C interview

Question

#include <stdio.h>
int main()
{
    int i;
    printf("%d", &i)+1;
    scanf("%d", i)-1;
}

a. Runtime error.

b. Runtime error. Access violation.

c. Compile error. Illegal syntax

d. None of the above

Comments

ohh i know that fact dear .

My qus was ?

printf("%d", &i)+1;is equivalent to printf("%d\n",printf("%d\n", &i)+1);

And what will be the output  of above program .Don,t consider scanf is there.

---

@Manojk sir,,both output will be different..

---

yes off-course both are different .Now coming to qus ,

And i think Here it is printing address of a.

Answer 1

printf("%d", &i)+1;

This prints the address of "i" but using "%d" causes the address to be printed in signed format which can give negative value. Also, this assumes integer size and pointer size are the same which might not be true. For address "%p" must be used. Anyway there is no error here as printf function returns integer (the no. of characters successfully printed) and this returned value is added with 1. In C language there is no restriction that a value must always be assigned or used. Simply writing 

123;

is also valid in C. 

scanf("%d", i)-1;

Like printf, scanf also returns an int. But it returns the no. of items successfully read. So, here also there is no syntax error. But scanf moves the read data to the memory location given to it as argument(s). Here, i is used as a memory location which most likely won't be valid. Hence, this causes runtime error- segmentation fault in Linux. 

Comments

@Arjun sir, it runs fine in ideone  (no runtime error)

Answer 2

I think no compile time and run time error. I t will print some garbage value