What might be the intention of printing a stream to itself? - c++

I am attempting to build a third-party C++ code base in which there are several places where an attempt is made to print an output stream to itself. This example demonstrates an expression of the same form:
#include <iostream>
int main(void) {
std::cout << std::cout << "Say what?" << std::endl;
return 0;
}
I find that without any options, g++ 4.8.5 accepts that code with zero complaints, but g++ 8.2.1 rejects it with a good old "no match for ‘operator<<’" error.
I understand the error. What I want to know is whether there is any reason to think that it was ever anything other than an error -- was there a C++ version or a widely used C++ compiler that would do something useful with such code? Although it accepts the code, g++ 4.8.5 is not such a compiler in my book because the program it builds evaluates the first << operation simply by outputting a hexadecimal number of unclear significance.
I have considered that it might be a simple typo, maybe magnified by copy & paste. For example, perhaps the second std::cout was an accidental duplicate of the first, or perhaps it was meant to be std::endl, instead. However, in a different source file in the same code base I see the same idiom of an output stream being printed to itself applied to objects of type std::stringstream, which suggests to me that it might be intentional.
My overall objective is to decide how best to fix the code.

What might be the intention of printing a stream to itself?
It is probably unintentional, or the intention itself was a mistake.
The difference between the GCC versions is the default dialect of C++ that they use. The newer GCC defaults to c++11 or newer. The program should compile with the newer compiler as long as you use a pre-C++11 dialect.
The change in C++11 was the removal of implicit conversion from std::basic_ios to void*. The purpose of this conversion was to check for failure state of the stream: null signified a failed stream while non-null signified a valid stream which allows the pattern if(stream >> input). In C++11 the conversion was replaced with explicit conversion to bool. Since the new conversion is explicit, it won't be applied to inserting a stream into stream. This was a backwards incompatible change which was presumably considered to not be a problem since the now-incompatible ways of using the conversion (such as the example) would have had no practical uses. In fact, it is useful to get a compilation error when doing something that has no practical use.
My overall objective is to decide how best to fix the code.
Probably just remove the insertion of cout into cout. If you're concerned that the output must remain same because it might be parsed by another program, then you can output any non-zero hex number to keep the output same for the sake of compatibility.

Related

What does "-flax-vector-conversions" exactly mean for ARM compiler?

I am tring to write a xxx.toolchain.cmake from arm-linux-gnueabihf gcc/g++ compiler.
What confused me is, whether should I use -flax-vector-conversions compilation flag or not. I read the doc/man page of the compiler, and it tells:
-flax-vector-conversions
Allow implicit conversions between vectors with differing numbers of elements and/or incompatible element types. This option should not be used for new code.
(via https://gcc.gnu.org/onlinedocs/gcc/C-Dialect-Options.html)
My two confusions:
What does "vectors" mean in this explanation? Is there any example illustrate this?
What does "new code" mean? Why "new code" should not use with this compilation option?
GCC offers vector extensions that are meant to provide a way to access SIMD instructions in a machine-independent way (as opposed to intrinsics). This involves special vector types defined with __attribute__((vector_size(n))) to help the compiler understand the packed multiple-element data types that SIMD instructions use. Note that this has nothing to do with C++'s std::vector container.
Consider the following code:
typedef short eight_short __attribute__((vector_size(16)));
typedef int four_int __attribute__((vector_size(16)));
eight_short v1;
four_int v2;
void foo(void) {
v2 = v1;
}
Here four_int and eight_short are vectors of the corresponding number of elements and types. They are both 16 bytes and thus suitable to store in a 128-bit SIMD register. Assigning one to the other is clearly meant to "reinterpret" (aka bit-cast), but it also violates type safety. Presumably older versions of the compiler used to accept such code, and there may be code like this out there, but the compiler authors want to discourage it. So such code now causes an error by default, but they provide the option -flax-vector-conversions that you can use when compiling old code like this to suppress the error.
"New code" means code you are writing for the first time, where you have a choice as to how to write it. For such code you are most likely expected to use v2 = reinterpret_cast<four_int>(v1);, and not use -flax-vector-conversions. Then the compiler will flag any place where you forgot to cast (since it could be a bug where you actually meant something else).
If you're compiling legacy code, your best bet would be to first try building without this option. If it builds successfully, then the option is not needed, so don't use it. If it gets errors about conversions of vector types, you could consider using this option, or else rewrite the code with explicit casts where needed.

What makes this usage of pointers unpredictable?

