CODEWITHSUNDEEP
c++program-entry-point
Arpit
Arpit

Reputation: 12797

Does Main() accept expression?

Why this code prints 1 instead of 5

Code:

main(int x=5) //this defn. is written intentionally to chec weather main accepts 
                expression or not. 
{
 printf("%d",x);  
}

Compiler used:minGW 3.2

EDIT

My point is weather x=5 executes or not. if not then why i don't get any error or warning.

Upvotes: 0

Views: 147

Answers (4)

Shafik Yaghmour
Shafik Yaghmour

Reputation: 158469

Update

Your main declaration is not valid, if we look at the C++ draft standard section 3.6.1 Main function paragraph 2 says(emphasis mine):

An implementation shall not predefine the main function. This function shall not be overloaded. It shall have a return type of type int, but otherwise its type is implementation-defined. All implementations shall allow both

— a function of () returning int and

— a function of (int, pointer to pointer to char) returning int

So main should adhere to one of these standard forms or implementation defined forms defined by compiler documentation.

gcc gives me a warning for this regardless of warning levels and in clang this is an error, so I am not sure why you do not see an error.

Orignal Answer

The first argument to main is the argument count usually denoted as argc for example:

int main(int argc, char *argv[])
{
}

and argv is an array of string which represents the arguments to your program, the first one being the command line.

Upvotes: 1

Albert
Albert

Reputation: 68140

Because the operating system expects this signature of main:

int main(int argc, char** argv);

argc is the amount of parameters. When it calls your main, it passes the amount of arguments (argc) as the first parameter, which is 1 (if you call your binary without arguments, you still get one argument: the binary filename, $0 in bash).

Note that this depends also on the C ABI. By the C/C++ standard, multiple signatures of main are allowed. So, depending on the compiler and the OS, there could be a different handling for main. What you are doing is not really defined behavior.

You should declare main like the expected - because that is what your OS expects and uses. Make another function for whatever you want to program.

Upvotes: -1

Pete Becker
Pete Becker

Reputation: 76275

In void f(int x = 5), the = 5 part is a default argument. You can call the function in two different ways:

f();  // uses default argument, as if f(5)
f(3); // explicit argument

Note that the decision to use the default argument is made at the point of the call, not at the point of the declaration. Regardless of whether int main(int x = 5, char *argv[]) is valid, the application's startup code (part of the compiler's library) won't know about the attempted default argument, so won't do anything with it. And don't try and get tricky by calling main from inside your program: that's not allowed.

Upvotes: 1

KevinDTimm
KevinDTimm

Reputation: 14376

because x is really argc (and your count of arguments is 1)

The signature for main is:

int main (int argc, char **argv)

with argc being a count of arguments
and argv being an array of those arguments

Upvotes: 10

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