Because you're printing p, not *p. scanf takes addresses of variable

int *p=20;
if u print like dis printf("%d",p);
o\p:- 20; how is it possible?
plz give me the explanation.

Question Posted / lylez00

Answer Posted / lylez00

Because you're printing p, not *p. scanf takes addresses of
variables as arguments, but printf does not.

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7 Yes

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