Basically, I am reading through this book right here and in Section 1.6: Some Differences between C and C++ it is stated:
Another subtle difference between C and C++ is that in a C++ program,
all functions must be prototyped.
I am sure that this is not true from all the C++ programs that I have written. Is this only true for some versions of C++? Is it also true for C?
It has been true of C++ since the beginning (although in C++ it's just called a "declaration", not a "prototype").
As C existed decades ago, it allowed you to call a function without declaring it. You could, however, declare a function if you wanted to--usually to tell the compiler that it had a return type different from what the compiler would deduce on its own. So, in C a function declaration looks something like this:
long f();
Note that empty parens there. That's what separates a function "declaration" from a function "prototype" (though a prototype is basically a superset of a declaration, so a prototype also declares the function in question). A prototype always has something inside the parens to indicate the number and type of parameters the function accepts, on this general order:
short g(int a, double b);
If it doesn't accept any parameters, you have to put in void to indicate that:
int h(void);
If you leave the parens empty, that (as noted above) means it's a function declaration instead of a prototype--and that means you're telling the compiler the function's return type, but you're not telling it anything about the number or type of parameters.
C++ (since before it was called C++, if I recall correctly) has only had one concept instead of the two in C. In C++ every function must be declared--and a declaration always includes the number of parameters, and the type of each. This is absolutely necessary to support (for one obvious example) function overloading, where the correct function to call is determined from the number and types of arguments you pass in the call.
A function definition in C++ also acts as a function declaration. Every function must be declared, but the declaration doesn't have to be separate from the definition.
In reasonably modern C, you normally get pretty much the same--that is, a "new" (i.e., not ancient) type function definition also acts as a prototype for that function. As C was originally defined, it included a syntax for a function definition that looked like this:
int f(a, b, c)
int a;
short b;
long c;
{
// function body here
}
This defines the function, but the compiler treats it only as a function declaration, not a prototype--that is, it tells the compiler the return type, but the number and types of parameters (even though they're specified) are not used by the compiler in the same way they would be with a function prototype. C++ has never used or supported this style of function definition though.
Related
I have couple questions regarding some C++ rules.
Why am I able to call a function/method from outside the class in the namespace when I include the return type? (look at the namespace test2::testclass2 in the code below) i.e. this works:
bool b = testclass1::foo<int>(2);
whereas this doesn't: - (it doesn't even compile - compiler throws that this is function redeclaration)
testclass1::foo<int>(2);
C++ complains that it is a function redeclaration. Is that so?
This line:
bool b = testclass1::foo<int>(2);
gets called first before anything else. Is this because static methods get created always first before anything else in C++?
Where can I find those rules? I have a few C++ books at home, so if someone would be kind enough to either point out a book (and chapter or page) or direct me to a website I would greatly appreciate it.
Here below is the sample (partial) code that I tested at home with Visual Studio 2008:
class testclass1
{
public:
testclass1(void);
~testclass1(void);
template<class A> static bool foo(int i)
{
std::cout <<"in static foo";
return true;
}
};
namespace test2
{
class testclass2
{
public:
testclass2(void);
~testclass2(void);
};
bool b = testclass1::foo<int>(2);
}
EDIT:
A few people mentioned that I need to call this inside the function and this will work without any problem.
I understand that; the only reason I asked this question is because I saw this code somewhere (in someone's elses project) and was wondering how and why this works. Since I never really seen anyone doing it before.
Also, this is used (in multiple places) as a way to call and instantiate a large number of classes like this via those function calls (that are outside). They get called first before anything else is instantiated.
C++ is not Python. You write statements in functions and execution starts from the main method. The reason bool b = ... happens to work is that it's defining the global variable b and the function call is merely the initialization expression.
Definitions can exist outside functions while other statements can only exist inside a function body.
Why am I able to call a function/method from outside the class in the namespace when I include the return type? (look at the namespace test2::testclass2)
Your declaration of b is not inside a function, so you are declaring a global variable. If you were inside a function's scope, your second statement would work, but outside a function it makes no sense.
This also answers your second question.
Of course, you wouldn't be allowed to call it this way (i.e. not as a method of an object) if it weren't a static member function.
