Answer required!

[karthikeyanck]

What will be the answer for this?

#include<stdio.h>
int strA = 1;
int *swappy(int *ptr);

void main(){
swappy(&strA);
}

int *swappy(int *ptr){
printf("\nThe value is: %d",*ptr);
*ptr++;
printf("\nThe value after post increment: %d",*ptr);
++*ptr;
printf("\nThe value after pre increment: %d",*ptr);
}

Please suggest the solution and explain why?

[Banfa]

What will be the answer for this?

Please suggest the solution and explain why?

The answer for what? The solution for what?

You have not described a problem or asked an answerable question but I will tell you for free that your program exhibits undefined behaviour.

[karthikeyanck]

Solved!

Sorry, I was trying to get the answer without initializing the pointer, solved now! BTW the question was what will printf print?

[JosAH]

Sorry, I was trying to get the answer without initializing the pointer, solved now! BTW the question was what will printf print?

Your pointer was initialized already in your original version. The execution of that thing still causes undefined behaviour; nothing is solved yet.

kind regards,

Jos

[karthikeyanck]

I meant the pointer wasn't initialzed to NULL, so i believe it should have been holding someother value if am not wrong..

Regards,

Karthik

[JosAH]

I meant the pointer wasn't initialzed to NULL, so i believe it should have been holding someother value if am not wrong.

That pointer is a parameter and it's initialized in your main function (line #6). I don't understand your remark/question.

kind regards,

Jos

[newb16]

[QUOTE=karthikey anck;3437214]What will be the answer for this?
[quote]
Change it to

int strA[3]={0,10,20};
...
void main(){
swappy(&(strA[0]));

compile and run.
In first *q++ it increments pointer beyond your original variable,
result of dereferencing is ignored, and the value of location where pointer points is printed.

[JosAH]

ps. better make that 'int main()'.

kind regards,

Jos