Geeksforgeeks

Question

What will be the output of following programs and why ?

1)
void foo(float *);
int main()
{
int i = 10, *p = &i;
foo((float*)&i);
return 0;
}
void foo(float *p)
{
printf("%f\n", *p);
}

 

2)

int main()
{
char arr[] = "geeksforgeeks";
char *ptr = arr;

while(*ptr != '\0')
++*ptr++;
printf("%s %s", arr, ptr);

getchar();
return 0;
}

3)

int main()
{
int c=5;
printf("%d\n%d\n%d", c, c <<= 2, c >>= 2);
getchar();
}
Output:4 4 4
I am not getting why the 3rd one is 4,because 5>>2 will give 1

Comments

See you are printing value of c 3 times.

so last value of c is 4 , so every place 4 is printing. there is one printf function which is printing value at last so at every place same value of c is printed, no matter what is c in starting and in between.

printf("%d\n%d\n%d", c, c <<= 2, c >>= 2); = 4,4,4 because last value of c is 4

printf("%d\n%d\n%d", c >>= 2, c, c <<= 2); = 5,5,5 because last value of c is 4

---

Thanks,now I got it :)

Answer 1

1) At first i taking the value 10. But when value of integer i converted to float, the alignment of int converted to float pointer. So, that is not same as change of value from float to int. Alignment change in pointer makes value 0 in float.So, it will print 0.000000

2)main point of this program is

(++(*(ptr++)));//According to precedence this will be execution order

Here increment operator first incrementing value of the pointer and then incrementing pointer index. So, it will print hfflt....

3) . It is tricky one. Here precedence and associativity plays main role. <<= is right to left associative. So, after assigning 5 in c, it first calculates right shift value i.e., c >>= 2  and then left shift value. So, first value of 5 becomes 1 and then it becomes 4. All of these things done in one sequence point. So, all C will point to same and final value 4. It will print 4 when the sequence point ends.

http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm

Comments

thanks :)

---

Is any explanation given there?

---

It was given but was unsatisfactory.

Srestha,can you please help me in below code :

#include<stdio.h>

int main()
{
    int a[] = { 1, 2, 3, 4, 5} ;
    int *ptr;
    ptr = a;
    printf(" %d ", a );
    printf(" %d ", ptr );
    printf(" %d ", &a );
    printf(" %d ", *( ptr + 1) );
    printf(" %d ", *( &a + 1) );
    return 0;
}

On executing the above code ptr and &a will have same value but *( ptr + 1) and  *( &a + 1) give different values,can you explain me why ?

---

In the last printf statement, you are writing *(&a+1) but the name of an array is itself the address of the array, so if you want to find the element at the 2nd position, do *(a+1)