Order of Evaluation of Arguments Using Std::Cout

Order of evaluation of arguments using std::cout

The evaluation order of elements in an expression is unspecified (except some very particular cases, such as the && and || operators and the ternary operator, which introduce sequence points); so, it's not guaranteed that test will be evaluated before or after foo(test) (which modifies it).

If your code relies on a particular order of evaluation the simplest method to obtain it is to split your expression in several separated statements.

std::cout statements evaluation order [duplicate]

Nothing did go wrong, but the pops get (in your case) evaluated from right to left if you put them all in one std::cout line. In general. the evaluation order is unspecified. For more detail on this see here.

So you correctly get the elements in reverse order of insertion and then -1 and then print them reversed.

C++: The order of std::cout [duplicate]

T get() { //Get the first value of the Queue
return (!first) ? first->value : NULL;
};

This is backwards. Drop the !. You are saying "if first is not non-null, (ie if first is null), use it.

That said, the order of evaluation of arguments to a function is unspecified (the compiler can evaluate the arguments in any order it feels like, as long as they are all done before the function itself starts). That is get(), pop() and pop() can be called in any order. Call them as separate statements:

int a = bag.get();
int b = bag.pop();
int c = bag.pop();
std::cout << a << b << c;

How to guarantee order of argument evaluation when calling a function object?

What about a silly wrapper class like this:

struct OrderedCall
{
template <typename F, typename ...Args>
OrderedCall(F && f, Args &&... args)
{
std::forward<F>(f)(std::forward<Args>(args)...);
}
};

Usage:

void foo(int, char, bool);

OrderedCall{foo, 5, 'x', false};

If you want a return value, you could pass it in by reference (you'll need some trait to extract the return type), or store it in the object, to get an interface like:

auto x = OrderedCall{foo, 5, 'x', false}.get_result();

What are the evaluation order guarantees introduced by C++17?

Some common cases where the evaluation order has so far been unspecified, are specified and valid with C++17. Some undefined behaviour is now instead unspecified.

i = 1;
f(i++, i)

was undefined, but it is now unspecified. Specifically, what is not specified is the order in which each argument to f is evaluated relative to the others. i++ might be evaluated before i, or vice-versa. Indeed, it might evaluate a second call in a different order, despite being under the same compiler.

However, the evaluation of each argument is required to execute completely, with all side-effects, before the execution of any other argument. So you might get f(1, 1) (second argument evaluated first) or f(1, 2) (first argument evaluated first). But you will never get f(2, 2) or anything else of that nature.

std::cout << f() << f() << f();

was unspecified, but it will become compatible with operator precedence so that the first evaluation of f will come first in the stream (examples below).

f(g(), h(), j());

still has unspecified evaluation order of g, h, and j. Note that for getf()(g(),h(),j()), the rules state that getf() will be evaluated before g, h, j.

Also note the following example from the proposal text:

 std::string s = "but I have heard it works even if you don't believe in it"
s.replace(0, 4, "").replace(s.find("even"), 4, "only")
.replace(s.find(" don't"), 6, "");

The example comes from The C++ Programming Language, 4th edition, Stroustrup, and used to be unspecified behaviour, but with C++17 it will work as expected. There were similar issues with resumable functions (.then( . . . )).

As another example, consider the following:

#include <iostream>
#include <string>
#include <vector>
#include <cassert>

struct Speaker{
int i =0;
Speaker(std::vector<std::string> words) :words(words) {}
std::vector<std::string> words;
std::string operator()(){
assert(words.size()>0);
if(i==words.size()) i=0;
// Pre-C++17 version:
auto word = words[i] + (i+1==words.size()?"\n":",");
++i;
return word;
// Still not possible with C++17:
// return words[i++] + (i==words.size()?"\n":",");

}
};

int main() {
auto spk = Speaker{{"All", "Work", "and", "no", "play"}};
std::cout << spk() << spk() << spk() << spk() << spk() ;
}

With C++14 and before we may (and will) get results such as

play
no,and,Work,All,

instead of

All,work,and,no,play

Note that the above is in effect the same as

(((((std::cout << spk()) << spk()) << spk()) << spk()) << spk()) ;

But still, before C++17 there was no guarantee that the first calls would come first into the stream.

References: From the accepted proposal:

Postfix expressions are evaluated from left to right. This includes
functions calls and member selection expressions.

Assignment expressions are evaluated from right to left. This
includes compound assignments.

Operands to shift operators are evaluated from left to right. In
summary, the following expressions are evaluated in the order a, then
b, then c, then d:

  1. a.b
  2. a->b
  3. a->*b
  4. a(b1, b2, b3)
  5. b @= a
  6. a[b]
  7. a << b
  8. a >> b

Furthermore, we suggest the following additional rule: the order of
evaluation of an expression involving an overloaded operator is
determined by the order associated with the corresponding built-in
operator, not the rules for function calls.

Edit note: My original answer misinterpreted a(b1, b2, b3). The order of b1, b2, b3 is still unspecified. (thank you @KABoissonneault, all commenters.)