I'm currently learning pointers and my professor provided this piece of code as an example:
//We cannot predict the behavior of this program!
#include <iostream>
using namespace std;
int main()
{
char * s = "My String";
char s2[] = {'a', 'b', 'c', '\0'};
cout << s2 << endl;
return 0;
}
He wrote in the comments that we can't predict the behavior of the program. What exactly makes it unpredictable though? I see nothing wrong with it.
The behaviour of the program is non-existent, because it is ill-formed.
char* s = "My String";
This is illegal. Prior to 2011, it had been deprecated for 12 years.
The correct line is:
const char* s = "My String";
Other than that, the program is fine. Your professor should drink less whiskey!
The answer is: it depends on what C++ standard you're compiling against. All the code is perfectly well-formed across all standards‡ with the exception of this line:
char * s = "My String";
Now, the string literal has type const char[10] and we're trying to initialize a non-const pointer to it. For all other types other than the char family of string literals, such an initialization was always illegal. For example:
const int arr[] = {1};
int *p = arr; // nope!
However, in pre-C++11, for string literals, there was an exception in §4.2/2:
A string literal (2.13.4) that is not a wide string literal can be converted to an rvalue of type “pointer to char”; [...]. In either case, the result is a pointer to the first element of the array. This conversion is considered only when there is an explicit appropriate pointer target type, and not when there is a general need to convert from an lvalue to an rvalue. [Note: this conversion is deprecated. See Annex D. ]
So in C++03, the code is perfectly fine (though deprecated), and has clear, predictable behavior.
In C++11, that block does not exist - there is no such exception for string literals converted to char*, and so the code is just as ill-formed as the int* example I just provided. The compiler is obligated to issue a diagnostic, and ideally in cases such as this that are clear violations of the C++ type system, we would expect a good compiler to not just be conforming in this regard (e.g. by issuing a warning) but to fail outright.
The code should ideally not compile - but does on both gcc and clang (I assume because there's probably lots of code out there that would be broken with little gain, despite this type system hole being deprecated for over a decade). The code is ill-formed, and thus it does not make sense to reason about what the behavior of the code might be. But considering this specific case and the history of it being previously allowed, I do not believe it to be an unreasonable stretch to interpret the resulting code as if it were an implicit const_cast, something like:
const int arr[] = {1};
int *p = const_cast<int*>(arr); // OK, technically
With that, the rest of the program is perfectly fine, as you never actually touch s again. Reading a created-const object via a non-const pointer is perfectly OK. Writing a created-const object via such a pointer is undefined behavior:
std::cout << *p; // fine, prints 1
*p = 5; // will compile, but undefined behavior, which
// certainly qualifies as "unpredictable"
As there is no modification via s anywhere in your code, the program is fine in C++03, should fail to compile in C++11 but does anyway - and given that the compilers allow it, there's still no undefined behavior in it†. With allowances that the compilers are still [incorrectly] interpreting the C++03 rules, I see nothing that would lead to "unpredictable" behavior. Write to s though, and all bets are off. In both C++03 and C++11.
†Though, again, by definition ill-formed code yields no expectation of reasonable behavior
‡Except not, see Matt McNabb's answer
Other answers have covered that this program is ill-formed in C++11 due to the assignment of a const char array to a char *.
However the program was ill-formed prior to C++11 also.
The operator<< overloads are in <ostream>. The requirement for iostream to include ostream was added in C++11.
Historically, most implementations had iostream include ostream anyway, perhaps for ease of implementation or perhaps in order to provide a better QoI.
But it would be conforming for iostream to only define the ostream class without defining the operator<< overloads.
The only slightly wrong thing that I see with this program is that you're not supposed to assign a string literal to a mutable char pointer, though this is often accepted as a compiler extension.
Otherwise, this program appears well-defined to me:
The rules that dictate how character arrays become character pointers when passed as parameters (such as with cout << s2) are well-defined.
The array is null-terminated, which is a condition for operator<< with a char* (or a const char*).
#include <iostream> includes <ostream>, which in turn defines operator<<(ostream&, const char*), so everything appears to be in place.
You can't predict the behaviour of the compiler, for reasons noted above. (It should fail to compile, but may not.)
If compilation succeeds, then the behaviour is well-defined. You certainly can predict the behaviour of the program.
If it fails to compile, there is no program. In a compiled language, the program is the executable, not the source code. If you don't have an executable, you don't have a program, and you can't talk about behaviour of something that doesn't exist.
So I'd say your prof's statement is wrong. You can't predict the behaviour of the compiler when faced with this code, but that's distinct from the behaviour of the program. So if he's going to pick nits, he'd better make sure he's right. Or, of course, you might have misquoted him and the mistake is in your translation of what he said.
As others have noted, the code is illegitimate under C++11, although it was valid under earlier versions. Consequently, a compiler for C++11 is required to issue at least one diagnostic, but behavior of the compiler or the remainder of the build system is unspecified beyond that. Nothing in the Standard would forbid a compiler from exiting abruptly in response to an error, leaving a partially-written object file which a linker might think was valid, yielding a broken executable.
Although a good compiler should always ensure before it exits that any object file it is expected to have produced will be either valid, non-existent, or recognizable as invalid, such issues fall outside the jurisdiction of the Standard. While there have historically been (and may still be) some platforms where a failed compilation can result in legitimate-appearing executable files that crash in arbitrary fashion when loaded (and I've had to work with systems where link errors often had such behavior), I would not say that the consequences of syntax errors are generally unpredictable. On a good system, an attempted build will generally either produce an executable with a compiler's best effort at code generation, or won't produce an executable at all. Some systems will leave behind the old executable after a failed build, since in some cases being able to run the last successful build may be useful, but that can also lead to confusion.
My personal preference would be for disk-based systems to to rename the output file, to allow for the rare occasions when that executable would be useful while avoiding the confusion that can result from mistakenly believing one is running new code, and for embedded-programming systems to allow a programmer to specify for each project a program that should be loaded if a valid executable is not available under the normal name [ideally something which which safely indicates the lack of a useable program]. An embedded-systems tool-set would generally have no way of knowing what such a program should do, but in many cases someone writing "real" code for a system will have access to some hardware-test code that could easily be adapted to the purpose. I don't know that I've seen the renaming behavior, however, and I know that I haven't seen the indicated programming behavior.

