New to C. So I saw this line of code
system->word[system->index++] = system->next_char;
Is it equivalent to:
system->word[system->index] = system->next_char;
index++;
What is the precedence for post increment? Does it only increment index's value by 1 after all the operations on the line are done executing?
Updating system->index is defined as a side effect that is not sequenced (is not specified to come before or after) the other operations in the statement. The update may occur before, during, or after other operations.
The fact that it is not sequenced is irrelevant as long as it is not used elsewhere in the statement, because, if it is not used elsewhere, then nothing the statement does can be affected by when the update occurs. (Note that, even if the update to system->index in memory is done before the value is used, the compiler is response for ensuring that the pre-update value is used.)
If the object being updated were used elsewhere in the statement in an unsequenced way (that is, no rule specifies which comes first, the update or the other use), then the behavior of the program would not be defined by the C standard.
This is not a consequence of precedence. Precedence determines the structure of how expressions are interpreted, not the sequencing of their operations.
Not.
system->word[system->index++] = system->next_char;
is equivalent to:
system->word[system->index] = system->next_char;
system->index++;
as index is a field on a struct pointed to by system. In case you have also a scalar variable named index you had had no errors but the wrong variable should have been incremented.
As a general rule, all the unary operators evaluate first on the right side of the variable, then the ones on the left (the right side operators have higher precedence thatn left side ones) and evaluate from closest to the operand, to farthest. So the operator closest to the operand evaluates first, then the one next in the right... so until there are no more right operators on the right, then the closest on the left side, and so on until there are no more left operators on the left.
So what happens if we have -x++ ? The ++ is evaluated first to the value of x and a post increment of the variable is schedule to occur, then the one on the left is evaluated, givin as result the value of x before incrementing, changed of sign, and the variable x gets its value incremented (only incremented, no sign change).
Another example: let's compute ++x->field: first of all ->field (the whole thing, the -> arrow and the field name) is considered a right unary operator, given a pointer gets the value in field named field fo structure pointed to by x. This value is incremented (and stored again in x->field) and the returned value is the final incremented one.
Another final example: let's compute *p++. The value of p is scheduled to be post incremented, but it's previous value is retained. Then the old value of p is used to access the value pointed to by it, giving the original pointed to value. Finally p is incremented and points to the value next to the value accessed.
Related
From C in a Nutshell:
An lvalue is an expression that designates an object, and it
can appear on the left side of an assignment operator.
An lvalue can always be resolved to the corresponding object’s
address, unless the object is a bit-field or a variable declared with the register storage class.
According to the second sentence from the quote, when the object is
a bit-value or a variable declared with the register storage class,
a lvalue can't be resolved to the corresponding object's address.
Why is that?
Given a lvalue which can't be resolved to the corresponding object's
address, how can the lvalue designate an object, and appear on the
left side of an assignment operator?
I think that the following three statements are equivalent:
a lvalue can't be resolved to the corresponding object's address,
a lvalue designate an object,
a lvalue can appear on the left side of an assignment operator.
and either of them can be used as a definition of lvalue. Am I
right?
Thanks.
The minimum addressability resolution is the char. If you want a bit field, that's one or more bits within a char so cannot have its own address, unless you were to allow addresses like 42.6 which would blow the heads off most coders :-)
Variables with register storage class generally can't be addressed because they're not necessarily stored anywhere in memory - that is, after all, what the register storage class means: try to keep this value in a register.
But, even though you cannot get an address for those objects, that doesn't mean you cannot assign to them. For objects held in registers, you just change the register.
And, for bit fields, you can just use the Boolean operations like and/or to manipulate parts of an addressable value.
Im toying with the problem of prefix evaluations, and I want to evaluate them using only a queue. Here is my pseudocode
while Q has more than 1 element
if the pattern operand, number, number occurs
op=dequeue(Q)
num1=dequeue(Q)
num2=dequeue(Q)
eval=evaluate(op,num1,num2)
enqueue(Q,eval)
else
elem=dequeue(Q)
enqueue(Q,elem)
I think my logic is correct for well formed prefix operations, but I'm not sure how to account for invalid prefix syntax, such as 10+4.
