What happens when we dont specify datatype of arguments in a function and pass parameters to it while calling it?

Question

Look at the following program.

int main()
{
    char a=65, ch ='c';
    printit(a,ch);
}

printit(a,ch)
{
    printf("a=%d   ch=%c",a,ch);
}

Even if the data type of the arguments is not specified in the function 'printit()', the result is shown on printf. I see correct answer when i compile it with gcc and run it.Why? Is it not necessary to specify the data type of arguments in C ? What is the default datatype of argument taken in the case shown above?

Answer 1

The only default datatype assumed in C is int as in the code above.

Newer versions of C++ prohibit implicit data typing and newer C++ compilers refuse to compile the code above.

Answer 2

Because you don't specify a prototype for printit(), compiler makes up implicit declaration:

int printit(int, int);

When later compiler sees the definition of printit() function without types for arguments, it uses that implicit declaration.

It is very dangerous technique - you basically prohibit type checking for this function.