when I execute this code why error occur?

Question

#include <stdio.h>
#include <stdlib.h>
int main ()
{
    int a=2,*p;
    p=&a;
    printf ("address of a=%u", p);
    return (0);
}

When I execute this code on Ubuntu it will print this error:

format '%u' expects argument of type 'unsigned int', 
but argument 2 has type 'int *' [-wformat].

Why this type of error even in code from a book? I know this is a stupid question but I'm very confused.

when you execute? or when you compile? and is that an error or a warning? — thang, Oct 12 '13 at 07:49
maybe that's what you should put in the post instead of saying "when I execute" and "print this error". — thang, Oct 12 '13 at 08:39
the answer to your question: Why this type of error even in code from a book? is because %u expects unsigned int, and you're passing in int*. this is a type mismatch. it is not an error because technically it is legit c syntax. you're just using printf incorrectly. — thang, Oct 12 '13 at 08:44
you can fix this by either casting to unsigned int or using a different format spec (like %08x). — thang, Oct 12 '13 at 08:45
ok, well you didn't tell us what compiler gave you that warning, and it still works. you just get a warning. — thang, Oct 12 '13 at 08:50
I give up. Going to bed. You should learn English before learning C. — thang, Oct 12 '13 at 08:55
Can u just tell me that how can I run this program without any warning.. now you get it or not — user2873459, Oct 12 '13 at 09:01

score 4 · Answer 1 · edited May 23 '17 at 12:21

4

when I execute this code on ubuntu it will print this error:

Naw. It prints this when you compile it.

And that's because %u is not suited for printing pointers. Use %p for that purpose. And do read the documentation.

edited May 23 '17 at 12:21

Community

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answered Oct 12 '13 at 07:51

then why it executes on turbo c ? – user2873459 Oct 12 '13 at 07:58
2

@user2873459 Because Turbo C is a crappy, outdated, stupid compiler which doesn't know anything about format string errors. – Oct 12 '13 at 07:58
ohk and when I minus two addresses using pointer like int *I,*j, a []={1, 2,3, 4}; I=&a[0]; j=&a[2]; printf ("%p", j-I); why it will agin give error – user2873459 Oct 12 '13 at 08:05
And when I assign base address of array in pointer like p=&a; compiler again give me a warning ? – user2873459 Oct 12 '13 at 08:19
@user2873459 Because the difference of two pointers is not a pointer. It's an integer of type `ptrdiff_t` which is to be printed using `%td`. Also, As to why it warns on `p = &a`: I don't know. It shouldn't, if `a` is really an `int` and `p` is an `int *`. – Oct 12 '13 at 08:29
Using % td compiler minus locations but I want to minus addresses ? – user2873459 Oct 12 '13 at 08:33
@user2873459 Um what? What "locations"? Aren't you subtracting pointers? – Oct 12 '13 at 08:35
I hve subtract two pointers but but using pointer %td it subtract locactions like in my previous code ehen I subtract j-I awns ws 2 so its subtract locations but I want asnw 8 and o will print after subtractting addresses – user2873459 Oct 12 '13 at 08:39
@user2873459 That's how pointer arithmetic works. You can multiply that by `sizeof(int)` but I don't see why you would need that. Read about pointer arithmetic, I can't teach you the entire language in a comment. – Oct 12 '13 at 08:40
I just read this program in book and when I execute this code then it prints an error thats why I asked this question – user2873459 Oct 12 '13 at 08:42
@user2873459 is that book "let us C" by any chance? If so (and not only then), throw it far away and google for a decent language tutorial. That book is terrible (it's not your fault of course, but you better don't use it.) – Oct 12 '13 at 08:44
@user2873459 K&R for the basics, maybe Learn C The Hard Way to clean up outdated practices found in K&R. There's a "definitive C book/tutorial list" here on Stack Overflow, you can read that question/its answers as well. – Oct 12 '13 at 08:57
@user2873459 Can you use the site search? – Oct 12 '13 at 09:01

when I execute this code why error occur?

1 Answers1