What is difference between void main() and void main(void)

Please is there anyway to write a code to print the prime numbers from 1 to 1000?

Just count prime numbers in a loop which terminates when they reach 1000

You first need the algorithm to calculate the prime numbers before printing with for loop

That algorithm is very common, you can find it very easily trough google

Another maybe easier solution would be to use the "Sieve of Erastosthenes" https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

perhaps anyone would like to check https://github.com/luckyycode/neko-webframework/

i like it that way

Do u want c++ code

I didn't ask for even number. Prime numbers.

No

Guys i have learned the c language perfectly but i m new to c++. Which books should i use?

So you're saying ()! = (void) but you can pass arguments to () but you can't get them but you're not passing arguments to (void) and you can't get them, I get your point about main() not equal to void main(void) because you can return int but main(void) != main() doesn't make sense to me.

So you're saying ()! = (void) but you can pass arguments to () but you can't get them but you're not passing arguments to (void) and you can't get them, I get your point about main() not equal to void main(void) because you can return int but main(void) != main() doesn't make sense to me.

This applies for function declarations In C++ foo(); function foo with no parameters foo(void); function foo with no parameters In C foo(); function foo with an unspecified numbers of parameters foo(void); function foo with no paramters This applies for function definitions in C and C++ foo(){} and foo(void){} mean a function with no parameter So when defining the function int main() {} it should be equivalent to int main(void) {} - since it's a definition As example for the unspecified parameters, take the following C code extern void foo(); void qaz() {} void bar() { foo(1, 3); // Valid! foo declared with () qaz(1, 3);. //Invalid! qaz defined as () }

This, the invalid case, would result in a compile error

void can’t return any thing

It's different because compiler thinks so but functionality wise it's same.

A function with void result type ends either by reaching the end of the function or by executing a return statement with no returned value. The void type may also appear as the sole argument of a function prototype to indicate that the function takes no arguments. Note that despite the name, in all of these situations, the void type serves as a unit type, not as a zero or bottom type (which is sometimes confusingly called the "void type"), even though unlike a real unit type which is a singleton, the void type lacks a way to represent its value and the language does not provide any way to declare an object or represent a value with type void.

Are we talking about declarations or definitions?

And are we talking about the return type or the parameters?

But if main is not passed any arguments and doesn't have capability to accept any arguments, if you say void main(void) it's same as void main()

Yes when defining main they are equivalent

I already know java

int i = 0; while(i<=1000) { cout<<i<<endl; i++; i=i+1; } return 0;

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So you're saying ()! = (void) but you can pass arguments to () but you can't get them but you're not passing arguments to (void) and you can't get them, I get your point about main() not equal to void main(void) because you can return int but main(void) != main() doesn't make sense to me.