Consider the following code:

$int p; void fun( ) {static int x = 1; static int y = 2; x = x + y; y = x - y; p = &x;} void main( ) {fun( ); fun( ); printf ("%d", p);}$

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Solution

The correct option is B 4
option (c)

$4 will be printed by *p$

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Consider the following code: int *p;void fun( ){static int x = 1;static int y = 2;x = x + y;y = x - y; p = &x;}void main( ){fun( );fun( );printf ("%d", *p);}

The correct option is B 4option (c) 4 will be printed by *p

Consider the following code:

$int p; void fun( ) {static int x = 1; static int y = 2; x = x + y; y = x - y; p = &x;} void main( ) {fun( ); fun( ); printf ("%d", p);}$

The correct option is B 4
option (c)

$4 will be printed by *p$