I have successfully installed MinGW on a Windows 7 32bit machine, and have tried to compile a simple program using either the command line or the MinGW console.
The code has an intentional error in a printf statement:
#include <stdio.h>
#include <stdlib.h>
int main( void )
{
printf("%d\n" , 3.14 ) ;
return 0 ;
}
The command gcc -Wall hello.c gives a correct warning: hello.c:7:2: warning: format '%d' expects argument of type 'int'...
But the command gcc -std=c99 -Wall hello.c doesn't give any warning.
Both create an executable a.exe ( that runs and gives the same result ).
(Interestingly a command gcc -std=gnu99 -Wall hello.c gives the warning.)
I don't know if this is a bug, or did the installation go wrong somehow, but both seem unlikely since the compiler works and successfully compiled a larger project( but the same warning of course omitted when using -std=c99 ).
I must be missing some information.
(ps: If someone has a new MinGW install, please test this.)
gcc version 4.8.1 (GCC)
Update 1:
Defining _GNU_SOURCE before including stdio.h removes the warning even with gcc -Wall hello.c.
Update 2( might be less relevant ):
Compiling
printf("%lf\n" , 3.14 ) ;
-std=c99 flag outputs: 0.000000
-std=gnu99 outputs: 3.140000
And compiling:
printf("%f\n" , 3.14 ) ;
-std=gnu99 and -std=c99 output: 3.140000
Update 3:
Functions that seem to be affected are: printf, fprintf, snprintf, sprintf.
The problem with the lack of warning when using the std=c99 option looks like it's because MinGW 4.8.1 preprocesses stdio.h a little different for the printf() family of functions when -std=c99 is used compared to when -std=gnu99 is used.
Note: I'm looking at MinGW 4.8.1 from TDM - I think other distributions might differ in these details.
MinGW has had some compatibility issues with formatting floating point values because of its historic reliance on msvcrt.dll for the C runtime and the fact that MSVC uses a 64-bit representation for long double while gcc uses a 96-bit (or 128-bit on x64) representation. See gcc: printf and long double leads to wrong output. [C - Type conversion messes up] for some details. More recent versions of MinGW have provided thier own implementation of the printf() family of functions (with a __mingw_ prefix on the name) in libmingwex.a to solve those problems.
The header files _mingw.h and stdio.h configure whether or not the libmingwex.a implementations or the msvcrt.dll implementations will be used.
It appears that if ANSI compliance is requested, MinGW will use the libmingwex.a implementations (there are a number of other ways to get this configuration too - look at the headers for details). Wiring up a user call to printf() to the __mingw_printf() implementation in libmingwex.a is done by stdio.h defining a static inline implementation of printf() that is a thin wrapper around a call to __mingw_vfprintf(). Apparently the -Wformat doesn't get applied to versions of printf() family functions that the compiler doesn't believe to be part of the library (a reasonable assumption - the compiler doesn't really know anything about those functions). This problem can be fixed by applying the appropriate function attribute (for example: __attribute__ ((format (printf, 1, 2)))) to the static inline wrapper functions.
The other problem you found, where printf("%lf\n", 3.14) prints 0.000000 when using std=c99, looks to be a bug in the libmingwex.a implementation of __mingw_vfprintf(). It seems that __mingw_vfprintf() mistakenly interprets "%lf" to mean the argument is a long double. I'm not too surprised by that - I always have to look up whether %lf means double or long double.
Related
I'm new to C and am learning C from Programming in C, 4th ed. by Stephen Kochan. On page 29, he writes $ is not a valid character for variable names. He is using the C11 standard.
I wrote the following code
#include <stdio.h>
int main (void)
{
int a$ = 1;
printf ("%i", a$);
return 0;
}
and ran it with the command gcc -std=c11 -pedantic practice.c -o practice.o && ./practice.o. My filename is practice.c.
The output is 1. Shouldn't the compiler give me a warning for using $? Isn't using $ sign for identifiers an extension that GCC provides?
I'm using GCC 8.2.0 in Ubuntu 18.10.
Edit:
Also, doesn't GCC not use the GNU extensions when I use -std=c11? That is what is written in the Appendix of the book (pg. no. 497).
I am getting an warning by using -std=c89 though.
You get a warning with -std=c89 -pedantic. C99 and later allow other implementation-defined characters in identifiers.
According to this : GCC Documentation
In GNU C, you may normally use dollar signs in identifier names. This
is because many traditional C implementations allow such identifiers.
