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 / shruti
The output would be "HELLO"..
though we are not returning the string, we are making
direct changes at the memory location..
so "bye" will be overwritten with "HELLO"
because we are using pointers, the dying pointer scenario
is not applicabe here..
Its a pointer, not a variable..
This function will work similar to -> swapping two numbers
using pointers..
juss check that prog if you fnd somewhere.. :-)
you will get the logic... :-)
Cheers...
--By the way a gud ques.. :-)
| Is This Answer Correct ? | 1 Yes | 4 No |
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