You can find the rules on e.g. Koenig lookup and template in the standard documentation -- good luck with navigating that! You're not mentioning which compiler you are testing, but I'm not entirely sure it's compliant!
As Mehrdad points out, you're declaring and initializing a global variable within the test2 namespace: this has nothing to do with static methods.
if you write this inside a function like below then it works without a problem. As mentioned above, you need to call these functions from within a function unless you are using the function to initialize a global variable ...
int main()
{
testclass1::foo<int>(2);
return 0;
}
1. First, a helpful correction: you said "...when I include the return type". I think you might be misunderstanding what the <int> part of testclass1::foo<int>(2) does. It doesn't (necessarily) specify the return type, it just provides a value for the template argument "A".
You could have chosen to use A as the return type, but you have the return type hard-coded to "bool".
Basically, for the function as you have written it you will always need to have the <> on it in order to call it. C++ does allow you to omit the <args> off the function when the type can be deduced from the function arguments; in order to get it to do that you have to use the type argument A in your function arguments. For instance if you declared the function this way instead then you could call it without the <>:
template<class A> static bool foo(A i);
In which case it you could call "foo(2)" and it would deduce A to be "int" from the number two.
On the other hand there isn't any way to make it deduce anything based on what you assign the function to. For template argument deduction it only looks at the arguments to the function, not what is done with the result of calling the function. So in:
bool b = testclass1::foo(2);
There is no way to get it to deduce "bool" from that, not even if you made A the return type.
So why doesn't the compiler just tell you "you needed to use <> on the function"? Even though you declared foo once as a template function, you could have also overloaded it with a non-template version too. So the compiler doesn't just automatically assume that you're trying to call a template function when you leave the <> off the call. Unfortunately having NOT assumed you were calling template-foo and not seeing any other declaration for foo, the compiler then falls back on an old C style rule where for a function that takes an int and returns an int, in a very old dialect of C you didn't need to declare that kind of before using it. So the compiler assumed THAT was what you wanted - but then it notices that template-foo and old-crufty-C-foo both take an int parameter, and realizes it wouldn't be able to tell the difference between them. So then it says you can't declare foo. This is why C++ compilers are notorious for giving bad error messages - by the time the error is reported the compiler may have gone completely off the rails and be talking about something that is three or four levels removed from your actual code!
2. Yes you're exactly right.
3. I find that the C++ references and whitepapers that IBM makes available online are the most informative. Here's a link to the section about templates: C++ Templates
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Possible Duplicate:
C++ Why put void in params?
What's the difference between these two declarations and which is used more commonly?
void function1();
and
void function2( void );
There is no difference in C++, where it is well defined that it represents 0 parameters.
However it does make one in C. A function with (void) means with no parameter, whereas () means with any number of parameters.
From http://publib.boulder.ibm.com/infocenter/comphelp/v8v101/index.jsp?topic=%2Fcom.ibm.xlcpp8a.doc%2Flanguage%2Fref%2Fparam_decl.htm
An empty argument list in a function definition indicates that a function
that takes no arguments. An empty argument list in a function declaration
indicates that a function may take any number or type of arguments. Thus,
int f()
{
...
}
indicates that function f takes no arguments. However,
int f();
simply indicates that the number and type of parameters is not known.
To explicitly indicate that a function does not take any arguments, you should
define the function with the keyword void.
There is no difference in C++.
The second declaration just explicitly says that function takes no parameter.
Second is more commonly used in C, First is the one that is more common in C++.
There is a difference in case of C because:
With (void), you're specifying that the function has no parameters, while with () you are specifying that the parameters are unspecified(unknown number of arguments).
However, if it was not a function declaration but a function definition, then even in C it is same as (void).
There is no difference. I'd say the first one is more common, clear and concise.
In C++, there is no difference, and the second form is only retained for C compatibility. The first form is preferred in C++.
In C, they mean different things.
The first form specifies a function which takes an unknown number of arguments, and the second is a function taking zero arguments.
Some very old (non-standard) C compiler might complain about the first one, so the second one should be more portable.
Apart from that, there is no difference.
