Decision Making
- Which of the following is the correct output for the program given below?
#include <stdio.h>
int main ( )
{
int x = 0, y = 1, z = 3;
* ( ( x ) ? &y : &x = x ? y :z;
printf ("%d %d %d\n", x, y, z);
return 0;
}
-
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Step 1: int x=0, y=1, z=3; here variable x, y, and z are declared as integer type and initialized to 0, 1, 3 respectively.
Step 2: *((x) ? &y : &x) = x ? y : z; The right side of the expression(x?y:z) becomes (0?1:3). Hence it return the value '3'.
The left side of the expression *((x) ? &y : &x) becomes *((0) ? &y : &x). Hence this contains the address of the variable x *(&x).
Step 3: *((x) ? &y : &x) = x ? y : z; Finally this statement becomes *(&x)=3. Hence the variable a has the value '3'.
Step 4: printf("%d, %d, %d\n", x, y, z); It prints "3, 1, 3".Correct Option: C
Step 1: int x=0, y=1, z=3; here variable x, y, and z are declared as integer type and initialized to 0, 1, 3 respectively.
Step 2: *((x) ? &y : &x) = x ? y : z; The right side of the expression(x?y:z) becomes (0?1:3). Hence it return the value '3'.
The left side of the expression *((x) ? &y : &x) becomes *((0) ? &y : &x). Hence this contains the address of the variable x *(&x).
Step 3: *((x) ? &y : &x) = x ? y : z; Finally this statement becomes *(&x)=3. Hence the variable a has the value '3'.
Step 4: printf("%d, %d, %d\n", x, y, z); It prints "3, 1, 3".
