Discussion
What will be the output of the program ?
#include <stdio.h>
int main()
{
static char mess[6][30] = {"Don't walk in front of me...",
"I may not follow;",
"Don't walk behind me...",
"Just walk beside me...",
"And be my friend." };
printf("%c, %c\n", *(mess[2]+9), *(*(mess+2)+9));
return 0;
}
Answer: B
Correct answer: k, k
mess[2] is "Don't walk behind me...", and index 9 counts D-o-n-'-t-space-w-a-l-k - giving k. The second expression, *(*(mess+2)+9), is the same thing written in pointer form, since mess[i] is *(mess+i). So both print k.
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