However, dollar signs in identifiers are not supported on a few target
machines, typically because the target assembler does not allow them.
So, $ is valid, but it's not a conforming way to code in C.
How a standard library like libc.a (static library) which is included using #include <stdio.h> in our main.c differ from user defined header file (cube.h) included in main.c with its implementation file (cube.c) in C ?
I mean both are header files but one's implementation is a static library (.a) and others is source file (.c) .
You would have the definition (implementation) in, say, cube.c
#include "cube.h"
int cube( int x ) {
return x * x * x;
}
Then we'll put the function declaration in another file. By convention, this is done in a header file, cube.h in this case.
int cube( int x );
We can now call the function from somewhere else, main.c for instance, by using the #include directive (which is part of the C preprocessor) .
#include "cube.h"
#include <stdio.h>
int main() {
int c = cube( 10 );
printf("%d", c);
...
}
Also if I included include guards in cube.h what would happen when I include cube.h in both main.c and cube.c . Where it will get included?
A programming language is not the same as its implementation.
A programming language is a specification (written on paper; you should read n1570, which practically is the C11 standard), it is not a software. The C standard specifies a C standard library and defines the headers to be #include-d.
(you could run your C program with a bunch of human slaves and without any computers; that would be very unethical; you could also use some interpreter like Ch and avoid any compiler or object or executable files)
How a standard library like libc.a (static library) which is included using #include <stdio.h> ... differs from a user file cube.c
The above sentence is utterly wrong (and makes no sense). libc.a does not #include -or is not included by- the <stdio.h> header (i.e. file /usr/include/stdio.h and other internal headers e.g. /usr/include/bits/stdio2.h). That inclusion happens when you compile your main.c or cube.c.
In principle, <stdio.h> might not be any file on your computer (e.g. #include <stdio.h> could trigger some magic in your compiler). In practice, the compiler is parsing /usr/include/stdio.h (and other included files) when you #include <stdio.h>.
Some standard headers (notably <setjmp.h>, <stdreturn.h>, <stdarg.h>, ....) are specified by the standard but are implemented with the help of special builtins or attributes (that is "magic" things) of the GCC compiler.
The C standard knows about translation units.
Your GCC compiler processes source files (grossly speaking, implementing translation units) and starts with a preprocessing phase (processing #include and other directives and expanding macros). And gcc runs not only the compiler proper (some cc1) but also the assembler as and the linker ld (read Levine's Linkers and Loaders book for more).
For good reasons, your header file cube.h should practically start with include guards. In your simplistic example they are probably useless (but you should get that habit).
You practically should almost always use gcc -Wall -Wextra -g (to get all warnings and debug info). Read the chapter about Invoking GCC.
You may pass also -v to gcc to understand what programs (e.g. cc1, ld, as) are actually run.
You may pass -H to gcc to understand what source files are included during preprocessing phase. You can also get the preprocessed form of cube.c as the cube.i file obtained with gcc -C -E cube.c > cube.i and later look into that cube.i file with some editor or pager.
You -or gcc- would need (in your example) to compile cube.c (the translation unit given by that file and every header files it is #include-ing) into the cube.o object file (assuming a Linux system). You would also compile main.c into main.o. At last gcc would link cube.o, main.o, some startup files (read about crt0) and the libc.so shared library (implementing the POSIX C standard library specification and a bit more) to produce an executable. Relocatable object files, shared libraries (and static libraries, if you use some) and executables use the ELF file format on Linux.
If you code a C program with several source files (and translation units) you practically should use a build automation tool like GNU make.
If I included include guards in cube.h what would happen when I include cube.h in both main.c and cube.c ?
These should be two different translation units. And you would compile them in several steps. First you compile main.c into main.o using
gcc -Wall -Wextra -g -c main.c
and the above command is producing a main.o object file (with the help of cc1 and as)
Then you compile (another translation unit) cube.c using
gcc -Wall -Wextra -g -c cube.c
hence obtaining cube.o
(notice that adding include guards in your cube.h don't change the fact that it would be read twice, once when compiling cube.c and the other time when compiling main.c)
At last you link both object files into yourprog executable using
gcc -Wall -Wextra -g cube.o main.o -o yourprog
(I invite you to try all these commands, and also to try them with gcc -v instead of gcc above).
Notice that gcc -Wall -Wextra -g cube.c main.c -o yourprog is running all the steps above (check with gcc -v). You really should write a Makefile to avoid typing all these commands (and just compile using make, or even better make -j to run compilation in parallel).
