How do I loop over a char pointer in C?

Question

I am trying to access all elements of this preinitialized char pointer in 'c'. Here is the code:

#include <stdio.h>
#include <stdlib.h>

void main()
{
    char **s="Hello";
    printf("%c",*(*&s+1));
}

This code should output "e", but doesn't. What am I doing wrong?

Also, how do I access all elements one by one?

Does this compile? Without warnings? And please edit and add a question. You’ve now only included a mention what you’re trying to do but not what’s wrong. Or what this “e” is — Sami Kuhmonen
– Sami Kuhmonen, Commented Nov 26, 2019 at 15:50
This gives me more warnings that there are lines of code. Here's a tip: if you want to print all chars of "Hello" individually with the %c specifier, you'll need a loop. — Blaze
– Blaze, Commented Nov 26, 2019 at 15:51
Always use your compiler's warnings and head them. With gcc, I use -Wall -Wextra -pedantic. This would have found the problem. — ikegami
– ikegami, Commented Nov 26, 2019 at 16:01
@SamiKuhmonen No, it is giving warning like warning: initialization from incompatible pointer type [-Wincompatible-pointer-types] char **s="Hello"; — Prince Kumar Singh
– Prince Kumar Singh, Commented Nov 27, 2019 at 3:09

dbush · Accepted Answer · 2019-11-26 15:51:46Z

3

The type of s is incorrect. The string constant "Hello" has array type which can decay to type char * but you're assigning it to a variable of type char **.

Change the type of s to char * and your code will output "e":

 char *s = "Hello";

Also, looking at this:

*(*&s+1)

When * comes right before & they cancel each other out. So you can simplify this to:

*(s+1)

Or equivalently:

s[1]

answered Nov 26, 2019 at 15:51

dbush

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Comments

JiHo Han · Accepted Answer · 2019-11-26 16:02:43Z

0

Assumed that you are not seeing the 'e' of the Hello, I highly recommend watching this post.

What to do when the pointer doesn't print the string properly

Just to summarize, when you reference the pointer s, it is only reading the H of Hello (since that is how the pointer of string works - it points to the first character of the string, like s[0]), and that results your code in printing out H only.

Most importantly, your pointer is double-pointer, which is incorrect.

Try it in this way.

#include <stdio.h>
#include <stdlib.h>
void main()
{
  char s[] = "Hello";
  printf("%s\n",s);
}

answered Nov 26, 2019 at 16:02

JiHo Han

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