This question already has answers here:
Undefined behavior and sequence points
(5 answers)
Closed 8 years ago.
Firstly, this is an attempt to understand how operator precedence works and not being used by me in any of my projects.
int *a=new int[3];
a[0]=3;a[1]=7;a[2]=11;
*a*=++*a**a++;
cout<<*(a-1)<<endl<<*a<<endl<<*(a+1);
Gives the following output,
4
112
11
I am expecting the following output,
64
7
11
Can someone please tell me how this is happening?
Thank you
Even by respecting operator precedences, you're not respecting sequence points and thus invoking undefined behavior.
Take a look at the cpp faq here: http://www.parashift.com/c++-faq/sequence-points.html
The C++ standard says (1.9p7),
At certain specified points in the execution sequence called sequence
points, all side effects of previous evaluations shall be complete and
no side effects of subsequent evaluations shall have taken place.
For example, if an expression contains the subexpression y++, then the
variable y will be incremented by the next sequence point. Furthermore
if the expression just after the sequence point contains the
subexpression ++z, then z will not have yet been incremented at the
moment the sequence point is reached.
The order of evaluation of operands of individual operators and subexpressions of individual expressions, and the order in which side effects take place, is unspecified
Additional info on sequence points: https://stackoverflow.com/a/4176333/1938163
Related
Similar codes for example (a+=1)%=7;, where a is an int variable.
We know that operator += or = is not a sequence point, therefore we have two side-effects between two adjcent sequence points. (we are using cpp98's sequence point rules here)
However, assignment operators like += or = guarantees to return the lvalue after assignment, which means the order of execution is to some degree "defined".
So is that an undefined behaviour ?
(a=1)=2 was undefined prior to C++11, as the = operator did not introduce a sequence point and therefore a is modified twice without an intervening sequence point. The same applies to (a+=1)%=7
The text was:
Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression.
It's worth mentioning that the description of the assignment operator is defective:
The result of the assignment operation is the value stored in the left operand after the assignment has taken place; the result is an lvalue.
If the result is an lvalue then the result cannot be the stored value (that would be an rvalue). Lvalues designate memory locations. This sentence seems to imply an ordering relation, but regardless of how we want to interpret it, it doesn't use the term "sequence point" and therefore the earlier text about sequence points applies.
If anything, that wording casts a bit of doubt on expressions like (a=1) + 2. The C++11 revision of sequencing straightened out all these ambiguities.
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 3 years ago.
Just to see how much I understood how the ++c/c++ operator works, I tried to run these C programs:
int c = 5;
c = c - c++;
printf("%d\n", c);
prints 1, I guess the logic is that the ++ is applied after the line of code where it's used, so c becomes = c - c which is 0, and on the "next line" it's increased by one. But it seems strange to me, I'd like to know more in detail what should happen with regards to the operators priority.
Now on to this:
int c = 5;
c = c - ++c;
printf("%d\n", c);
this one prints 0, and I can't really understand why. If right hand values are parsed from left to right, I guess it would read c which is 5, then ++c which is 6 as it should be applied immediately. Or does it calculate the ++c before the whole right hand value calculation, so that it's actually doing 6 - 6 because the increment also involves the first calling of c?
For C++ (all versions, explanation applies to C++11 and later):
Both have undefined behavior, meaning that not only is the value that it will return unspecified, but that it causes your whole program to behave in an undefined manner.
The reason for this is that evaluation order inside an expression is only specified for certain cases. The order in which expressions are evaluated does not follow the order in the source code and is not related to operator precedence or associativity. In most cases the compiler can freely choose in which order it will evaluate expressions, following some general rules (e.g. the evaluation of an operator is sequenced after the value computation of its operands, etc.) and some specific ones (e.g. &&'s and ||'s left-hand operands are always sequenced before their right-hand operands).
In particular the order in which the operands of - are evaluated is unspecified. It is said that the two operands are unsequenced relative to one another.
This in itself means that we won't know whether c on the left-hand side of c - [...] will evaluate to the value of c before or after the increment.
There is however an even stricter rule forbidding the use of a value computation from a scalar object (here c) in a manner unsequenced relative to a side effect on the same scalar object. In your case both ++c and c++ cause side effects on c, but they are unsequenced with the use of the value on the left hand side of c - [...]. Not following this rule causes undefined behavior.
