Program without semicolon compiles fine in C not in C++ why? - c++

I am using Orwell Dev C++ IDE. Recently I tested following simple program in which I forgot to put semicolon (;) but still it compiles fine in C but not in C++. Why? What is the reason?
// C program compiles & runs fine, even ; missing at end of struct
#include <stdio.h>
struct test
{ int a,b}; // missing semicolon
int main()
{
struct test d={3,6};
printf("%d",d.a);
return 0;
}
[Warning] no semicolon at end of struct or union [enabled by default]
// Following is compilation error in C++
#include <stdio.h>
struct test
{ int a,b}; // missing semicolon
int main()
{
struct test d={3,6};
printf("%d",d.a);
return 0;
}
[Error] expected ';' at end of member declaration
I also tried same C program in codeblocks 13.12 IDE but it shows following error message
error: no semicolon at end of struct or union.
Why different error messages given by different implementations?

The semicolon is required by both languages. Specifically, C specifies the declaration of one or more structure members as
struct-declaration:
specifier-qualifier-list struct-declarator-list ;
and C++ specifies the declaration of one or more class member variables as
member-declaration:
attribute-specifier-seq<opt> decl-specifier-seq<opt> member-declarator-list<opt> ;
both of which require a semicolon at the end.
You'll have to ask the compiler writers why their C++ compiler is more strict than their C compiler. Note that the language specifications only require a "diagnostic" if a program is ill-formed, so it's legitimate either to issue a warning and continue compiling as if the semicolon were present, or to issue an error and stop.
It looks like your IDE is using GCC as its compiler; in which case you could use -Werror to convert warnings into errors, if you'd prefer stricter diagnostics.

Related

List initialization returns semicolon error

I'm trying to compile the following C++ code on Visual Studio Code, using the Mac clang compiler.
#include <iostream>
int main() {
int x { 5 };
std::cout << x;
return 0;
}
However, this returns an error, on the line of the list initialization: int x{ 5 };. Specifically, it says I need to insert a semicolon after the x.
I don't get what's wrong with this code, it works fine on an online compiler. How do I fix this?
Running man clang in the Terminal and skimming through, I found this:
The default C++ language standard is gnu++14.
UPDATE: I ran clang++ main.cpp in the compiler and it returned that semicolon error. This isn't a problem with VSCode, so I'll remove that tag.
Here's the error:
main.cpp:3:10: error: expected ';' at end of declaration
int x { 5 };
^
;
1 error generated.

Permitted usage context of declarations

The following code
#include <iostream>
#include <memory>
#include <ios>
using std::cout;
using std::endl;
using std::unique_ptr;
using std::make_unique;
using std::boolalpha;
template<typename T>
struct alloc{
alloc();
unique_ptr<T> operator() (void){
return(auto up = make_unique<T>(NULL));
}
};
int main (void){
auto up = alloc<int>()();
cout << boolalpha << ((up) ? 1 : 0) << endl;
return 0;
}
when compiled gives the following error:
g++ -ggdb -std=c++17 -Wall -Werror=pedantic -Wextra -c code.cpp
code.cpp: In member function ‘std::unique_ptr<_Tp> alloc<T>::operator()()’:
code.cpp:14:16: error: expected primary-expression before ‘auto’
return(auto up = make_unique<T>(NULL));
^~~~
code.cpp:14:16: error: expected ‘)’ before ‘auto’
make: *** [makefile:20: code.o] Error 1
There is an earlier question on SO reporting the same error:
C++17 std::optional error: expected primary-expression before 'auto'
The following is a snippet from the accepted answer to the above question:
Declarations are not expressions. There are places where expressions
are allowed, but declararions are not.
So my questions based on the compilation error I get are:
a) Is the use of a declaration in a return statement not permitted by the standard?
b) What are the permitted contexts for declarations?
Note: I had deliberately used the auto keyword in the return statement to reproduce this error. This error had originally appeared in a larger code base.
TIA
Is the use of a declaration in a return statement not permitted by the standard?
Indeed it isn't. We need only examine the grammar production at [stmt.jump]/1
Jump statements unconditionally transfer control.
jump-statement:
break ;
continue ;
return expr-or-braced-init-listopt ;
goto identifier ;
There is no production that turns an "expr-or-braced-init-list" into any sort of statement, so no declaration statement either. There is also no production that turns it into any other sort of declaration (such as a function, namespace or class). So you cannot declare anything in the return statement's operand.
What are the permitted contexts for declarations?
Almost anywhere an expression isn't required explicitly. The very definition of a translation unit in C++ (one file being translated) is a sequence of declarations per [basic.link]/1.
A program consists of one or more translation units linked together. A
translation unit consists of a sequence of declarations.
translation-unit:
declaration-seqopt
Different declarations have different structure. Somes such as namespaces, may contain more declarations. Others such as functions may contain statements, which themselves may be declaration statements of certain things. But most importantly, the standard makes clear where a statement may appear, and where only an expression is permitted.