The first one is used more commonly in user code, quite simply because it's shorter.
actually there is no difference .if you have not any parameters to pass to the method user void or empty parentheses .notice that it just fro passed parameters.if your method has not any returned value you have to use void keyword.the first one is more common in C#
So one of my previous exams had this question, and till now I've been reading that you don't need a declaration in any of the languages?
Which is right? Will C++ give an error if there's no declaration, or will it run?
In a discussion that involves both C and C++ "function declaration" is a rather vague term. These languages are significantly different in this regard.
In C++ language there's only one kind of function declaration: declaration with all parameter types and return type. Such declarations are necessary because C++ language supports function overloading. In order to choose which function to call the compiler needs to know everything about the function and needs to know which overloaded versions of the function are available. If you "forget" to declare some overloaded version, it will not be considered by overload resolution. That is at least one of the reasons function declarations are necessary in C++.
In C language there are two kinds of function declarations: non-prototype declarations and prototype declarations (or simply prototypes). A prototype in C is pretty similar to C++ declaration - it includes all parameter types. Prototypes have always been required in standard C for variadic functions (functions with ... parameters). For non-variadic functions prototype declarations are not required even today. But starting from C99 at least non-prototype declarations are required for all other functions. In older C89/90 version of the language function declarations for non-variadic functions were not required.
So, that should basically answer your question. In C++ function declarations are required because language features rely on them critically. In modern C function declarations are also required just to make the code safer. In older versions of C function declarations were not required mostly simply because the language was defined to work without them.
Function declarations in C are not mandatory for legacy / backwards compatability reasons - if they were made mandatory then some old / legacy code somewhere would stop compiling.
I'd guess that they are mandatory in C++ becasuse C++ isn't a strict superset of C and so can make the sensible choice of making them mandatory.
You should always declare them however - see this question Must declare function prototype in C?
FYI in C99 function declarations are now mandatory.
Function declarations are mandatory in C. Prototypes, however, are optional, except in the cases of variadic functions and functions whose argument types would be altered by default promotions.
I can't answer for C.
In C++, for functions:
void foo(); is a declaration.
void foo() { .. } is a definition and a declaration.
You need at least one declaration in scope before you call foo().
Much the same is true for data variables.
extern "C" int func(int *, Foo);
This is from a sample code from class. But I don't understand how the interface is working, one with no variable name, the other with no type. How is that going to work?
When declaring functions you don't need to specify a parameter name, just a type. Foo in this case is a type.
extern "C" tells the compiler it should use a C-style symbol, which more or less means it won't be using name mangling (which C++ uses to allow multiple functions share a name, but use different parameter sets or namespaces).
one with no variable name, the other with no type. How is that going to work?
In the function declaration (and even in definition), variable names are optional, And in your case, Foo is a type, it's not a variable name!
The following program is completely valid even though function f mentions no parameter names!
int f(int)
{
cout << "f(int) is called";
}
int main()
{
f(100);
}
This is a function declaration. You don't need to have a variable name.
The 2nd does have a type, it's Foo.
This is just a prototype. That is to say, it's what's needed to call the function, but not the code that says what the function actually does.
All the compiler needs to know to generate the calling code is the types of the arguments of the function, the function name, and the return type. It doesn't need to know the names of the arguments.
The second argument is a Foo. That's not the name, that's the type.
By using extern "C" you can link a C++ program to C functions.
In your example above, it will turn off name mangling for func so that you can link to code compiled by a C compiler.
C++ compilers need name mangling to differentiate between different functions with the same name. Name mangling is the rule according to which C++ changes function names into function signatures before invoking the linker.
Your assumption is not correct: both parameters have their type specified, and none of them has the name specified. In this case Foo is a type (a struct?) already defined somewhere.
That the parameters have no names is not a problem because this is the declaration of a function: it only serves to let the compiler know the signature (number and types of parameters, plus return type) of the function. It doesn't matter how the formal parameters are named (or if they are named at all). That information is only useful if you are about to write the function body.
I have couple questions regarding some C++ rules.
Why am I able to call a function/method from outside the class in the namespace when I include the return type? (look at the namespace test2::testclass2 in the code below) i.e. this works:
bool b = testclass1::foo<int>(2);
whereas this doesn't: - (it doesn't even compile - compiler throws that this is function redeclaration)
testclass1::foo<int>(2);
C++ complains that it is a function redeclaration. Is that so?