Finally you can run your executable using ./yourprog (but read about PATH), but you should learn how to use gdb and try gdb ./yourprog.
Where it cube.h will get included?
It will get included at both translation units; once when running gcc -Wall -Wextra -g -c main.c and another time when running gcc -Wall -Wextra -g -c cube.c. Notice that object files (cube.o and main.o) don't contain included headers. Their debug information (in DWARF format) retains that inclusion (e.g. the included path, not the content of the header file).
BTW, look into existing free software projects (and study some of their source code, at least for inspiration). You might look into GNU glibc or musl-libc to understand what a C standard library really contains on Linux (it is built above system calls, listed in syscalls(2), provided and implemented by the Linux kernel). For example printf would ultimately sometimes use write(2) but it is buffering (see fflush(3)).
PS. Perhaps you dream of programming languages (like Ocaml, Go, ...) knowing about modules. C is not one.
TL;DR: the most crucial difference between the C standard library and your library function is that the compiler might intimately know what the standard library functions do without seeing their definition.
First of all, there are 2 kinds of libraries:
The C standard library (and possibly other libraries that are part of the C implementation, like libgcc)
Any other libraries - which includes all those other libraries in /usr/lib, /lib, etc.., or those in your project.
The most crucial difference between a library in category 1 and a library in category 2 library is that the compiler is allowed to assume that every single identifier that you use from category 1 library behaves as if it is the standard library function and behaves as if in the standard and can use this fact to optimize things as it sees fit - this even without it actually linking against the relevant routine from the standard library, or executing it at the runtime. Look at this example:
% cat foo.c
#include <math.h>
#include <stdio.h>
int main(void) {
printf("%f\n", sqrt(4.0));
}
We compile it, and run:
% gcc foo.c -Wall -Werror
% ./a.out
2.000000
%
and correct result is printed out.
So what happens when we ask the user for the number:
% cat foo.c
#include <math.h>
#include <stdio.h>
int main(void) {
double n;
scanf("%lf\n", &n);
printf("%f\n", sqrt(n));
}
then we compile the program:
% gcc foo.c -Wall -Werror
/tmp/ccTipZ5Q.o: In function `main':
foo.c:(.text+0x3d): undefined reference to `sqrt'
collect2: error: ld returned 1 exit status
Surprise, it doesn't link. That is because sqrt is in the math library -lm and you need to link against it to get the definition. But how did it work in the first place? Because the C compiler is free to assume that any function from standard library behaves as if it was as written in the standard, so it can optimize all invocations to it out; this even when we weren't using any -O switches.
Notice that it isn't even necessary to include the header. C11 7.1.4p2 allows this:
Provided that a library function can be declared without reference to any type defined in a header, it is also permissible to declare the function and use it without including its associated header.
Therefore in the following program, the compiler can still assume that the sqrt is the one from the standard library, and the behaviour here is still conforming:
% cat foo.c
int printf(const char * restrict format, ...);
double sqrt(double x);
int main(void) {
printf("%f\n", sqrt(4.0));
}
% gcc foo.c -std=c11 -pedantic -Wall -Werror
% ./a.out
2.000000
If you drop the prototype for sqrt, and compile the program,
int printf(const char * restrict format, ...);
int main(void) {
printf("%f\n", sqrt(4));
}
A conforming C99, C11 compiler must diagnose constraint violation for implicit function declaration. The program is now an invalid program, but it still compiles (the C standard allows that too). GCC still calculates sqrt(4) at compilation time. Notice that we use int here instead of double, so it wouldn't even work at runtime without proper declaration for an ordinary function because without prototype the compiler wouldn't know that the argument must be double and not the int that was passed in (without a prototype, the compiler doesn't know that the int must be converted to a double). But it still works.
% gcc foo.c -std=c11 -pedantic
foo.c: In function ‘main’:
foo.c:4:20: warning: implicit declaration of function ‘sqrt’
[-Wimplicit-function-declaration]
printf("%f\n", sqrt(4));
^~~~
foo.c:4:20: warning: incompatible implicit declaration of built-in function ‘sqrt’
foo.c:4:20: note: include ‘<math.h>’ or provide a declaration of ‘sqrt’
% ./a.out
2.000000
This is because an implicit function declaration is one with external linkage, and C standard says this (C11 7.1.3):
[...] All identifiers with external linkage in any of the following subclauses (including the future library directions) and errno are always reserved for use as identifiers with external linkage. [...]
and Appendix J.2. explicitly lists as undefined behaviour:
[...] The program declares or defines a reserved identifier, other than as allowed by 7.1.4 (7.1.3).
