What this means compiler looks for closest possible match ,then other
alternatives.In this func(char*) is executed if that is not there then
func(const char*) is executed.
Comment the func(char*) and check.There is no overloading in c as far i
know.
On Tue, Nov 22, 2011 at 10:11 PM, Jagannath Prasad Das
<jpdasi...@gmail.com>wrote:
> Run this one:
>
> #include<stdio.h>
> void fun(const char *p)
> {
> printf("const\n");
> }
> void fun(char *P)
> {
> printf("simple\n");
> }
> int main(void)
> {
> char str[] = "funcheck";
> //const char str1[] =
> "constcheck";
>
> fun(str);
> fun(str);
>
> }
>
>
>
> On Tue, Nov 22, 2011 at 7:13 PM, MJ <mayurdj...@gmail.com> wrote:
>
>> Hi
>> This type of function overloading works in c .
>> Execute following code and it will call function fun as per the
>> function parameter
>>
>> code snip-in
>> ---------------------
>> #include<stdio.h>
>> void fun(const char *p)
>> {
>> printf("funsfsdafsf1\n");
>> }
>> void fun(char *P)
>> {
>> printf("fun2\n");
>> }
>> void main()
>> {
>> char str[] = "funcheck";
>> const char str1[] = "constcheck";
>> fun(str);
>> fun(str1);
>> }
>> ---------------------
>>
>> output
>> ------------
>> fun2
>> funsfsdafsf1
>> ----------------
>>
>> Mayur
>>
>>
>> On Nov 21, 6:30 pm, Anil Arya <anilarya...@gmail.com> wrote:
>> > dude....compile using g++
>> >
>> > On 11/21/11, atul anand <atul.87fri...@gmail.com> wrote:
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > > this is what i am getting after executing code meintion in the above
>> link.
>> >
>> > > (using gcc compiler )
>> >
>> > > temp.c:12: error: conflicting types for 'fun'
>> > > temp.c:7: error: previous definition of 'fun' was here
>> >
>> > > On Mon, Nov 21, 2011 at 3:09 PM, UTKARSH SRIVASTAV
>> > > <usrivastav...@gmail.com>wrote:
>> >
>> > >>http://www.ideone.com/JQFNh
>> > >> CHECK THE OUTPUT THERE IS NO AMBIGUITY BUT PLEASE EXPLAIN THE RESULTS
>> >
>> > >> On Sun, Nov 20, 2011 at 11:00 PM, Akash Coder
>> > >> <akash.coder.g...@gmail.com>wrote:
>> >
>> > >>> it wont work
>> >
>> > >>> even this wont
>> >
>> > >>> function(char a[])
>> > >>> function (char *)
>> > >>> coz the two forms are interchangeable
>> >
>> > >>> On Sun, Nov 20, 2011 at 10:12 PM, saurabh singh
>> > >>> <saurab...@gmail.com>wrote:
>> >
>> > >>>> wont work......Thre is no way compilers gonna guess which function
>> to
>> > >>>> call.The const qualifier only guarantees that the value at the
>> address
>> > >>>> wont
>> > >>>> be altered inside the function.That has nothing to do with calling,
>> >
>> > >>>> On Sun, Nov 20, 2011 at 9:50 PM, rahul vatsa
>> > >>>> <vatsa.ra...@gmail.com>wrote:
>> >
>> > >>>>> yes, it will work.
>> >
>> > >>>>> On Sun, Nov 20, 2011 at 9:12 PM, Akash Coder <
>> > >>>>> akash.coder.g...@gmail.com> wrote:
>> >
>> > >>>>>> no it wont work ... const is not a datatype. its a qualifier
>> >
>> > >>>>>> On Sun, Nov 20, 2011 at 7:49 PM, rahul sharma <
>> rahul23111...@gmail.com
>> > >>>>>> > wrote:
>> >
>> > >>>>>>> void fun(char *)
>> > >>>>>>> void fun(const char *)
>> >
>> > >>>>>>> is this overloading works or these are same type of
>> arguments??????
>> >
>> > >>>>>>> --
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>> >
>> > >>>> --
>> > >>>> Saurabh Singh
>> > >>>> B.Tech (Computer Science)
>> > >>>> MNNIT ALLAHABAD
>> >
>> > >>>> --
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>> > >>> --
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>> >
>> > >> --
>> > >> *UTKARSH SRIVASTAV
>> > >> CSE-3
>> > >> B-Tech 3rd Year
>> > >> @MNNIT ALLAHABAD*
>> >
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>> >
>> > --
>> > *Anil Arya,
>> > Computer Science *
>> > *Motilal Nehru National Institute of Technology,Allahabad .
>> > *
>>
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