Including #include files at the end - c

I had an interview question where the interviewer asked me what error would we get from the below output:
int main()
{
printf("hello world");
return 0;
}
#include <stdio.h>
I answered "no error" and it will display the output.
can anyone help me with this???
Please note"the missing angular brackets is intentionally done by me so dont bother about that"

It depends on the compiler.
Most C compilers will probably accept that code (perhaps with a warning) and produce an executable that prints the expected output.
Under C90 rules, the behavior of the printf call is undefined; it's invalid to call a variadic function with no visible prototype. Variadic functions can have a different calling convention from ordinary functions, and you have to let the compiler know that printf is variadic so it can generate correct code for the call.
Under C99 and later rules, calling any function with no visible declaration (which may or may not be a prototype) is a constraint violation, requiring at least a compile-time warning.
The standard doesn't hint at what happens if you call printf without the required prototype, but in practice most compilers will handle it "correctly".
The missing '\n' at the end of the output means that the program's behavior is undefined if the implementation requires a newline at the end of the output; whether that's required or not is implementation-defined. In any case, adding a newline is a good idea.
The #include <stdio.h> at the end of the source file should be useless but harmless.
I'm assuming that the source file actually contains #include <stdio.h> and not #include stdio.h; the latter would be a syntax error.
(Practically speaking, of course, the #include <stdio.h> should be at the top. In a professional setting, the output is irrelevant, since the program will never survive a code review.)

You will get an error for missing quotes or <> in the filename for the #include. It should be:
#include <stdio.h>
Apart from that, it should compile with a warning about an implicit declaration to printf(). On Clang, it gives me:
test.c:3:5: warning: implicitly declaring library function 'printf' with type 'int (const char *, ...)'
printf("hello world");
^

When your main() function is being declared, the compiler will run into the first line of main() and it will have no idea what printf() is. You will get an error about the compiler complaining about an undeclared function.
Assuming, of course, the missing < and > was a mistake introduced when you copied your question over.

There are two errors here. The first is that the include directive must happen before the code that requires that header information, in this case printf() is declared in stdio.h, and the second is that the filename in the include directive must be enclosed in angle brackets, <>, or quotes, "".

Related

getch undefined while compiling for the second time

I am using visual studio 2017 .
First I wrote the following code :
void main()
{
printf("abcdefgh %d hjhjh %d", 5, 6);
getch();
}
It ran perfectly fine .
But after that I modified the code to the following :
void main()
{
char abc[100];
strcpy_S(abc, "premraj");
printf("%s", abc);
printf("abcdefgh %d hjhjh %d", 5, 6);
getch();
}
But now I am getting an error with getch stating that "'getch' undefined, assuming extern returning int"
But this new code has been built on the existing code which recognized getch perfectly , how can it not recognize getch the second time ?
I checked out the following question :
getch() is working without conio.h - how is that possible?
which also carried a similar problem but here with modifications only I got this error .
There is an informative answer there by user named "Fatal Error" but still I would like to know about this intriguing phenomenon that is coming in after modifications . What can be the reason behind this ?
P.S : The following was my header file declarations for the first time :
#include <stdio.h>
and the following for the second time :
#include <stdio.h>
#include <string.h>
Once upon a time, if you called a function which the compiler had never heard of, like this:
#include <stdio.h>
int main()
{
int x = foo();
printf("%d\n", foo);
}
Anyway, if you did that, the compiler quietly assumed that foo() was a function returning int. That is, the compiler behaved just as if you had typed
extern int foo();
somewhere before you called foo.
But in, I think, C99, the language was changed. It was no longer legal to call a function you had not explicitly declared. Because there was lots and lots of code out there that was written under the previous set of rules, most compilers did not immediately begin rejecting the old-style code. Some continued to quietly assume that unrecognized functions returned int. Others -- like yours -- began noisily assuming that unrecognized functions returned int, emitting warnings along the lines of "'foo' undefined, assuming extern returning int".
It sounds like your question is that some time ago, your code containing calls to getch() was accepted without warning, but today, you're getting the warning "'getch' undefined, assuming extern returning int". What changed?
One possibility is that your code changed slightly. If your code used to contain the line
#include <conio.h>
somewhere, that file would have contained a declaration along the lines of
extern int getch();
and this would have goven the compiler the declaration that it needed, and you would not have gotten the warning. But if today your code does not contain that #include line, that explain why the warning started cropping up.
It's also possible that your compiler has changed somehow. It's possible you're using a new version of the compiler, that's more fussy, that has gone from quietly assuming, to normally assuming. Or, it's possible that your compiler options have changed. Many compilers can be configured to accept different variants of the language, corresponding to the different versions of the language standards that have been released over the years. For example, if some time ago your compiler was compiling for language standard "C89", but today, it's doing "C99" or "C11", that would explain why it's now being noisy with this warning.
The change in the compiler could be a change in the defaults as configured by a system administrator, or a change in the way you're invoking the compiler, or a change in your project's Makefile, or a change in the language settings in your IDE, or something like that.
A few more points:
getch is not a Standard C function; it's specific to Windows. Your program would be more portable, in general, if you didn't use it. Are you sure you need it? (I know what it's for; what I don't know if there's some other way of keeping your program's output window on the screen after if exits.)
You should get in the habit of declaring main() as int, not void. (void will work well enough, but it's not correct, and if nothing else, you'll get lots of negative comments about it.)
I think there's something wrong with your call to strcpy_S, too,