Clash with identifier 'clock' from time.h

The following program
#include <ctime>
struct clock {};
int main() {
clock c;
}
fails to compile on both g++ 5.4 and clang 3.8 (Ubuntu 64-bit).
g++ output
clock.cpp: In function ‘int main()’:
clock.cpp:6:11: error: expected ‘;’ before ‘c’
clock c;
^
clang output
clock.cpp:6:5: error: must use 'struct' tag to refer to type 'clock' in this scope
clock c;
^
struct
/usr/include/time.h:189:16: note: struct 'clock' is hidden by a non-type declaration of 'clock' here
extern clock_t clock (void) __THROW;
^
1 error generated.
The diagnostics vary a little in form, but are related to the same issue. There is a clash with the standard C function clock and the struct of the same name defined in the program. The relevant declaration from time.h:
extern clock_t clock (void) __THROW;
The question is: shouldn't such symbols be in the std namespace, since the program includes <ctime>? Interestingly, that very declaration sits a couple of lines after a macro which reads __BEGIN_NAMESPACE_STD. In addition, in <ctime>, one can see:
namespace std
{
using ::clock_t;
using ::time_t;
using ::tm;
using ::clock;
...
}
Is there some kind of bug here?
Thank you.
The question is: shouldn't such symbols be in the std namespace...
Yes, and they are. Unfortunately, the C++ standard also allows implementations to put names from C-library derived headers in the global namespace. In this case, you get std::clock and ::clock.
This applies to all the <c*> C++ headers with a corresponding <*.h> version in C.

Code compiles in ideone but not with gcc

I wrote the following code:
#include <iostream>
using namespace std;
int main()
{
int v()
return 0;
}
I ran it in ideone, and it compiled successfully. I have the same code in file test1.cpp on my computer, I ran g++ test1.cpp and I got the following error:
./test1.cpp: In function ‘int main()’:
./test1.cpp:7:2: error: a function-definition is not allowed here before ‘return’
Why dose this happen? is this a bug?
I'm using linux mint, gcc version 4.7.
You are missing a semi-colon here:
int v()
^
should be:
int v() ;
which is a function declaration, not clear that was what was intended though. If you want to initialize v then the following would work:
int v(0) ;
or in C++11:
int v{0} ;
This is commonly known as C++'s most vexing parse. When you do something like
int f();
the compiler reads this as a function prototype, declaring a function f that returns an int. If you're using C++11, you should instead do
int f{}; // f initialized to 0
if you're not using C++11, make sure to initialize the variable right away.
You forgot the semicolon after
int v();
Ideone is using gcc 4.8.1 for your code (as you can see in your own link) while you are using 4.7
There are several difference regarding C++ 11 implementation, and apparently it is affected by the line that looks like a function delcaration.

Overkilling "crosses initialization of variable" error in C++?

I noticed that g++ complain a bit too strictly about crossed initialization and I wonder why these false-positive errors could not be removed just by looking at the SSA form of the program while compiling.
Let me give a very simple example:
#include <cstdlib>
int main ()
{
goto end;
int i = 0; // unused variable declaration
end:
return EXIT_SUCCESS;
}
When compiled with g++ -Wall -Wextra -o example1 example1.cc (g++ 4.8.1), the compiler gives the following error message:
example1.cc: In function ‘int main()’:
example1.cc:10:2: error: jump to label ‘end’ [-fpermissive]
end:
^
example1.cc:6:8: error: from here [-fpermissive]
goto end;
^
example1.cc:8:7: error: crosses initialization of ‘int i’
int i = 0;
^
example1.cc:8:7: warning: unused variable ‘i’ [-Wunused-variable]
So, it will raise an error where there is actually no risk because the variable is unused (the compiler obviously has both information and cannot combine it to deduce that the error is a false-positive).
More strange, I hoped that LLVM was more efficient at analyzing a program. So, I tried clang++ (LLVM) on this simple example with clang++ -Wall -Wextra -o example1 example1.cc (clang++ 3.4). And, I got about the same error message:
example1.cc:8:7: warning: unused variable 'i' [-Wunused-variable]
int i = 0;
^
example1.cc:6:3: error: goto into protected scope
goto end;
^
example1.cc:8:7: note: jump bypasses variable initialization
int i = 0;
^
1 warning and 1 error generated.
So, I am pretty sure that I am missing something important here, a problem that make the detection of this false-positive harder than I though. But, I do not know what is it. Or maybe, the C++ specification specifically says that it must be like this.
If somebody has an idea, feel free to share !
Edit: I also compiled the exact same code in C (gcc or clang), and it went fine just with the warning about i being an unused variable. So, it reinforce the fact that this is more likely linked to the specification of C++ and not a problem detecting this issue at compile time.
There is nothing wrong with the compilers. Your code is ill-formed according to the standard.
In your particular case, the requirement of the standard may not be necessary and the jump could be allowed and the compiler could create valid code. However, this is only because the initialisation of the variable int i has no side effects.
You can make your code valid by simply enclosing the jumped section in its own scope:
#include <cstdlib>
int main ()
{
goto end;
{
int i = 0; // unused variable declaration
}
end:
// cannot use i here, as it's not defined.
return EXIT_SUCCESS;
}
This is disallowed because potentially you'd call destructors for objects that aren't properly constructed. Admittedly, int doesn't have constructor or destructor, but it's making it "fair" for all types of objects. And technically, something at label end: could be using i, and by making the rule strict, it prevents the machine having to check every single code-path (which becomes a "halting problem").