Order of evaluation of arguments in function calling?
order of evaluation of function arguments in c
the order of evaluation can be changed by using in an expression
c++ order of evaluation if statement
c order of evaluation
operand evaluation order
c++ function argument order of evaluation
order of passing arguments in c
I am studying about undefined behavior in C and I came to a statement that states that
there is no particular order of evaluation of function arguments
but then what about the standard calling conventions like
_stdcall, whose definition said (in a book) that arguments are evaluated from right to left.
Now I am confused with these two definitions one, in accordance of UB, states different than the other which is in accordance of the definition of calling convention. Please justify the two.
As Graznarak's answer correctly points out, the order in which arguments are evaluated is distinct from the order in which arguments are passed.
An ABI typically applies only to the order in which arguments are passed, for example which registers are used and/or the order in which argument values are pushed onto the stack.
What the C standard says is that the order of evaluation is unspecified. For example (remembering that
printf returns an
the C standard says that the two messages will be printed in some order (evaluation is not interleaved), but explicitly does not say which order is chosen. It can even vary from one call to the next, without violating the C standard. It could even evaluate the first argument, then evaluate the second argument, then push the second argument's result onto the stack, then push the first argument's result onto the stack.
An ABI might specify which registers are used to pass the two arguments, or exactly where on the stack the values are pushed, which is entirely consistent with the requirements of the C standard.
But even if an ABI actually requires the evaluation to occur in a specified order (so that, for example, printing
"second\n" followed by
"first\n" would violate the ABI) that would still be consistent with the C standard.
What the C standard says is that the C standard itself does not define the order of evaluation. Some secondary standard is still free to do so.
Incidentally, this does not by itself involve undefined behavior. There are cases where the unspecified order of evaluation can lead to undefined behavior, for example:
printf("%d %d\n", i++, i++); /* undefined behavior! */
Order of evaluation - cppreference.com, Both compilers for the ARM7 apparently evaluate function call arguments from left to right, Function calls that depend on the order of argument evaluation are � In C++17, the order of evaluation of function arguments remains unspecified. However, each function argument is completely evaluated, and the calling object is guaranteed evaluated before any function arguments are.
_stdcall merely specify that the arguments are pushed onto the stack in right-to-left order, not that they are evaluated in that order. Think about what would happen if calling conventions like
pascal changed the order that the arguments were evaluated.
If evaluation order were modified by calling convention, you would have to know the calling convention of the function you're calling in order to understand how your own code would behave. That's a leaky abstraction if I've ever seen one. Somewhere, buried in a header file someone else wrote, would be a cryptic key to understanding just that one line of code; but you've got a few hundred thousand lines, and the behavior changes for each one? That would be insanity.
I feel like much of the undefined behavior in C89 arose from the fact that the standard was written after multiple conflicting implementations existed. They were maybe more concerned with agreeing on a sane baseline that most implementers could accept than they were with defining all behavior. I like to think that all undefined behavior in C is just a place where a group of smart and passionate people agreed to disagree, but I wasn't there.
I'm tempted now to fork a C compiler and make it evaluate function arguments as if they're a binary tree that I'm running a breadth-first traversal of. You can never have too much fun with undefined behavior!
Evaluating Function Arguments, In C++17, the order of evaluation of function arguments remains unspecified. However, each function argument is completely evaluated, and the calling object is� 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++.
Argument evaluation and argument passing are related but different problems.
Arguments tend to be passed left to right, often with some arguments passed in registers rather than on the stack. This is what is specified by the ABI and
The order of evaluation of arguments before placing them in the locations that the function call requires is unspecified. It can evaluate them left to right, right to left, or some other order. This is compiler dependent and may even vary depending on optimization level.
C++, It is compiler dependent in C. It is never safe to depend on the order of evaluation of side effects. For example, a function call like below may� There is no guarantee (in either the C or the C++ standard language definitions) that the increments will be evaluated in any particular order. Either increment might happen first. func might get the arguments `2, 3′, or it might get `3, 2′, or even `2, 2′.
Check the book you mentioned for any references to "Sequence points", because I think that's what you're trying to get at. Basically, a sequence point is a point that, once you've arrived there, you are certain that all preceding expressions have been fully evaluated, and its side-effects are sure to be no more.
For example, the end of an initializer is a sequence point. This means that after:
bool foo = !(i++ > j);
You are sure that
i will be equal to
i's initial value +1, and that
foo has been assigned
false. Another example:
int bar = i++ > j ? i : j;
Is perfectly predictable. It reads as follows: if the current value of
i is greater than
j, and add one to
i after this comparison (the question mark is a sequence point, so after the comparison,
i is incremented), then assign
i (NEW VALUE) to
bar, else assign
j. This is down to the fact that the question mark in the ternary operator is also a valid sequence point.
All sequence points listed in the C99 standard (Annex C) are:
The following are the sequence points described in 188.8.131.52: — The call to a function, after the arguments have been evaluated (184.108.40.206). — The end of the first operand of the following operators: logical AND && (6.5.13); logical OR || (6.5.14); conditional ? (6.5.15); comma , (6.5.17). — The end of a full declarator: declarators (6.7.5); — The end of a full expression: an initializer (6.7.8); the expression in an expression statement (6.8.3); the controlling expression of a selection statement (if or switch) (6.8.4); the controlling expression of a while or do statement (6.8.5); each of the expressions of a for statement (220.127.116.11); the expression in a return statement (18.104.22.168). — Immediately before a library function returns (7.1.4). — After the actions associated with each formatted input/output function conversion specifier (7.19.6, 7.24.2). — Immediately before and immediately after each call to a comparison function, and also between any call to a comparison function and any movement of the objects passed as arguments to that call (7.20.5).