I.e. if the program did actually have its own sqrt then the behaviour is simply undefined, because the compiler can assume that the sqrt is the standard-conforming one.
I was reading a book on GCC. It said only the standard library is linked by default for any C program. Since the pow() is not in the standard library, I will have to link to it using the -lm flag. However, when I compiled, I simply used:
gcc hello.c -o hello
and it still worked..
And there is another similar problem, the book also said that if you have printf("%f\n", 4); in your C program and if you compile WITHOUT -Wall option, no warning will be issued. However, I tried compiling it without the -Wall option but I still got a warning:
hello.c:6:2: warning: format ‘%f’ expects argument of type ‘double’,
but argument 2 has type ‘int’ [-Wformat]
Why is this? The book said I have to supply -lm and -Wall in order to make my program compiled and get the warning but I did not use either of them but I still got my program compiled and got the warning?
Thank you!
A bit more information based on some experimentation.
Consider this program:
#include <stdio.h>
#include <math.h>
int main(void) {
#ifdef CONSTANT
double x = pow(2.0, 10.0);
#else
double expon = 10.0;
double x = pow(2.0, expon);
#endif
printf("x = %f\n", x);
return 0;
}
On Ubuntu, when I compile it with
$ gcc c.c -o c
it complains about an undefined reference to pow; adding -lm corrects it.
But this:
$ gcc -DCONSTANT c.c -o c
compiles and links without error, replacing the pow() call with a constant 1024.0. But this:
$ gcc -fno-builtin -DCONSTANT c.c -o c
complains again about the undefined reference to pow.
Conclusion: gcc uses the built-in implementation of pow only when the result can be determined at compile time. Otherwise it generates an explicit call to the pow() function, which requires linking with -lm.
Note also that this depends on the way the C library is organized (the library is provided separately; it's not part of gcc). On Cygwin, which uses newlib rather than the glibc used on Ubuntu, the -lm option is not needed; apparently the math routines are an integral part of the standard library rather than being provided separately. (gcc on Cygwin still accepts the -lm option.)
GCC supplies several standard library functions as built-ins:
GCC provides a large number of built-in functions other than the ones mentioned above. Some of these are for internal use in the processing of exceptions or variable-length argument lists and will not be documented here because they may change from time to time; we do not recommend general use of these functions.
The remaining functions are provided for optimization purposes.
If you look at the list of built-ins, you'll see that pow is one of them.
If you add -fno-builtin to your compiler options, you should get the linker error that you're expecting.
I have the following C code:
#include <math.h>
int main(int argc, char ** argv)
{
double mydouble = 100.0;
double whatever = round(mydouble);
return (int) whatever;
}
When I compile this, I get the warnings:
round_test.c: In function ‘main’:
round_test.c:6: warning: implicit declaration of function ‘round’
round_test.c:6: warning: incompatible implicit declaration of built-in function ‘round’
I'm rusty with C, but I thought that the #include brought a declaration for round() into scope. I've checked my ANSI standard (C99 is the only copy I have) which confirms that the round() function exists in the math.h header. What am I missing here?
Edit: The compiler is GCC 4.3.2 on Ubuntu (intrepid, IIRC). Running gcc -E gives:
$ gcc -E round_test.c | grep round
# 1 "round_test.c"
# 1 "round_test.c"
# 2 "round_test.c" 2
double whatever = round(mydouble);
so the definition obviously isn't being found in the headers.
I see you're using gcc.
By default, gcc uses a standard similar to C89. You may want to "force" it to use the C99 standard (the parts it complies with)
gcc -std=c99 -pedantic ...
Quote from GCC Manual
By default, GCC provides some
extensions to the C language that on
rare occasions conflict with the C
standard. See Extensions to the C
Language Family. Use of the -std
options listed above will disable
these extensions where they conflict
with the C standard version selected.
You may also select an extended
version of the C language explicitly
with -std=gnu89 (for C89 with GNU
extensions) or -std=gnu99 (for C99
with GNU extensions). The default, if
no C language dialect options are
given, is -std=gnu89; this will change
to -std=gnu99 in some future release
when the C99 support is complete. Some
features that are part of the C99
standard are accepted as extensions in
C89 mode.
