Difference between revisions of "Undefined Value"

Revision as of 15:47, 20 August 2014

<syntaxhighlight lang="c" name="undefined_value" >

1. include<stdio.h>

int main() {

   int x=10, y;
   y = (x++) + (++x);
   printf("%d %d %d %d ", y, x++, x, ++x);

} </syntaxhighlight>

a) 22,10,11,13

b) 22,11,11,11

c) 12,10,11,13

d) 22 13 13 13

Solution by Arjun Suresh

None of the choices are correct as per C standard. This is because the statement

y = (x++) + (++x);

can causes undefined behavior, so does the statement

printf("%d %d %d %d ", y, x++, x, ++x);

C standard says that the side effects of an operation (for ++a; modification of a is a side-effect and return of a+1 is main effect) need to be completed only before the next sequence point. This relaxation is given so that the compiler would be able to generate the most optimal code (which run faster). But as a consequence, programmer shouldn't do a read and a write from a memory location between two consecutive sequence points or otherwise the result would be undefined.

In the statement,

y = (x++) + (++x);

There are 2 reads to the memory location of x and 2 writes to the same between two consecutive sequence points. So, the result of this statement has no guarantee as per C standard.

Similarly, the statement

printf("%d %d %d %d ", y, x++, x, ++x);

is undefined. In a function call in C, the order in which the argument expressions are processed is not defined. The only thing defined is that all arguments must be processed before the next sequence point- which here is the start of execution of the called function.

Undefined value means compiler can give any value. i.e.; different compilers or even different versions of the same compiler can give different answers. As per C standard, no programmer should write this code. This is different from compiler defined, in which case the output is clearly defined by the compiler and programmer can write those code as long as he is aware of the compiler. Example of compiler defined code is sizeof(int).

Still undefined

<syntaxhighlight lang="c" name="undefined" >

#include<stdio.h>

int main() {

   int x=10;
   printf("%d %d %d\n", x++, x, ++x);

} </syntaxhighlight>

asuresh@parambara:~$ gcc undefined.c 
asuresh@parambara:~$ ./a.out
11 12 12
asuresh@parambara:~$ icc undefined.c 
asuresh@parambara:~$ ./a.out
10 12 12
asuresh@parambara:~$ clang undefined.c 
asuresh@parambara:~$ ./a.out
10 11 12

Is this Problem with ++? NO

<syntaxhighlight lang="c" name="undefined" >

#include<stdio.h>

int main() {

   int x=10;
   printf("%d %d %d\n", x=1, x, x=2);

} </syntaxhighlight>

asuresh@parambara:~$ gcc -Wall undefined.c
undefined.c: In function ‘main’:
undefined.c:6:35: warning: operation on ‘x’ may be undefined [-Wsequence-point]
    printf("%d %d %d\n", x=1, x, x=2);
                                  ^
undefined.c:6:35: warning: operation on ‘x’ may be undefined [-Wsequence-point]
asuresh@parambara:~$ ./a.out
1 1 1
asuresh@parambara:~$ icc -Wall undefined.c
asuresh@parambara:~$ ./a.out
1 2 2
asuresh@parambara:~$ clang -Wall undefined.c
asuresh@parambara:~$ ./a.out
1 1 2

Precedence Rule Violated? No

Undefined behaviour is happening not because C is violating precedence rule. Precedence (and associativity which is used when precedence of operators are same) rule is used to specify the operands for operations. But the actual execution order might be different. For example:

x = a*b+c*d;

After applying precedence rule, we get

x = ((a*b) + (c*d));

Thus the first $*$ has operands $a$ and $b$, the second $*$ has operands $c$ and $d$, the $+$ has operands returned by the result of two $*$ and $=$ has left operand $x$ and right operand the result of $+$. But, during execution $(a*b)$ or $(c*d)$ can occur first as both these won't be violating the precedence rule.

Similarly, if we take

y = x++ * ++x;

After applying precedence rule, we get

y = ((x++) * (++x));

$x$++ or ++$x$ can happen first. Suppose ++$x$ happened first. It'll surely return $x+1$. But whether this incremented $x$ is used by $x$++ is not guaranteed. This increment can happen to $x$ ANYTIME before the next sequence point, which here is the ; at the end of the expression. This makes this code produce undefined behaviour.

Other Examples for Undefined behaviour

int i = 2;
i = i++;
printf("%d\n", i);
Possible outputs: {3, 2}

int i = 2;
a[i] = i++;
Possible Results: a[2] = 2, a[3] = 2

int i = 2, j;
j = (i++) * (++i);
printf("%d", j);
Possible outputs: {6, 9}

int i = 2;
printf("%d %d", i, i++);
Possible outputs: {(2,3), (2,2)}