Doubt in char pointers

Question

Hi,

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 displays the rest of the characters?

Thanks.

Answer 1

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 2

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 3

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 4

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 5

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 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.