-3 
void skip(char *msg)
{
  puts(msg+6);
}

char *message="Don't call me";
skip(message);

My doubt is why we don't use puts(*(msg+6)) to display text from 7th character onward;
according to me (msg+6) refers to memory and *(msg+6) content

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4
  • 5
    Yes, but puts wants the 'memory' (or rather, a pointer). *(msg + 6) is but a single character, not a string. Commented Sep 2, 2014 at 19:23
  • then what about printf("%s",msg+6); Commented Sep 2, 2014 at 19:25
  • @sandy: int printf(const char *restrict format, ...); Commented Sep 2, 2014 at 19:28
  • @sandy Same story, printf expects a (character-)pointer for every %s in the format. I don't understand what you're not understanding... Commented Sep 2, 2014 at 19:28

3 Answers 3

Reset to default
0

*msg is essentially a reference to a single char, not to the string of char's. due to this char * and char[] are essentially the same thing and you don't need to dereference character pointer in C because compiler automatically print fully string from given base address upto '\0' not get. you can also refer this for more info .

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2 Comments

@vikash Can u please elaborate ur statement...char * and char [] are essentially same
@sandy This will help you stackoverflow.com/questions/20347170/char-array-and-char-array
0 
#include <stdio.h>

void skip(char *msg) {
    puts(msg + 6);
}
int main() {
    char *message = "Don't call me";
    skip(message);
}
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1 Comment

Please consider adding an explanation.
-1

This is what you can find in puts manual:

int puts(const char *s);

as you can see it also expects pointer to a content as a parameter, rather than an actual value.

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