see simple example:
int a = 0;
int b = (a ++ , a + 1); // result of b is UB or well defined ? (c++03).
This was changed in c++11/c++14 ?
The result is well defined and has been since C++98. The comma operator introduces a sequence point (or a "sequenced before" relationship in later C++s) between the the write and the second read of a and I don't see any other potential reasons for undefined behavior.
Related
main.cpp
const int& f(int& i ) { return (++i);}
int main(){
int i = 10;
int a = i++ + i++; //undefined behavior
int b = f(i) + f(i); //but this is not
}
compile
$ g++ main.cpp -Wsequence-point
statement int a = i++ + i++; is undefined behaviour.
statement int b = f(i) + f(i); is not undefined .
why?
statement int b = f(i) + f(i); is not undefined . why?
No, the second statement will result in unspecified behavior. You can confirm this here. As you'll see in the above linked demo, gcc gives the output as 23 while msvc gives 24 for the same program.
Pre-c++11, we use Sequence_point_rules:
Pre-C++11 Undefined behavior
Between the previous and next sequence point, the value of any object in a memory location must be modified at most once by the evaluation of an expression, otherwise the behavior is undefined.
Pre-C++11 Rules
3) There is a sequence point after the copying of a returned value of a function and before the execution of any expressions outside the function.
In int a = i++ + i++;, i is modified twice
In int b = f(i) + f(i);, there are sequence point with function call. i is modified only once between the sequence call. so no UB.
Note though that order of evaluation is unspecified so (evaluation of result) f(i) might happens after or before the second f(i), which might lead to different result depending of optimization/compiler and even between call.
Since C++11, we use "Sequenced before" rules, which is a similar way disallows i++ + i++ but allows f(i) + f(i).
I have a pointer which is defined as follows:
A ***b;
What does accessing it as follows do:
A** c = b[-1]
Is it an access violation because we are using a negative index to an array? Or is it a legal operation similar to *--b?
EDIT Note that negative array indexing has different support in C and C++. Hence, this is not a dupe.
X[Y] is identical to *(X + Y) as long as one of X and Y is of pointer type and the other has integral type. So b[-1] is the same as *(b - 1), which is an expression that may or may not be evaluated in a well-formed program – it all depends on the initial value of b! For example, the following is perfectly fine:
int q[24];
int * b = q + 13;
b[-1] = 9;
assert(q[12] == 9);
In general, it is your responsibility as a programmer to guarantee that pointers have permissible values when you perform operations with them. If you get it wrong, your program has undefined behaviour. For example:
int * c = q; // q as above
c[-1] = 0; // undefined behaviour!
Finally, just to reinforce the original statement, the following is fine, too:
std::cout << 2["Good morning"] << 4["Stack"] << 8["Overflow\n"];
I vaguely remember reading somewhere that it is undefined behaviour if multiple operands in a compound expression modify the same object.
I believe an example of this UB is shown in the code below however I've compiled on g++, clang++ and visual studio and all of them print out the same values and can't seem to produce unpredictable values in different compilers.
#include <iostream>
int a( int& lhs ) { lhs -= 4; return lhs; }
int b( int& lhs ) { lhs *= 7; return lhs; }
int c( int& lhs ) { lhs += 1; return lhs; }
int d( int& lhs ) { lhs += 2; return lhs; }
int e( int& lhs ) { lhs *= 3; return lhs; }
int main( int argc, char **argv )
{
int i = 100;
int j = ( b( i ) + c( i ) ) * e( i ) / a( i ) * d( i );
std::cout << i << ", " << j << std::endl;
return 0;
}
Is this behaviour undefined or have I somehow conjured up a description of supposed UB that is not actually undefined?
I would be grateful if someone could post an example of this UB and maybe even point me to where in the C++ standard that it says it is UB.
No. It is not. Undefined behavior is out of question here (assuming the int arithmetic does not overflow): all modifications of i are isolated by sequence points (using C++03 terminology). There's a sequence point at the entrance to each function and there's a sequence point at the exit.
The behavior is unspecified here.
Your code actually follows the same pattern as the classic example often used to illustrate the difference between undefined and unspecified behavior. Consider this
int i = 1;
int j = ++i * ++i;
People will often claim that in this example the "result does not depend on the order of evaluation and therefore j must always be 6". This is an invalid claim, since the behavior is undefined.
