passing array of characters to function in c

Question

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

Yes, but `puts` wants the 'memory' (or rather, a pointer). `*(msg + 6)` is but a single character, not a string. — Kninnug, Sep 02 '14 at 19:23
@sandy Same story, `printf` expects a (character-)pointer for every `%s` in the format. I don't understand what you're not understanding... — Kninnug, Sep 02 '14 at 19:28

score 0 · Answer 1 · edited May 23 '17 at 10:33

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 .

edited May 23 '17 at 10:33

Community

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answered Sep 02 '14 at 19:31

Vikas Verma

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@vikash Can u please elaborate ur statement...char * and char [] are essentially same – sandy Sep 03 '14 at 07:08
@sandy This will help you http://stackoverflow.com/questions/20347170/char-array-and-char-array – Vikas Verma Sep 03 '14 at 15:58

Engineer · Answer 2 · 2021-04-03T17:24:17.787

0

#include <stdio.h>

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

edited Apr 03 '21 at 17:24

answered Apr 03 '21 at 17:18

Engineer

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Please consider adding an explanation. – Frightera Apr 03 '21 at 19:47

score -1 · Accepted Answer · answered Sep 02 '14 at 19:26

-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.

answered Sep 02 '14 at 19:26

Jovik

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passing array of characters to function in c

3 Answers3