Therefore your compiler is allowed to output whatever it wants and you should avoid writing code like that.
For a detailed list of all the evaluation order rules of C++, see cppreference.com. Note that they changed somewhat with the different C++ versions, making more and more previously undefined or unspecified behavior defined. None of these changes apply to your particular case though.
c = c - c++;
In C, this is a very bad idea(a). You are not permitted to modify and modify/use the same object without an intervening sequence point, and that subtraction operator is not a sequence point.
The things that are sequence points can be found in Annex C of the ISO standard.
(a) Technically, the behaviour of each operation (the evaluation of c1 and c++, and the assignment to c) is well defined but the sequencing is either unsequenced or indeterminate. In the former case, actions from each part can interleave while, in the latter, they do not interleave but you don't know in which order the two parts will be done.
However, the standard C11 6.5/2 also makes it clear that a sequencing issue using the same variable is undefined behaviour:
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. If there are multiple allowable orderings of the subexpressions of an expression, the behavior is undefined if such an unsequenced side effect occurs in any of the orderings.
Bottom line, it's not something you should be doing.
Let me present a example :
a = ++a;
The above statement is said to have undefined behaviors ( I already read the article on UB on SO)
but according precedence rule operator prefix ++ has higher precedence than assignment operator =
so a should be incremented first then assigned back to a. so every evaluation is known, so why it is UB ?
The important thing to understand here is that operators can produce values and can also have side effects.
For example ++a produces (evaluates to) a + 1, but it also has the side effect of incrementing a. The same goes for a = 5 (evaluates to 5, also sets the value of a to 5).
So what you have here is two side effects which change the value of a, both happening between sequence points (the visible semicolon and the end of the previous statement).
It does not matter that due to operator precedence the order in which the two operators are evaluated is well-defined, because the order in which their side effects are processed is still undefined.
Hence the UB.
Precedence is a consequence of the grammar rules for parsing expressions. The fact that ++ has higher precedence than = only means that ++ binds to its operand "tighter" than =. In fact, in your example, there is only one way to parse the expression because of the order in which the operators appear. In an example such as a = b++ the grammar rules or precedence guarantee that this means the same as a = (b++) and not (a = b)++.
Precedence has very little to do with the order of evaluation of expression or the order in which the side-effects of expressions are applied. (Obviously, if an operator operates on another expression according to the grammar rules - or precedence - then the value of that expression has to be calculated before the operator can be applied but most independent sub-expressions can be calculated in any order and side-effects also processed in any order.)
why it is UB ?
Because it is an attempt to change the variable a two times before one sequence point:
++a
operator=
Sequence point evaluation #6: At the end of an initializer; for example, after the evaluation of 5 in the declaration int a = 5;. from Wikipedia.
You're trying to change the same variable, a, twice. ++a changes it, and assignment (=) changes it. But the sequence point isn't complete until the end of the assignment. So, while it makes complete sense to us - it's not guaranteed by the standard to give the right behavior as the standard says not to change something more than once in a sequence point (to put it simply).
It's kind of subtle, but it could be interpreted as one of the following (and the compiler doesn't know which:
a=(a+1);a++;
a++;a=a;
This is because of some ambiguity in the grammar.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Undefined Behavior and Sequence Points
Pleae explain the behaviour of following statements
int b=3;
cout<<b++*++b<<endl;
How will it be calculated?
The behavior here is undefined. See this question
Relevant standard quote:
§5/4.1 Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression.
The most common sequence point is the end of a statement.
Also worth noting from the standard:
§5.2.2/8 The order of evaluation of arguments is unspecified.
The standard says this is undefined. The compiler is free to evaluate the statements in any order is sees fit as long as it follows the operator precedence rules. This results in UB:
b++ * ++b; // b is modified more than once
The behavior will be undefined as told by others.
The output depends upon the implementation of compiler.
But as per the standard it should be undefined.
AS this is undefined behaviour, one can't tell the end result. The result depends on the implementation.
Undefined behavior, Compiler is free to evaluate this expression in any order because of the same precedence of the operators. Consider using
(b++)*(++b)
instead
A few days back there was a discussion here about whether the expression
i = ++i + 1
invokes UB
(Undefined Behavior) or not.
