main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”,
“violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
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Answer Posted / susie
Answer :
ck
Explanation:
In this problem we have an array of char pointers pointing
to start of 4 strings. Then we have ptr which is a pointer
to a pointer of type char and a variable p which is a
pointer to a pointer to a pointer of type char. p hold the
initial value of ptr, i.e. p = s+3. The next statement
increment value in p by 1 , thus now value of p = s+2. In
the printf statement the expression is evaluated *++p causes
gets value s+1 then the pre decrement is executed and we get
s+1 – 1 = s . the indirection operator now gets the value
from the array of s and adds 3 to the starting address. The
string is printed starting from this position. Thus, the
output is ‘ck’.
| Is This Answer Correct ? | 5 Yes | 0 No |
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