Is it correct to drop #include <stdio.h> in C?

If I use printf, scanf, puts or some other function in C (not C++) and don't write an include line, can it be treated as unspecified or undefined behaviour?
As I remember, C didn't require porotope declaration at all, but it was recommended to have them to allow compiler to make type casts on calling. And prototypes for printf and other such functions are not required still, not sure about custom functions.
PS: This question relates to discussion in comments of https://codegolf.stackexchange.com/a/55989/32091.
Is it correct to drop #include in C?
No, it's not correct. Always include it if you use a stdio.h function like printf.
C has removed implicit declarations (with C99) and the includes are required. The only other alternative is to have a visible prototyped declaration for printf.
Moreover even when C had implicit declarations, implicit declarations are not OK for variable argument functions; so in C89 not adding a stdio.h include and not having a visible prototype (for printf example) is undefined behavior.
For professional development, no.
For codegolfing, it is ok.
If you don't declare a function, the compiler automatically generates one, which may or may not match its real declaration. If it doesn't, it may or may not produce segfault or a software bug. gcc also gives in this case a warning.
No and Yes
stdio.h has an explicit declaration of the functions you want to use, such things are prohibited if it was a C++ compiler (g++ for example).
Since c++ requires explicit declarations of all functions, but any proper C compiler will create an implicit declaration of those functions, compile the code into object file, and when linked with the standard library, it will find a definition of those functions that accidentally will match the implicit declaration, probably gcc will give you a warning.
So if you are writing software that you want to be maintainable and readable, that's not an option to drop it, however for fast prototyping or code challenges, this might not be that important.
Technically you can skip #include in many cases. But for some functions the compiler cannot generate correct function call without prototype. E.g. if a parameter is double and you put 0 - with prototype it will be converted and stored as double value in stack and otherwise there will be int which will produce wrong calculations.