This line:
bool b = testclass1::foo<int>(2);
gets called first before anything else. Is this because static methods get created always first before anything else in C++?
Where can I find those rules? I have a few C++ books at home, so if someone would be kind enough to either point out a book (and chapter or page) or direct me to a website I would greatly appreciate it.
Here below is the sample (partial) code that I tested at home with Visual Studio 2008:
class testclass1
{
public:
testclass1(void);
~testclass1(void);
template<class A> static bool foo(int i)
{
std::cout <<"in static foo";
return true;
}
};
namespace test2
{
class testclass2
{
public:
testclass2(void);
~testclass2(void);
};
bool b = testclass1::foo<int>(2);
}
EDIT:
A few people mentioned that I need to call this inside the function and this will work without any problem.
I understand that; the only reason I asked this question is because I saw this code somewhere (in someone's elses project) and was wondering how and why this works. Since I never really seen anyone doing it before.
Also, this is used (in multiple places) as a way to call and instantiate a large number of classes like this via those function calls (that are outside). They get called first before anything else is instantiated.
C++ is not Python. You write statements in functions and execution starts from the main method. The reason bool b = ... happens to work is that it's defining the global variable b and the function call is merely the initialization expression.
Definitions can exist outside functions while other statements can only exist inside a function body.
Why am I able to call a function/method from outside the class in the namespace when I include the return type? (look at the namespace test2::testclass2)
Your declaration of b is not inside a function, so you are declaring a global variable. If you were inside a function's scope, your second statement would work, but outside a function it makes no sense.
This also answers your second question.
Of course, you wouldn't be allowed to call it this way (i.e. not as a method of an object) if it weren't a static member function.
You can find the rules on e.g. Koenig lookup and template in the standard documentation -- good luck with navigating that! You're not mentioning which compiler you are testing, but I'm not entirely sure it's compliant!
As Mehrdad points out, you're declaring and initializing a global variable within the test2 namespace: this has nothing to do with static methods.
if you write this inside a function like below then it works without a problem. As mentioned above, you need to call these functions from within a function unless you are using the function to initialize a global variable ...
int main()
{
testclass1::foo<int>(2);
return 0;
}
1. First, a helpful correction: you said "...when I include the return type". I think you might be misunderstanding what the <int> part of testclass1::foo<int>(2) does. It doesn't (necessarily) specify the return type, it just provides a value for the template argument "A".
You could have chosen to use A as the return type, but you have the return type hard-coded to "bool".
Basically, for the function as you have written it you will always need to have the <> on it in order to call it. C++ does allow you to omit the <args> off the function when the type can be deduced from the function arguments; in order to get it to do that you have to use the type argument A in your function arguments. For instance if you declared the function this way instead then you could call it without the <>:
template<class A> static bool foo(A i);
In which case it you could call "foo(2)" and it would deduce A to be "int" from the number two.
On the other hand there isn't any way to make it deduce anything based on what you assign the function to. For template argument deduction it only looks at the arguments to the function, not what is done with the result of calling the function. So in:
bool b = testclass1::foo(2);
There is no way to get it to deduce "bool" from that, not even if you made A the return type.
So why doesn't the compiler just tell you "you needed to use <> on the function"? Even though you declared foo once as a template function, you could have also overloaded it with a non-template version too. So the compiler doesn't just automatically assume that you're trying to call a template function when you leave the <> off the call. Unfortunately having NOT assumed you were calling template-foo and not seeing any other declaration for foo, the compiler then falls back on an old C style rule where for a function that takes an int and returns an int, in a very old dialect of C you didn't need to declare that kind of before using it. So the compiler assumed THAT was what you wanted - but then it notices that template-foo and old-crufty-C-foo both take an int parameter, and realizes it wouldn't be able to tell the difference between them. So then it says you can't declare foo. This is why C++ compilers are notorious for giving bad error messages - by the time the error is reported the compiler may have gone completely off the rails and be talking about something that is three or four levels removed from your actual code!
2. Yes you're exactly right.
3. I find that the C++ references and whitepapers that IBM makes available online are the most informative. Here's a link to the section about templates: C++ Templates