Explain the strange output ''

Question

I have written the code in C language,

#include<stdio.h>

int main(void)
{
    char c[]="Suck It Big";
    char *p=c;
    printf("%c\n",p);
    printf("%c\n",p[3]);
    printf("%c\n",++p);
    printf("%d\n",p);
    printf("%d\n",p[3]);
}

The output to this code i get is:

I copied the strange characters on line 1 and 3 of the output and pasting it on the editor gives this " DLE ". Can anybody explain the meaning of this.

Answer 1

All printf() calls you use are incorrect, except for the second one, because either the relative argument or the used conversion specifier is wrong.

This invokes Undefined Behavior :

Quote from C18, 7.21.6.1/9 - "The fprintf function":

" If a conversion specification is invalid, the behavior is undefined.288) If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined. "

---

printf("%c\n",p);

When you attempt to print the value of the object the pointer points to, you have to use the dereference operator ( * ) preceding the pointer object. Else you attempt to print the value of the pointer - the address of the object the pointer point to. And due to this operation, you using the wrong conversion specifier of %d instead of %p to print the value of a pointer.

---

The corrected program is:

#include<stdio.h>

int main(void)
{
    char c[]= "Suck It Big";
    char *p = c;
    printf("%c\n", *p);              // Prints the first element of array c.
    printf("%c\n", p[3]);            // Prints the fourth element of array c
    printf("%c\n", *(++p));          // Prints the second element of array c
    printf("%p\n", (void*) p);       // Prints the address held in p / first element of c.
    printf("%p\n", (void*) &p[3]);   // Prints the address of the fourth element of c.
}

Note, that the cast to void* is necessary to make the program conforming to the C standard.

Output:

S
k
u
0x7fff1d3a133d  // Address defined by system
0x7fff1d3a1340  // Address defined by system