Question about char pointers - c

Question

I am confused with this program.
#include <stdio.h>
int main(void)
{
char* ptr="MET ADSD";
*ptr++;
printf("%c\n", ptr);
ptr++;
printf("%c\n", ptr);
}
Here's the output.
ET ADSD
T ADSD
My question is how does the pointer display the rest of the characters?

Answer 1

You are passing a wrong combination of parameters to printf: the %c format specification requires a char parameter, not a char*. So the result is undefined - in your case printf seems to print the whole char array, but this is just by pure chance. Use either
printf("%c\n", *ptr);
or
printf("%s\n", ptr);

Answer 2

You are actually trying to print as a char %c a pointer value char*. This is false. But then, I really don't understand why it prints all the chars. Are you sure you didn't use %s instead of %c?

Answer 3

The * operator as lower precedence over the ++ operator. Thus in your example the both lines
*ptr++;
ptr++;
have the same effect.
And you are using the wrong types in your printf statement.
Change
printf("%c\n", ptr);
to
printf("%s\n", ptr);
or
printf("%c\n", *ptr);
depending on what you want to output.
Btw, turning on compiler warnings helps in that case. E.g. the GCC prints:
d.c: In function ‘main’:
d.c:7: warning: value computed is not used
d.c:8: warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘char *’
d.c:11: warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘char *’
d.c:12: warning: control reaches end of non-void function

Answer 4

In C a string is a array of chars, and an array is simply a pointer to the first location of memory of the array.
So defining
char* ptr="MET ADSD";
you are declaring and initializing an array of chars, a string, by using a pointer to char,
The next trick comes if you consider this two factors:
pointers arithmetic in that using the operator ++ on a pointer increments its value, the memory address it is pointing to
char size which is almost everywhere 1 byte
So you are scaling the array along of two positions, and you print that by using %s and passing the pointer to it
EDIT I guess you put %c mistakenly in the example

Answer 5

If you are expecting 'E' as first and 'T' as second output. Give it like
#include <stdio.h>
int main(void)
{
char* ptr="MET ADSD";
*ptr++;
printf("%c\n", *ptr);
ptr++;
printf("%c\n", *ptr);
}

Answer 6

printf( ..., ptr) is passing the pointer, not the char it points to.
The correct version of the program would be:
#include <stdio.h>
int main(int argc, char* argv[])
{
const char* ptr = "MET ADSD";
ptr++;
printf("%c\n", *ptr);
ptr++;
printf("%c\n", *ptr);
return 0;
}
which will print
E
T
The output you are seeing makes no sense unless you are using %s. - %c is going to convert the value of ptr into an integer, truncate the int to 8bits (the width of a char), and print that character to the output. Not a string of characters.