C program without header

I write "hello world" program in C.
void main()
{ printf("Hello World"); }
// note that I haven't included any header file
The program compiles with warning as
vikram#vikram-Studio-XPS-1645:~$ gcc hello.c
hello.c: In function ‘main’:
hello.c:2:2: warning: incompatible implicit declaration of built-in function ‘printf’
vikram#vikram-Studio-XPS-1645:~$ ./a.out
Hello Worldvikram#vikram-Studio-XPS-1645:~$
How is this possible? How does the OS link a library without including any header?
The compiler builds your source file with a reference to a function called printf(), without knowing what arguments it actually takes or what its return type is. The generated assembly contains a push of the address of the string "Hello World" in the static data area of your program, followed by a call to printf.
When linking your object file into an executable, the linker sees a reference to printf and supplies the C standard library function printf(). By coincidence, the argument you have passed (const char*) is compatible with the declaration of the real printf(), so it functions correctly. However, note that the printf() that your program implicitly declares has return type int (I think), which the standard printf() also has; but if they differed, and you were to assign the result of calling printf() to a variable, you would be in the land of undefined behaviour and you would likely get an incorrect value.
Long story short: #include the correct headers to get the correct declarations for functions you use, because this kind of implicit declaration is deprecated, because it is error-prone.
The printf function is in the C library (libc in your case) which is linked implicitly (actually gcc has a printf builtin but it's outside the point).
Including the header doesn't bring in any functions for the linker, it simply informs the compiler about their declarations (i.e. "what they look like").
Obviously you should always include headers otherwise you force the compiler into making assumptions about what the functions look like.
In C, if you use a standard library function, you have to include the standard header where the function is declared. For printf you have to include stdio.h header file.
In C89 (and GNU C89 which is the language by default on gcc), a function declaration can be sometimes omitted because there is a feature called implicit function declaration: when a function identifier foo is used and the function has not been declared, the implementation would use this declaration:
/* foo is a function with an unspecified number of arguments */
extern int foo();
But this declaration is OK only for functions that return int with an unspecified but fixed number of arguments. If the function accepts a variable number of arguments (like printf) such program would invoke an undefined behavior.
Here is what C89/C90 says:
(C90, 6.7.1) "If a function that accepts a variable number of arguments is defined without a parameter type list that ends with the ellipsis notation, the behavior is undefined.
So gcc is kind enough to compile even in C89 and GNU C89: a compiler could refuse to compile.
Also note that
void main() { ... }
is not a valid definition for main (at least on hosted implementations which is probably your case).
If your main function doesn't take any argument use this valid definition:
int main(void) { ... }
The header usually1 contains only function declarations, symbolic constants, and macro definitions; it doesn't usually include function definitions.
All stdio.h gives you is the prototype declaration for printf:
int printf(const char * restrict format, ...); // as of C99
The implementation of printf is in a separate library file that your code links against.
Your code "works" for two reasons:
Under C89 and earlier versions, if the compiler sees a function call
before a declaration or definition of that function, it will assume
that the function returns int and takes an unspecified number of
parameters;
The implementation of printf returns an int, and you passed in
an argument that just happens to be compatible with what the
implementation of printf expects for the first argument.
And to echo what everyone else says, use int main(void) or int main(int argc, char **argv); unless your compiler documentation explicitly lists void main() as a legal signature, using it will invoke undefined behavior (which means everything from your code running with no apparent issues to crashing on exit to failing to load completely).
I say "usually"; I've run across some headers that contained code, but those were usually written by people who didn't know what they were doing. There may be very rare occasions where putting code in a header is justified, but as a rule it's bad practice.
hello.c:2:2: warning: incompatible implicit declaration of built-in function ‘printf’
To deal with this warning, you should include the header file (stdio.h). You're accidently using an old feature of C that has been deprecated since 1999.
Also, the fact that the link doesn't fail simply means that the standard C library is linked in by default. Whether or not you have included the relevant header is immaterial.

Why cast "extern puts" to a function pointer "(void(*)(char*))&puts"?