Right now, my algorithm would dequeue the 10 and then enqueue it at the end, which would then become proper prefix and be evaluated, but I do not want it to. Is there some pre condition to make sure the syntax is correct?
If you have a function to pop the first element of the queue without dequeuing it, you could use it in your else path with a while(first element not operand) to dequeue the invalid elements. This would guarantee you at least, that the first item in your queue is an operand.
I was answering a postgres question yesterday, and also came across a postgres thread (here) where they describe the following error:
ERROR: operator does not exist: text = text[]
HINT: No operator matches the given name and argument type(s). You
might need to add explicit type casts.
The error seems to appear whenever an ARRAY string type is fed to ANY without using = ANY. This seems completely strange since based on language, logic, and sql conventions, usually you have (e.g. IN):
variable FUNCTION(set)
instead of.
variable = FUNCTION(set) , unless ofcourse operator is a summation/count operation returning one result :)
It would make more senseto have variable ANY(Set/Array) instead of variable=ANY(Set/Array). Similar example is the IN function.
Can anyone explain what is going on here?
IN (...) is basically equivalent to = ANY (ARRAY[...])
Crucially, ANY is not a function. It's syntax defined by the SQL standard, and is no more a function than GROUP BY or the OVER clause in a window function.
The reason that = is required before ANY is that ANY can apply to other operators too. What it means is "Test the operator to the left against every element in the array on the right, and return true if the test is true for at least one element."
You can use > ANY (ARRAY[...]) or whatever. It's a general purpose operator that isn't restricted to =. Notably useful for LIKE ANY (albeit with somewhat bad performance).
There is ALL too, which does much the same thing but returns true only if all results are true.
For example,
int x = 10;
we say that "x designates an int object which stores 10". But what does "designate" exactly mean? Does it mean x behaves like a label which refers to the whole chunk of memory?
x is an identifier.
There is an int object (i.e. a region of storage) containing the value 10.
The identifier x is associated with that int object.
The C standard uses the English word designate to express the relationship between an identifier and the object it identifies. You could say the same thing in several different ways. I said "associate" just now, there are many words we could choose. "x is a label for this region of memory" would be another way.
Note: designating is not limited to identifiers. Other expressions can designate an object too. For example *(&x) also designates the same object, as does *(&x + 0).
When an expression designates an object, the expression may be used to either assign a value to the object, or retrieve the value from the object. (The same syntax covers both of those cases; it depends on context whether the value is read or written).
The word lvalue means an expression that might designate an object (according to the above definition of 'designate').
I want to know if it is possible to use the return value of a statement in C.
For example: If I start with a=0; then I know that the expression a++ will return 0 while a+=1 will return 1. But talking about statements, is there any semantic difference between a++; and a+=1;?
At first, they both look like they have similar behavior and, so, the only difference could be the existence of a way of using the return values of statements. Is it possible?
Statements don't have return values. Expressions have values (unless of type void). Some expressions also have side effects.
a++ has a value equal to the original value of a and a+=1 has a value of one greater than the original value of a.
Assigning a new value to a is a side effect. Both expressions have the same side effect.
There's nothing more complicated about any of this. Do note though, that depending on a side effect within the same statement (not just the same subexpression) produces undefined behavior, unless there is a sequence point. Multiple side effects within the same statement, and side effect dependencies, are a more advanced topic so it is better to keep assignment statements simple.
You want to compare ++a and (a+=1) which will both return 1
a++ means return the old value and then change a
++a means add one to a, and return the new value
(a+=1) means add one to a and return the new value (but technically it can be slower)
On its own line, ++a, a++ and a+=1 are all the same thing
if you call: foo(a++) and foo(a+=1) foo will be passed the different values.
Technically ++a can be faster than a+=1, but any level of optimization should result in the same thing