However, (as @Yakk points out) and this is important: Even when b1, b2, b3 are non-trivial expressions, each of them are completely evaluated and tied to the respective function parameter before the other ones are started to be evaluated. The standard states this like this:

§5.2.2 - Function call 5.2.2.4:

. . .
The postfix-expression is sequenced before each expression in the
expression-list and any default argument. Every value computation and
side effect associated with the initialization of a parameter, and the
initialization itself, is sequenced before every value computation and
side effect associated with the initialization of any subsequent
parameter.

However, one of these new sentences are missing from the GitHub draft:

Every value computation and side effect associated with the
initialization of a parameter, and the initialization itself, is
sequenced before every value computation and side effect associated
with the initialization of any subsequent parameter.

The example is there. It solves a decades-old problems (as explained by Herb Sutter) with exception safety where things like

f(std::unique_ptr<A> a, std::unique_ptr<B> b);

f(get_raw_a(), get_raw_a());

would leak if one of the calls get_raw_a() would throw before the other
raw pointer was tied to its smart pointer parameter.

As pointed out by T.C., the example is flawed since unique_ptr construction from raw pointer is explicit, preventing this from compiling.*

Also note this classical question (tagged C, not C++):

int x=0;
x++ + ++x;

is still undefined.

order of function call inside cout

As stated here: https://en.cppreference.com/w/cpp/language/eval_order

Order of evaluation of any part of any expression, including order of
evaluation of function arguments is unspecified (with some exceptions
listed below). The compiler can evaluate operands and other
subexpressions in any order, and may choose another order when the
same expression is evaluated again.

There is no concept of left-to-right or right-to-left evaluation in
C++. This is not to be confused with left-to-right and right-to-left
associativity of operators: the expression a() + b() + c() is parsed
as (a() + b()) + c() due to left-to-right associativity of operator+,
but the function call to c may be evaluated first, last, or between
a() or b() at run time

In your line

cout << sum(3) << sum(2)

the order of the two operator<< calls depends on the operator you use (here << so left-to-right), but the evaluation of each subexpression, namely sum(3) and sum(2) has no defined order and depends on the mood (most optimized compile approach usually) of your compiler.

For info here is a list of operators associativity: https://en.cppreference.com/w/cpp/language/operator_precedence

Problem with order of evaluation in parameter pack expansion

For example, we can use an intermediate std::tuple to force the order of evaluation:

template<typename TR, typename ... TArgs>
TR Bar(std::istream &s, TR f(TArgs...) )
{
std::tuple<TArgs...> args{Parse<TArgs>(s)...};
return std::apply(f, std::move(args));
}

In contrast to function arguments, the order of evaluation of arguments in a braced list is fixed by their order in that list, [dcl.init.list/4]:

Within the initializer-list of a braced-init-list, the initializer-clauses, including any that result from pack expansions, are evaluated in the order in which they appear. That is, every value computation and side effect associated with a given initializer-clause is sequenced before every value computation and side effect associated with any initializer-clause that follows it in the comma-separated list of the initializer-list. [ Note: This evaluation ordering holds regardless of the semantics of the initialization; for example, it applies when the elements of the initializer-list are interpreted as arguments of a constructor call, even though ordinarily there are no sequencing constraints on the arguments of a call. ]

How not specify an exact order of evaluation of function argument helps C & C++ compiler to generate optimized code?

I think the whole premise of the question is wrong:

How not specify an exact order of evaluation of function argument helps C & C++ compiler to generate optimized code?

It is not about optimizing code (though it does allow that). It is about not penalizing compilers because the underlying hardware has certain ABI constraints.

Some systems depend on parameters being pushed to stack in reverse order while others depend on forward order. C++ runs on all kinds of systems with all kinds on constraints. If you enforce an order at the language level you will require some systems to pay a penalty to enforce that order.

The first rule of C++ is "If you don't use it then you should not have to pay for it". So enforcing an order would be a violation of the prime directive of C++.

C++ output order [duplicate]

h.extract_min() obviously changes the variable h. So, here you have multiple changes of the same variable within one statement. Unfortunately, C++ standard does not specify the execution order in the statements, so, different calls to h.extract_min() are not necessarily executed in left-to-right manner (in fact, you have right-to-left order, but it's just luck).

To have correct output and readable code, consider using loops:

for (int i = 0; i < 4; ++i) {
cout << h.extract_min() << ' ';
}

Order of Evaluation for passing function arguments - Order of operations for F1( int F2( int& x ), int x )

The evaluation order of function arguments is unspecified. When you write:

TakeValues(modifyRef(Q), Q);

you are relying upon the fact that modifyRef(Q) to be evaluated before Q. But the evaluation order of function arguments is unspecified - it is not necessarily the case that modifyRef(Q) will be sequenced before Q nor vice versa.

In this case, Q (the second argument) gets evaluated first. So we read 9, and initialize the parameter Y with it. Then we evaluate modifyRef(Q), which zeros out Q and returns it, which leads to initializing X with 0.



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