Answer / susie

Answer :

10

Explanation:

The variable i is a block level variable and the visibility
is inside that block only. But the lifetime of i is lifetime
of the function so it lives upto the exit of main function.
Since the i is still allocated space, *j prints the value
stored in i since j points i.

Answer / vishu

the answer is that
int i varibale is part of int*j block code ,but outside the
block of code i variable also show their existanse.if we
write a code after the int*j block of code .
int*h
{
h=&i
}
printf("%d",*h);

}

Answer / anand

j=10

Answer / bipin chandra sai.s

actually j has beeen assigned the addresss of i so the ans
will be the value present in the address location 10

Answer / jerome.s,final year eee,adhipa

There i-is initialised by 10.
and j-also initialised by address of i.
so *j is the value in the address of j.
therefore,
*j=i=10.
OUTPUT:
10

Answer / sivakrishna

j=10

Answer / ashish p

The answer is undefined.
int *j;
{ //prolog
int i=10;
j = &i;
}//epilog

in the above code , at the prolog level the variables are
pushed into a un-named function space on the stack. Whereas
at epilog level the variable i dies.
J contains address of valid memory location but invalid
contents. Since i's memory is release back, any other
program can claim it and over-ride the contenets. Unless
then if we try to print the content using J it will give us
the value 10.
Which is not recommended it is something like returning
reference to the local variable in a function.

Answer / govind verma

i think ans will be 10 because here is the concept of dagling pointer......

Answer / hameennaveen

error

main() { char *p = “ayqm”; printf(“%c”,++*(p++)); }

29 Answers   IBM, TCS, UGC NET, Wipro,

main() { int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m); }

1 Answers  

write a program for area of circumference of shapes

0 Answers  

main() { register int a=2; printf("Address of a = %d",&a); printf("Value of a = %d",a); }

3 Answers  

#include <stdio.h> #define a 10 main() { #define a 50 printf("%d",a); }

2 Answers  

main() { int c=- -2; printf("c=%d",c); }

1 Answers   TCS,

main() { int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }

3 Answers   GNITC,

Predict the Output: int main() { int *p=(int *)2000; scanf("%d",2000); printf("%d",*p); return 0; } if input is 20 ,what will be print

2 Answers  

find simple interest & compund interest

2 Answers  

Write a procedure to implement highlight as a blinking operation

2 Answers  

main() { char *p = "hello world"; p[0] = 'H'; printf("%s", p); } a. Runtime error. b. “Hello world” c. Compile error d. “hello world”

5 Answers   HCL,

main() { char i=0; for(;i>=0;i++) ; printf("%d\n",i); }

2 Answers