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main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
- 29 Answers
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- IBM, TCS, UGC NET, Wipro, I also Faced
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Answers were Sorted based on User's Feedback
Answer / kirit vaghela
p is char pointer that store the address of string "ayqm"
that means p have the address of 'a'.
if we write printf("%c",*p);
we get the output 'a';
now the *(p++) is post-increment that means it print the
value than increment the address by 1.
so *(p++) is print the value 'a'.
at finally ++*(p++) is increment the ascii value of 'a' by 1.
so it become 'b'.
the final output is 'b'.
| Is This Answer Correct ? | 88 Yes | 6 No |
Answer / vikas mathur
correct answer is :- b
why because
p is a char pointer that holds the base address of
string"ayqm". so p points to address where 'a' is stored.
p++ is a post increnment statement so p will still points
to a but after this printf statement p will start pointing
to 'y'.
now *(p++) will return the values at address pointed by p
which is 'a'. and ++*(p++) means ++'a' which returns 'b'
so correct answer is b
| Is This Answer Correct ? | 13 Yes | 3 No |
Answer / anu-priya
the answer is b..
char *p="ayqm"; p a character pointer points to the base add
of string ie a...
printf("%c",*p); will print a....
printf("%c",*(p++)); will print a....
when we post increment the p then it will giv answer a..but
it wil increment by 1
printf("%c",++*(p++));
here it is ++a will print b....
| Is This Answer Correct ? | 9 Yes | 4 No |
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