Can you print contents in a pointer pointing to an array? - c

Question

I am playing around with this code from one of my lecture slides and one of my questions is when printing out the array why can't we use the pointer instead of just printing "a", In the last line of the print statement
printf("\n message after decryption: %s\n", a);
I don't see why we cant use a pointer to print out the array. Can someone explainto me why this can't be done?
printf("\n message after decryption: %s\n", *q);
#include <stdio.h>
#define NUM 78
int main()
{
int i = 0;
char a[] = "Hello CSE 240";
printf("\n message: %s\n ", a);
while(a[i] != '\0'){a[i] = *(a+i)+1;i++;}
printf("\n message after encryption: %s\n ", a);
char *q = a;
while(*q != '\0'){*q = *q-1;q++;}
printf("\n message after decryption: %s\n", a);
}

Answer 1

You want to print out the content of the array after the q modification loop but it after all the increment it points to the null termination character i.e. '\0' - hence using q in printf will print this only - message after decryption: - modified array/string will not be printed.
To do it with q you need to reassign q to the first element of the array: q=a after the while loop ends.
and then this statement will work as you wish:
printf("\n message after decryption: %s\n", q);
It is good that you have changed the print statement with the suitable format specifier.

Answer 2

#include <stdio.h>
#define NUM 78
int main()
{
int i = 0;
char a[] = "Hello CSE 240";
printf("\n message: %s\n ", a);
while(a[i] != '\0'){a[i] = *(a+i)+1;i++;}
printf("\n message after encryption: %s\n ", a);
char *q = a;
while(*q != '\0'){*q = *q-1;q++;
//decrypting the value of a
}//end of this is pointing to null
q=a;//repointing to point a
printf("\n message after decryption: %c\n", *q);//H is printed
printf("\n message after decryption: %s\n", q);//Hello CSE 240 is printed
}