Skip to main content
What is wrong if I declare main something like this?

const int MYMAX_PARA=10;
int main(int argc, char *(*argv)[MYMAX_PARA], char *env[])
printf("%s %s",(*argv)[1],env[2]);

When I want to pass a double by reference I use 'pointer to double'.
When I want to pass a structure by reference I use 'pointer to structure'.
Then why not 'pointer to array' when passing array by reference?
Why is there and exception in case of arrays?
I try to think why array-to-pointer 'decay' occurs in C/C++,
I think of above example.

To me, the answer is about simplicity of coding. In above
example, argv++ won't give you next argument but rather next
array of arguments. To get next argument you need to do (*argv)[2]
(*argv)[3] and so on. So decay simplifies your coding and avoids
pointer to array syntax.

This rules out char *(*argv)[].

And BTW, third parameter in main is 'recommended' by standard.
Standard demands:
int main() { /* ... */ }
AND
int main(int argc, char* argv[]) { /* ... */ }
be allowed.

An excerpt from C++98 standards section 3.6.1 [basic.start.main]
"[Note: it is recommended that any further (optional) parameters be added after argv. ]"

It is upto you to discover what the third para means on your favorite platform.

Comments

Anonymous said…
Enchanting blog. Your site was off the chain and I
will return! I peep the web for blogs just like this
one.
golden slot mobile

Popular Content

Multi-dimensional arrays in C++11

What new can be said about multi-dimensional arrays in C++? As it turns out, quite a bit! With the advent of C++11, we get new standard library class std::array. We also get new language features, such as template aliases and variadic templates. So I'll talk about interesting ways in which they come together. It all started with a simple question of how to define a multi-dimensional std::array. It is a great example of deceptively simple things. Are the following the two arrays identical except that one is native and the other one is std::array? int native[3][4]; std::array<std::array<int, 3>, 4> arr; No! They are not. In fact, arr is more like an int[4][3]. Note the difference in the array subscripts. The native array is an array of 3 elements where every element is itself an array of 4 integers. 3 rows and 4 columns. If you want a std::array with the same layout, what you really need is: std::array<std::array<int, 4>, 3> arr; That's quite annoying for...

Want speed? Use constexpr meta-programming!

It's official: C++11 has two meta-programming languages embedded in it! One is based on templates and other one using constexpr . Templates have been extensively used for meta-programming in C++03. C++11 now gives you one more option of writing compile-time meta-programs using constexpr . The capabilities differ, however. The meta-programming language that uses templates was discovered accidently and since then countless techniques have been developed. It is a pure functional language which allows you to manipulate compile-time integral literals and types but not floating point literals. Most people find the syntax of template meta-programming quite abominable because meta-functions must be implemented as structures and nested typedefs. Compile-time performance is also a pain point for this language feature. The generalized constant expressions (constexpr for short) feature allows C++11 compiler to peek into the implementation of a function (even classes) and perform optimization...

Covariance and Contravariance in C++ Standard Library

Covariance and Contravariance are concepts that come up often as you go deeper into generic programming. While designing a language that supports parametric polymorphism (e.g., templates in C++, generics in Java, C#), the language designer has a choice between Invariance, Covariance, and Contravariance when dealing with generic types. C++'s choice is "invariance". Let's look at an example. struct Vehicle {}; struct Car : Vehicle {}; std::vector<Vehicle *> vehicles; std::vector<Car *> cars; vehicles = cars; // Does not compile The above program does not compile because C++ templates are invariant. Of course, each time a C++ template is instantiated, the compiler creates a brand new type that uniquely represents that instantiation. Any other type to the same template creates another unique type that has nothing to do with the earlier one. Any two unrelated user-defined types in C++ can't be assigned to each-other by default. You have to provide a...