confusing char pointer

if we initialise char pointer as
char *p="Hello";

compiler will search for 6 bytes and will return address of first
byte to pointer p.Then how we can print value of p as
printf("%s",p);
instead of
printf("%s",*p);

p is a char*, so the correct format string in printf is %s
Eg printf( "string is %s\n", p );

*p is a char, so the correct format string in printf is %c
Eg printf( "First char of string is %c\n", *p );

I assume that %s will print a char at each address of 'p', starting from the first, until '\0' is found. That is why you pass the location (p) not the contents (*p)

Actually, with the code

Code:

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char *p = "Hello";

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the compiler sets aside some memory for the string "Hello", along with all other string constants used in the program. p is then assigned the address of the first char in the string, that is, 'H'.

we do not give address of variable while printing
[code]
char a[]="Hello";
char *p="Hello";
printf("%s",a);
printf("%s",p);//Why we are passing address of first constant string ie.H instead of *p
[\code]

Quote:

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//Why we are passing address of first constant string ie.H instead of *p

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Because you're printing a string of characters, not a single character. If you give printf the value of the address using *p, then how would you propose getting to the next address?

More examples

Code:

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char a[]="Hello";
char *p="Hello";
printf("%s",a);
printf("%s",p);
printf("%s",&a[0]);
printf("%s",&p[0]);

printf("%s",a+2);
printf("%s",p+2);
printf("%s",&a[2]);
printf("%s",&p[2]);

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One that always tickles me is this:

Code:

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char s[] = "hippo", *p = s+2;

printf("%c\n", p[-1]);

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