Why do streams still convert to pointers in C++11?

The canonical way to read lines from a text file is:
std::fstream fs("/tmp/myfile.txt");
std::string line;
while (std::getline(line, fs)) {
doThingsWith(line);
}
(no, it is not while (!fs.eof()) { getline(line, fs); doThingsWith(line); }!)
This works beacuse std::getline returns the stream argument by reference, and because:
in C++03, streams convert to void*, via an operator void*() const in std::basic_ios, evaluating to the null pointer value when the fail error flag is set;
see [C++03: 27.4.4] & [C++03: 27.4.4.3/1]
in C++11, streams convert to bool, via an explicit operator bool() const in std::basic_ios, evaluating to false when the fail error flag is set
see [C++11: 27.5.5.1] & [C++11: 27.5.5.4/1]
In C++03 this mechanism means the following is possible:
std::cout << std::cout;
It correctly results in some arbitrary pointer value being output to the standard out stream.
However, despite operator void*() const having been removed in C++11, this also compiles and runs for me in GCC 4.7.0 in C++11 mode.
How is this still possible in C++11? Is there some other mechanism at work that I'm unaware of? Or is it simply an implementation "oddity"?
I'm reasonably certain this is not allowed/can't happen in a conforming implementation of C++11.
The problem, of course, is that right now, most implementations are working on conforming, but aren't there completely yet. At a guess, for many vendors, this particular update is a fairly low priority. It improves error checking, but does little (or nothing) to enable new techniques, add new features, improve run-time efficiency, etc. This lets the compiler catch the error you've cited (some_stream << some_other_stream) but doesn't really make a whole lot of difference otherwise.
If I were in charge of updating a standard library for C++11, I think this would be a fairly low priority. There are other changes that are probably as easy (if not easier) to incorporate, and likely to make a much bigger difference to most programmers.
To use one of the examples you gave, if I were in charge of updating the VC++ standard library to take advantage of the compiler features added in the November CTP, my top priority would probably be to add constructors to the standard container types to accept initialization_lists. These are fairly easy to add (I'd guess one person could probably add and test them in under a week) and make quite an obvious, visible difference in what a programmer can do.
As late as GCC 4.6.2, the libstdc++ code for basic_ios is evidently still C++03-like.
I'd simply put this down to "they haven't gotten around to it yet".
By contrast, the libc++ (LLVM's stdlib implementation) trunk already uses operator bool().
This was a missed mini-feature buried in a pre-existing header. There are probably lots of missing error of omission and commission in pre-2011 components.
Really, if anyone comes up with things like this in gcc then it would do a world of good to go to Bugzilla and make a bug report. It may be a low priority bug but if you start a paper trail
I'll go out on a limb and extend this idea to all the other C++ compilers: clang, Visual Studio,etc.
This will make C++ a better place.
P.S. I entered a bug in Bugzilla.

