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f(char *p)
{
p=(char *)malloc(sizeof(6));
strcpy(p,"HELLO");
}
main()
{
char *p="BYE";
f(p)
printf("%s",p);
}
what is the output?
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Answer Posted / yathish m yadav
the output is "hello".
here we are overwriting pointer *p thrice.
that is in the function we get an piece of memory from
malloc and assigned to p,
in the statement strcpy(p,"hello");
the malloc memory is lost and the compiler creates an char
array and copies the string "hello" and it makes the
character array as constant.
| Is This Answer Correct ? | 0 Yes | 2 No |
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