#include <stdio.h>
int main(int argc, char *argv[])
{
char a, *pc, c[9];
int i, *pk, k[9];
a='z';
pc=&(c[8]);
pk=&(k[0]);
for (i=0; i<9; i++)
{
*pc=a-(char)i;
pc--;
*pk=(int)a-i;
pk++;
}
return 0;
}//end of main
Answer the below questions with the above code
Variable |
Start address |
End address |
a |
||
Pc |
||
c |
||
i |
||
pk |
||
K |
Variable |
Value |
a |
|
pc |
Use an expression involving c to represent pc’s value |
c |
list the contents of the entire array |
i |
|
pk |
Use an expression involving k to represent pk’s value |
k |
list the contents of the entire array |
C.False
D. True
#include <stdio.h> int main(int argc, char *argv[]) { char a, *pc, c[9]; int i, *pk, k[9]; a='...
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