What this means, in essence is that any expression that is not a followed by a sequence point can invoke undefined behaviour, like, for example:
printf("%d, %d and %d\n", i++, i++, i--);
In this statement, the sequence point that applies is "The call to a function, after the arguments have been evaluated". After the arguments are evaluated. If we then look at the semantics, in the same standard under 22.214.171.124, point ten, we see:
10 The order of evaluation of the function designator, the actual arguments, and subexpressions within the actual arguments is unspecified, but there is a sequence point before the actual call.
That means for
i = 1, the values that are passed to printf could be:
1, 2, 3//left to right
But equally valid would be:
1, 0, 1//evaluated i-- first //or 1, 2, 1//evaluated i-- second
What you can be sure of is that the new value of
i after this call will be 2.
But all of the values listed above are, theoretically, equally valid, and 100% standard compliant.
But the appendix on undefined behaviour explicitly lists this as being code that invokes undefined behaviour, too:
Between two sequence points, an object is modified more than once, or is modified and the prior value is read other than to determine the value to be stored (6.5).
In theory, your program could crash, instead of printinf
1, 2, and 3, the output
"666, 666 and 666" would be possible, too
Evaluation strategy, C has no evaluation order of function parameters. Given a function call expression [code ]f(a, b, c)[/code], the expressions [code ]f[/code] (that's� I have studied that the order of evaluation of arguments in a function call is completely compiler dependent,hence its upto the compiler whether it evaluates it from right to left,or from left to
so finally i found it...yeah. it is because the arguments are passed after they are evaluated.So passing arguments is a completely different story from the evaluation.Compiler of c as it is traditionally build to maximize the speed and optimization can evaluate the expression in any way. so the both argument passing and evaluation are different stories altogether.
What is evaluation order of function parameters in C?, In C++17, it is now specified that the expression giving what to call (the code on the left of the ( of the function call) goes before the arguments,� Since the order of evaluation for function arguments is unspecified, a compiler can evaluate the arguments to a call in any order, as long as it evaluates all the arguments before jumping to the function. In theory, a compiler could use even a random number generator to determine the order of evaluation for each call.
What is the evaluation order of function parameters in C?, Order of evaluation of the operands of any C++ operator, including the order of evaluation of function arguments in a function-call expression, and the order of� 10 There is a sequence point after the evaluations of the function designator and the actual arguments but before the actual call. Every evaluation in the calling function (including other function calls) that is not otherwise specifically sequenced before or after the execution of the body of the called function is indeterminately sequenced with respect to the execution of the called function.
Quick Q: Why doesn't c++ have a specified order for evaluating , The order of evaluation for function call arguments is not specified. In the following example: method(sample1, batch.process--, batch.process); the argument batch� We pass different arguments into some functions. Now one questions may come in our mind, that what the order of evaluation of the function parameters. Is it left to right or right to left? To check the evaluation order we will use a simple program. Here some parameters are passing. From the output, we can find how they are evaluated. Example
Order of evaluation, Order of evaluation of the operands of any C operator, including the order of evaluation of function arguments in a function-call expression, and the order of evaluation of the subexpressions within any expression is unspecified (except where noted below).
- It seems that you are doing your Ph.D on UB.
- Read the standard, throw out the book :-)
- @Marichyasana: I see no indication that the book is incorrect.
- Your implementation might make any guarantees it wants to with regard to UB, as the standard explicitly allows anything to happen. No contradiction found yet. Just be doubly sure the book has it right, because calling conventions only describe the interface between caller and callee, having no business dictating to either how it gets (or doesn't) the desired result. Still, even if the book should be right, I would strongly suggest you refrain from using any such guarantees, as they are non-portable, and quite likely not even followed by all compilers on the platform.
- 'The wonderful thing about standards is that there are so many of them to choose from'. Not an original saying, but apposite. It would help if you identified the references you're trying to understand. The
_stdcallconventions are not part of the C standard; they belong to Microsoft. It is probable that these define the sequence/order in which parameters are placed on the stack for a function call (and I believe they also relate to which code reinstates the stack pointer after a function call), but even that need not determine the order of evaluation.
- but how your last example is UB
- Multiple assignments to a single variable without a sequencing point between is undefined.
- @Bayant_singh, it is UB because you don't know the order of increments and the time of storing the incremented values. Assuming
i==0prior to the instruction all results (0,0), (0,1) and (1,0) are possible.
- @CiaPan: Since the behavior is undefined, all results (2,3), ("foo", "bar"), and "I'm sorry, Dave, I'm afraid I can't do that" are also possible.
- @VinodPatidar: As far as the ISO C standard is concerned, that's completely wrong. The standard doesn't even mention the word "stack", and does not specify the order of evaluation of arguments or operands apart from a few operators (
?:). Even for implementations that use a stack, there's no necessary relationship between the order in which arguments are evaluated and the order in which their values appear in memory. Optimizing compilers can and do violate your assumptions. Recommended reading: section 3 of the comp.lang.c FAQ.