Finally the conclusion was made that it invokes UB as the value of 'i' is changing more than once between two sequence points.
I was involved in a discussion with Johannes Schaub in that same thread. According to him
i=(i,i++,i)+1 ------ (1) /* invokes UB as well */
I said (1) does not invoke UB because the side effects of the previous subexpressions are cleared by the comma operator ',' between i and i++ and between i++ and i.
Then he gave the following explanation:
"Yes the sequence point after i++ completes all side effects before it, but there is nothing that stops the assignment side effect overlapping with the side effect of i++.The underlying problem is that the side effect of an assignment is not specified to happen after or before the evaluation of both operands of the assignment, and so sequence points cannot do anything with regard to protecting this: Sequence points induce a partial order: Just because there is a sequence point after and before i++ doesn't mean all side effects are sequenced with regard to i.
Also, notice that merely a sequence point means nothing: The order of evaluations isn't dictated by the form of code. It's dictated by semantic rules. In this case, there is no semantic rule saying when the assignment side effect happens with regard to evaluating both of its operands or subexpressions of those operands".
The statement written in "bold" confused me. As far as I know:
"At certain specified points in the execution sequence called sequence points,all side effects of previous evaluations shall be complete and no side effects of subsequent evaluations shall have taken place."
Since,comma operators also specify execution order the side effect of i++ have been cancelled when we reach the last i.He(Johannes) would have been right had the order of evaluation been not specified(but in case of comma operator it is well specified).
So I just want to know whether (1) invokes UB or not?. Can someone give another valid explanation?
Thanks!
The C standard says this about assignment operators (C90 6.3.16 or C99 6.5.16 Assignment operators):
The side effect of updating the stored value of the left operand shall occur between the previous and the next sequence point.
It seems to me that in the statement:
i=(i,i++,i)+1;
the sequence point 'previous' to the assignment operator would be the second comma operator and the 'next' sequence point would be the end of the expression. So I'd say that the expression doesn't invoke undefined behavior.
However, this expression:
*(some_ptr + i) = (i,i++,i)+1;
would have undefined behavior because the order of evaluation of the 2 operands of the assignment operator is undefined, and in this case instead of the problem being when the assignment operator's side effect takes place, the problem is you don't know whether the value of i used in the left handle operand will be evaluated before or after the right hand side. This order of evaluation problem doesn't occur in the first example because in that expression the value of i isn't actually used in the left-hand side - all that the assignment operator is interested in is the "lvalue-ness" of i.
But I also think that all this is sketchy enough (and my understanding of the nuances involved are sketchy enough) that I wouldn't be surprised if someone can convince me otherwise (on either count).
i=(i,i++,i)+1 ------ (1) /* invokes UB as well */
It does not invoke undefined behaviour. The side effect of i++ will take place before the evaluation of the next sequence point, which is denoted by the comma following it, and also before the assignment.
Nice language sudoku, though. :-)
edit: There's a more elaborate explanation here.
I believe that the following expression definitely has undefined behaviour.
i + ((i, i++, i) + 1)
The reason is that the comma operator specifies sequence points between the subexpressions in parentheses but does not specify where in that sequence the evaluation of the left hand operand of + occurs. One possibility is between the sequence points surrounding i++ and this violates the 5/4 as i is written to between two sequence points but is also read twice between the same sequence points and not just to determine the value to be stored but also to determine the value of the first operand to the + operator.
This also has undefined behaviour.
i += (i, i++, i) + 1;
Now, I am not so sure about this statement.
i = (i, i++, i) + 1;
Although the same principals apply, i must be "evaluated" as a modifiable lvalue and can be done so at any time, but I'm not convinced that its value is ever read as part of this. (Or is there another restriction that the expression violates to cause UB?)
The sub-expression (i, i++, i) happens as part of determining the value to be stored and that sub-expression contains a sequence point after the storage of a value to i. I don't see any way that this wouldn't require the side effect of i++ to be complete before the determination of the value to be stored and hence the earliest possible point that the assignment side effect could occur.
After this sequnce point i's value is read at most once and only to determine the value that will be stored back to i, so this last part is fine.