I'm looking at example abo3.c from Insecure Programming and I'm not grokking the casting in the example below. Could someone enlighten me?
int main(int argv,char **argc)
{
extern system,puts;
void (*fn)(char*)=(void(*)(char*))&system;
char buf[256];
fn=(void(*)(char*))&puts;
strcpy(buf,argc[1]);
fn(argc[2]);
exit(1);
}
So - what's with the casting for system and puts? They both return an int so why cast it to void?
I'd really appreciate an explanation of the whole program to put it in perspective.
[EDIT]
Thank you both for your input!
Jonathan Leffler, there is actually a reason for the code to be 'bad'. It's supposed to be exploitable, overflowing buffers and function pointers etc. mishou.org has a blog post on how to exploit the above code. A lot of it is still above my head.
bta, I gather from the above blog post that casting system would somehow prevent the linker from removing it.
One thing that is not immediately clear is that the system and puts addresses are both written to the same location, I think that might be what gera is talking about “so the linker doesn’t remove it”.
While we are on the subject of function pointers, I'd like to ask a follow-up question now that the syntax is clearer. I was looking at some more advanced examples using function pointers and stumbled upon this abomination, taken from a site hosting shellcode.
#include <stdio.h>
char shellcode[] = "some shellcode";
int main(void)
{
fprintf(stdout,"Length: %d\n",strlen(shellcode));
(*(void(*)()) shellcode)();
}
So the array is getting cast to a function returning void, referenced and called? That just looks nasty - so what's the purpose of the above code?
[/EDIT]
Original question
User bta has given a correct explanation of the cast - and commented on the infelicity of casting system.
I'm going to add:
The extern line is at best weird. It is erroneous under strict C99 because there is no type, which makes it invalid; under C89, the type will be assumed to be int. The line says 'there is an externally defined integer called system, and another called puts', which is not correct - there are a pair of functions with those names. The code may actually 'work' because the linker might associate the functions with the supposed integers. But it is not safe for a 64-bit machine where pointers are not the same size as int. Of course, the code should include the correct headers (<stdio.h> for puts() and <stdlib.h> for system() and exit(), and <string.h> for strcpy()).
The exit(1); is bad on two separate counts.
It indicates failure - unconditionally. You exit with 0 or EXIT_SUCCESS to indicate success.
In my view, it is better to use return at the end of main() than exit(). Not everyone necessarily agrees with me, but I do not like to see exit() as the last line of main(). About the only excuse for it is to avoid problems from other bad practices, such as functions registered with atexit() that depend on the continued existence of local variables defined in main().
/usr/bin/gcc -g -std=c99 -Wall -Wextra -Wmissing-prototypes -Wstrict-prototypes -Wold-style-definition -c nasty.c
nasty.c: In function ‘main’:
nasty.c:3: warning: type defaults to ‘int’ in declaration of ‘system’
nasty.c:3: warning: type defaults to ‘int’ in declaration of ‘puts’
nasty.c:3: warning: built-in function ‘puts’ declared as non-function
nasty.c:8: warning: implicit declaration of function ‘strcpy’
nasty.c:8: warning: incompatible implicit declaration of built-in function ‘strcpy’
nasty.c:10: warning: implicit declaration of function ‘exit’
nasty.c:10: warning: incompatible implicit declaration of built-in function ‘exit’
nasty.c: At top level:
nasty.c:1: warning: unused parameter ‘argv’
Not good code! I worry about a source of information that contains such code and doesn't explain all the awfulness (because the only excuse for showing such messy code is to dissect it and correct it).
There's another weirdness in the code:
int main(int argv,char **argc)
That is 'correct' (it will work) but 100% aconventional. The normal declaration is:
int main(int argc, char **argv)
The names are short for 'argument count' and 'argument vector', and using argc as the name for the vector (array) of strings is abnormal and downright confusing.
Having visited the site referenced, you can see that it is going through a set of graduated examples. I'm not sure whether the author simply has a blind spot on the argc/argv issue or is deliberately messing around ('abo1' suggests that he is playing, but it is not helpful in my view). The examples are supposed to feed your mind, but there isn't much explanation of what they do. I don't think I could recommend the site.
Extension question
What's the cast in this code doing?
#include <stdio.h>
char shellcode[] = "some shellcode";
int main(void)
{
fprintf(stdout,"Length: %d\n",strlen(shellcode));
(*(void(*)()) shellcode)();
}
This takes the address of the string 'shellcode' and treats it as a pointer to a function that takes an indeterminate set of arguments and returns no values and executes it with no arguments. The string contains the binary assembler code for some exploit - usually running the shell - and the objective of the intruder is to get a root-privileged program to execute their shellcode and give them a command prompt, with root privileges. From there, the system is theirs to own. For practicing, the first step is to get a non-root program to execute the shellcode, of course.
Reviewing the analysis
The analysis at Mishou's web site is not as authoritative as I'd like:
One, this code uses the extern keyword in the C language to make the system and puts functions available. What this does (I think) is basically references directly the location of a function defined in the (implied) header files…I get the impression that GDB is auto-magically including the header files stdlib.h for system and stdio.h for puts. One thing that is not immediately clear is that the system and puts addresses are both written to the same location, I think that might be what gera is talking about “so the linker doesn’t remove it”.
Dissecting the commentary:
The first sentence isn't very accurate; it tells the compiler that the symbols system and puts are defined (as integers) somewhere else. When the code is linked, the address of puts()-the-function is known; the code will treat it as an integer variable, but the address of the integer variable is, in fact, the address of the function - so the cast forces the compiler to treat it as a function pointer after all.
The second sentence is not fully accurate; the linker resolves the addresses of the external 'variables' via the function symbols system() and puts() in the C library.
GDB has nothing whatsoever to do the compilation or linking process.
The last sentence does not make any sense at all. The addresses only get written to the same location because you have an initialization and an assignment to the same variable.
This didn't motivate me to read the whole article, it must be said. Due diligence forces me onwards; the explanation afterwards is better, though still not as clear as I think it could be. But the operation of overflowing the buffer with an overlong but carefully crafted argument string is the core of the operation. The code mentions both puts() and system() so that when run in non-exploit mode, the puts() function is a known symbol (otherwise, you'd have to use dlopen() to find its address), and so that when run in exploit mode, the code has the symbol system() available for direct use. Unused external references are not made available in the executable - a good thing when you realize how many symbols there are in a typical system header compared with the number used by a program that includes the header.
There are some neat tricks shown - though the implementation of those tricks is not shown on the specific page; I assume (without having verified it) that the information for getenvaddr program is available.
The abo3.c code can be written as:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char **argv)
{
void (*fn)(char*) = (void(*)(char*))system;
char buf[256];
fn = (void(*)(char*))puts;
strcpy(buf, argv[1]);
fn(argv[2]);
exit(1);
}
Now it compiles with only one warning with the fussy compilation options I originally used - and that's the accurate warning that 'argc' is not used. It is just as exploitable as the original; it is 'better' code though because it compiles cleanly. The indirections were unnecessary mystique, not a crucial part of making the code exploitable.
Both system and puts normally return int. The code is casting them to a pointer that returns void, presumably because they want to ignore whatever value is returned. This should be equivalent to using (void)fn(argc[2]); as the penultimate line if the cast didn't change the return type. Casting away the return type is sometimes done for callback functions, and this code snippet seems to be a simplistic example of a callback.
Why the cast for system if it is never used is beyond me. I'm assuming that there's more code that isn't shown here.