However in this example
int inc(int &i) { return ++i; }
int i = 1;
int j = inc(i) * inc(i);
the behavior is formally only unspecified. Namely, the order of evaluation is unspecified. However, since the result of the expression does not depend on the order of evaluation at all, j is guaranteed to always end up as 6. This is an example of how generally dangerous unspecified behavior combination can lead to perfectly defined result.
In your case the result of your expression does critically depend on the order of evaluation, which means that the result will be unpredictable. Yet, there's no undefined behavior here, i.e. the program is not allowed to format your hard drive. It is only allowed to produce unpredictable result in j.
P.S. Again, it might turn out that some of the evaluation scenarios for your expression lead to signed integer overflow (I haven't analyzed them all), which by itself triggers undefined behavior. So, there's still a potential for unspecified behavior leading to undefined behavior in your expression. But this is probably not what your question is about.
No its not undefined behavior.
But it does invoke unspecified behavior.
This is because the order that sub-expressions are evaluated is unspecified.
int j = ( b( i ) + c( i ) ) * e( i ) / a( i ) * d( i );
In the above expression the sub expressions:
b(i)
c(i)
e(i)
a(i)
d(i)
Can be evaluated in any order. Because they all have side-effects the results will depend on this order.
If you divide up the expression into all sub-expressions (this is pseudo-code)
Then you can see any ordering required. Not only can the above expressions be done in any order, potentially they can be interleaved with the higher level sub-expressions (with only a few constraits).
tmp_1 = b(i) // A
tmp_2 = c(i) // B
tmp_3 = e(i) // C
tmp_4 = a(i) // D
tmp_5 = d(i) // E
tmp_6 = tmp_1 + tmp_2 // F (Happens after A and B)
tmp_7 = tmp_6 * tmp_3 // G (Happens after C and F)
tmp_8 = tmp_7 / tmp_4 // H (Happens after D and G)
tmp_9 = tmp_8 * tmp_5 // I (Happens after E and H)
int j = tmp_9; // J (Happens after I)
It isn't undefined behavior but it has unspecified results: The only modified object is i through the references passed to the functions. However, the call to the functions introduce sequence points (I don't have the C++ 2011 with me: they are called something different there), i.e. there is no problem of multiple changes within an expression causing undefined behavior.
However, the order in which the expression is evaluated isn't specified. As a result you may get different results if the order of the evaluation changes. This isn't undefined behavior: The result is one of all possible orders of evaluation. Undefined behavior means that the program can behave in any way it wants, including producing the "expected" (expected by the programmer) results for the expression in question while currupting all other data.
int main() {
int a = 10;
int b = a * a++;
printf("%i %i", a, b);
return 0;
}
Is the output of the above code undefined behavior?
No in
int b = a * a++;
the behavior is undefined, so the result can be anything - that's not what "implementation dependent" means.
You might wonder why it's UB here since a is modified only once. The reason is there's also a requirement in 5/4 paragraph of the Standard that the prior value shall be accessed only to determine the value to be stored. a shall only be read to determine the new value of a, but here a is read twice - once to compute the first multiplier and once again to compute the result of a++ that has a side-effect of writing a new value into a. So even though a is modified once here it is undefined behavior.
I recently learned about the , operator and the fact that it introduces a sequence point.
I also learned that the following code led to undefined behavior:
i = ++i;
Because i was modified twice between two sequence points.
But what about the following codes ?
i = 0, ++i;
i = (0, ++i);
While I know the rules, I can't get to a conclusion. So is it defined behavior or not ?
edit: Just as #paxdiablo mentions, defined or not, this is really a bad practice which should be avoided. This question is asked solely for educational purposes and better understanding of the "rules".
Yes. = has higher precedence than ,, so this expression is equivalent to (i = 0), ++i. , is a sequence point, so it's guaranteed that the ++i occurs after the assignment.
I'm not sure whether i = (0, ++i) is defined though. My guess would be no; there's no sequence point between the increment and the assignment.
i = 0, ++i;
As the other answer pointed out it is not Undefined Behaviour.
i = (0, ++i);
The behaviour is undefined in this case because there is no sequence point between ++i and assignment to i.
i = (0, ++i, 0)
The behaviour is well defined1 in C++03, IMHO.
1 See extended discussion for a similar expression.