

What I have tried:

I have tried in Dev c , GCC and Turbo c compiler.

Posted 2-Mar-17 20:31pm

Member 13035169

Updated 2-Mar-17 23:04pm

nv3

v2

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Suvendu Shekhar Giri 3-Mar-17 2:48am

Your question title should not contain code pieces/the whole program.

[no name] 3-Mar-17 3:05am

->Which one is correct and why ?
Please explain in words the algorithm you are implementing with this code. This code has Undefined behavior[^]

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Jochen Arndt · Answer 1 · 2017-03-02T21:25:00

You have a logical Short-circuit evaluation - Wikipedia[^] operation of the form

a || b

The boolean result of the first expression

C++

z = 0

is the value of the left operand z after the assignment. When that is zero, the boolean result is false. You can check this:

int z = 0;
int a = (z = 0) ? 1 : 0;
int b = (z = 1) ? 1 : 0;
printf("%d %d\n", a, b);

This should print 0 1 with all compilers.

Once the boolean result of the left term has been evaluated, the second operation should be only executed when the first was true:

C++

int b = (z = 0) ? 1 : 0;
if (b)
    b = (z++) ? 1: 0;

Then z will be still zero because the increment operation is not executed.
[EDIT2]
The above is wrong (thank you NV)!
It should be off course if (!b)
[/EDIT2]

So compilers executing always the second operation may be not conforming to the C standard. I wrote "maybe" because I'm not really sure in this special case where the logical result is not used.

You might check it this way (all optimisations disabled; e.g. -O0 with GCC):

C++

if (z = 0 || z++) printf("true: %d\n", z); // Never reached else printf("false: %d\n", z); // Always executed

A compiler that prints "1" here is not conforming to the C standard.
[EDIT]
Note also that optimising compilers might detect that the result of the logical operation is always false and assign zero to z or even skip the line

z = 0 || z++;

because z has been already set to zero in the previous line.
[/EDIT]

[EDIT2]
Solution 3 is pointing out the problem.
The behaviour is undefined as indicated by the warning.
Try this:

int z = 0;
z = 0 || z++;
printf ("z: %d\n", z);

(z = 0) || (z++);
printf ("z: %d\n", z);

With GCC, it will print

z: 0
z: 1

[/EDIT2]

CPallini · Answer 2 · 2017-03-02T21:26:00

gcc -Wall mytest.c
mytest.c: In function ‘main’:
mytest.c:6:7: warning: operation on ‘z’ may be undefined [-Wsequence-point]
     z = 0 || z++;

That is you hit a dark corner of the C programming language. See, for instance Sequence point - Wikipedia[^].
As a side note, enabling all the warnings (-Wall with gcc may prove itself useful, after all).

nv3 · Answer 3 · 2017-03-02T20:58:00

Solution 1

What did you expect from the statements:

C++

int z = 0;
z = 0 || z++;

This is again a question of whether the post-increment is performed before or after the assignment to z.

Posted 2-Mar-17 20:58pm

nv3

Comments

Jochen Arndt 3-Mar-17 3:43am

It is about short-circuit evaluation and the increment should never be executed.

nv3 3-Mar-17 4:40am

Hi Jochen! I don't really agree. As the first operand (0) is false, the second operand (z++) must be evaluated and the expression result is always 0.

The actual question is: Will the post-decrement be executed before or after the assignment to z. In the first case, the post-increment to z is executed, then the expression result (0) is assigned to z -- thus overwriting the effect of the post-increment. In the second case, the assignment is first executed, then the post-increment will afterwards increment z to the value of 1.

To my knowledge, the C standard is not clear on the moment on which the post-increment will be executed. And I admit, I personally don't care. Nobody in his clear mind will write a statement like this in production code. Thus, again one of these questions that are purely academic.

Jochen Arndt 3-Mar-17 4:57am

You are right!
Silly mistake (it is an OR operation).

I will update my solution.

nv3 3-Mar-17 11:12am

It happens to the best of us :-)

Patrice T · Answer 4 · 2017-03-02T23:04:00

you are in gray zone because C compiler can rewrite your code to optimize.
This line is unpredictable:

C++

z = 0 || z++;

Because you have a postincrement on a variable that is used 2 times in the line.
you get 0 when the code translate to:

C++

tmp = 0 || z;
z++;
z=tmp;

you get 1 when the code translate to:

C++

tmp = 0 || z;
z=tmp;
z++;

It depend on the compiler.

Not understanding basic C problem

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