Why we do not use Printf and Scanf in C++? [duplicate]

This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Which I/O library do you use in your C++ code?
in C++ why we go for COUT and CIN
Because classes can overload the insertion operator <<
They cannot edit printf and add %Foo to know how to print a Foo class.
You can use them.
But the major problem is that they are not safe.
A lot of C code had problems because the format specifiers did not match the arguments and resulted in crashes etc.
// Example
char c;
scanf("%s",&c); // quite an easy mistake. Usually leads to a crash or something worse.
The C++ variants are type safe (in that the compiler guarantees that the type you are inputting into is the correct type at compile time) and the runtime does the work of actually moving the data.
A secondary reason is that it is a lot easier to define output functions in C++ (operator >> and <<) than in C (There was never a standardized way of naming streaming functions).
The third advantage is that a stream is not limited to console or file. There are several other types of stream that automatically work with the operator >> and << thus allowing you to generalize the streaming (or serialization) of objects in different circumstances.
Along with the ability to overload insertion and extraction operators for classes of your choice, cin and cout are type-safe, which scanf and printf are not -- e.g., a mismatch between the conversion format and actual type with printf is fairly common:
printf("%d\n", 1.0);
printf("%f\n", 1);
Neither of these is at all likely to produce useful output.
The old stdio functionality is in C++ for compatibility (and partially because the original C++ (cfront) was just a pre-processor for C so they were available anyway).
The newer I/O model found in C++ is a lot more flexible, allowing you to add types at will, which can be printed in a way defined by that type.
In other words, in an object oriented way.
Just like complex or rational numbers numbers (or address book entries or dictionary entries or any other non-basic type for that matter) should be self contained in a class that knows how to add, subtract or otherwise manipulate them separately from the base language, the printing of those objects should also be left to the class as well.
If you're worried about the verbosity of the C++ way, there's nothing stopping you from adding a fmt() method, returning a string, to your own data types to get around that problem, something like:
std::cout << name.fmt('left',15) // [Pax Diablo ]
<< " " << age.fmt(4) // [ 45]
<< " " << score.fmt(3,2) // [100.00]
<< std::endl;
for sizes and/or justifications.
It's more coding on your part within the class but it is an option if you're really adverse to the setfill and setw stuff.
However, since you'll almost certainly have your formatting code in a function for each type of output (if you're smart), it'll be isolated in one area. I'd just suck it up and learn the C++ way of doing things.
The C++ syntax is supposed to be easier but, frankly, I never adopted it. Besides, I do Windows programming, not console, so neither cout or printf are of much use. But sprintf and it's variations are.

Creating a cout function in C?

I assume most C++ compilers are written in assembly. Which makes them different languages entirely (I could be wrong). That being said if I were going to create a cout style function for plain old C, how would I do it? cout has some very impressive features take this snippet for example:
// endl not only prints a new line but also flushes the stream
cout << "Hello World!" << endl;
Which I'm pretty sure translates to this in C:
printf("Hello World!\n");
fflush(1); //stdout = 1
Next order of business, the << operators. In C++ this would be easy (operator overloading), but I can't think of a single way to do this in C.
It may help in thinking about this to translate between the "<<" operator syntax and the "operator<<" function syntax. Your C++ example is equivalent to this bit of C++ code:
operator<< ( operator<< (cout, "Hello World!"), endl);
The first thing that you should notice here is that there is not actually a lot of cleverness in cout at all. What is clever is the operator<< function -- specifically, the version of the operator<< function that takes a stream object (which is what cout is, but many other things are too) as its first argument. Or, even more precisely, the range of operator<< functions that take a stream object as the first argument, and take a particular thing as the second argument -- there's one for each type of object that you can put into the cout stream. You can see one of the C++ tricks in that syntax, too; the operator<< functions on stream objects always return the stream object that they were given, thereby allowing chaining of this nature.
In order to put C++ code into linkers and system ABIs that expect C-like function syntax, most C++ compilers "mangle" the function names, in order to encode in them the type of arguments that they have. (Also, of course, the "<<" isn't a valid C-like function name.) So, if you looked at the generated assembly for this bit of function, you'd see that the names of the two functions were different from each other -- they'd have suffixes indicating the argument types. You could do something like that manually:
operator_lshift__stream__endl(
operator_lshift__stream__string(cout, "Hello World!"), endl);
And there you've got something that you could implement in C.
That's right, because C does not have operator overloading, you cannot change the behaviour of the << operator, it will always do a bit shift, so there is no way to write 'cout' with the precise semantics it has in C++, in C.
Out of interest, g++ (GNU C++ Compiler) is written mostly in C.
C is actually a popular implementation language for C++ compilers and standard libraries (so is C++ itself, actually -- a concept sometimes known as self-hosting or bootstrapping a language), and you can study the whole sources of a rich, complex C++ standard library (plus extensions) here (sorry, this is gcc 3 -- can't find a gcc 4 source tree that's just as easily browseable online, though of course you can easily download those sources and study them on your local machine).
Personally, I would suggest starting instead with a good book, such as this one -- the sources will be much more meaningful to you once you've got a good grasp of all the obscure nooks and crannies of C++'s iostreams (as a bonus, the book also takes you on a guided tour through locales -- hold on to your hat!-).