Why do I get a warning every time I use malloc?

If I use malloc in my code:
int *x = malloc(sizeof(int));
I get this warning from gcc:
new.c:7: warning: implicit declaration of function ‘malloc’
new.c:7: warning: incompatible implicit declaration of built-in function ‘malloc’
You need to add:
#include <stdlib.h>
This file includes the declaration for the built-in function malloc. If you don't do that, the compiler thinks you want to define your own function named malloc and it warns you because:
You don't explicitly declare it and
There already is a built-in function by that name which has a different signature than the one that was implicitly declared (when a function is declared implicitly, its return and argument types are assumed to be int, which isn't compatible with the built-in malloc, which takes a size_t and returns a void*).
You haven't done #include <stdlib.h>.
You need to include the header file that declares the function, for example:
#include <stdlib.h>
If you don't include this header file, the function is not known to the compiler. So it sees it as undeclared.
Make a habit of looking your functions up in help.
Most help for C is modelled on the unix manual pages.
Using :
man malloc
gives pretty useful results.
Googling man malloc will show you what I mean.
In unix you also get apropos for things that are related.
Beside the other very good answers, I would like to do a little nitpick and cover something what is not discussed yet in the other answers.
When you are at Linux, To use malloc() in your code,
You don´t actually have to #include <stdlib.h>.
(Although the use of stdlib.h is very common and probably every non-toy-program should include it either way because it provides a wide range of useful C standard library functions and macros)
You could also #include <malloc.h> instead.
But please note that the use of malloc.h is deprecated and it makes your code non-portable. If you want to use malloc() you should always and ever (except for explicit reasons to do otherwise) #include <stdlib.h>.
The reasons why, are best explained in the answers to this question:
difference between <stdlib.h> and <malloc.h>

Resources