Something must be wrong with your gcc installation, system headers, or compilation options.
Try compiling with -E. That will show you what the preprocessor output -- including which headers are being included and what's in them. On my Ubuntu Linux system it's about 1000 lines of output, including this:
extern double round (double __x) __attribute__ ((__nothrow__)) __attribute__ ((__const__));
You need to tell gcc that you want C99, and that you want to link in libm:
gcc -std=c99 -lm round_test.c
The code you type compiles cleanly on MacOS X 10.5.8 with GCC 4.0.1. If prodded with options '-Wall -Wextra', it complains about unused parameters argc and argv - not material.
Have you looked in <math.h> on your machine?
Have you tried with options such as '-stc=c99'?
C99 was the answer, but the full story is a little more complicated. The reason I'd been playing with this at all was that I was trying to compile a library written for Windows, which had its own "optimised" definition of round(). I got a linker error telling me that the definition conflicted with the built-in, so I removed the definition (and declaration). Once I'd done that I started to get the "implicit declaration error".
It seems that the default compile mode (without the -std=c99 flag) is neither conforming C89 nor C99: if it were conforming C89, you should be able to provide a custom definition of round() without conflicting, and if it were conforming C99 the declaration should be in math.h.
you need to link with the math library. So when you compile, be sure to add the -lm flag.
I do "#include <stdlib.h>" at the top of the source.
Example compilation:
/usr/bin/colorgcc -std=c99 -fgnu89-inline -g -Wall -I/usr/include -I./ -I../ -I../../ -I../../../ -I../../../../ -O3 -o f8 f8.c
In file included from f8.c:7:
ctype-cmp.c: In function ‘randomized’:
ctype-cmp.c:48: warning: implicit declaration of function ‘random’
ctype-cmp.c: In function ‘main’:
ctype-cmp.c:153: warning: implicit declaration of function ‘srandom’
ais#xcalibur:t$
When I turn off -std=c99, the function isfinite() can not be found. So I do want to use -std=c99 for this and other reasons. Is there some trick I'm missing?
man srandom says that the function is not part of C99 but part of POSIX.
Activate _BSD_SOURCE or _XOPEN_SOURCE >= 500 or any other suitable feature test macro that declares the srandom/random function (see man feature_test_macros and man srandom).
This one has good chances, but you need to figure out the macros that are defined/not defined implicitly thereby too by reading the manpages above.
/usr/bin/colorgcc -std=c99 -D_XOPEN_SOURCE=600 -fgnu89-inline -g -Wall
-I/usr/include -I./ -I../ -I../../ -I../../../ -I../../../../ -O3 -o f8 f8.c
Yes, there is a trick you are missing: you can use -std=gnu99 instead of -std=c99.
-std=c99 #defines __STRICT_ANSI__, which /usr/include/features.h interprets as "do not enable anything outside the C standard by default" (without it, you get at least both _SVID_SOURCE and _BSD_SOURCE). -std=gnu99, on the other hand, means "C99 plus GNU extensions" (the gcc default is currently -std=gnu89, its C89 equivalent, which is why you needed to specify something to get the new C99 features).
As an alternative, you can enable the feature test macros (as mentioned in #litb's answer). Looking at /usr/include/stdlib.h in my system, it expects one of __USE_SVID, __USE_XOPEN_EXTENDED, or __USE_BSD. /usr/include/features.h tells me that the feature test macros which would enable these are:
_SVID_SOURCE (enables __USE_SVID)
_BSD_SOURCE (enables __USE_BSD)
_XOPEN_SOURCE with a value of at least 500 (enables __USE_XOPEN_EXTENDED)
_XOPEN_SOURCE_EXTENDED (also enables __USE_XOPEN_EXTENDED)
_GNU_SOURCE (enables everything, including the four feature test macros above)
For new programs where you are not too concerned about potential name collisions with new functions from future standards, using both -std=gnu99 and -D_GNU_SOURCE is a good idea. It allows you to use all the new standard features and GNU extensions, which combined with some sort of fallback (for instance, autoconf-style feature tests) gives the most flexibility.
References:
Options Controlling C Dialect
Feature Test Macros
I've created random numbers using gcc in CodeBlocks under Ubuntu 9.10 (with compiler options: -std=gnu99 -D_GNU_SOURCE) So this worked for me:
This is my code I had played with:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
enum computer {keyboard, CPU, screen, printer};
int main(void)
{
enum computer comp;
time_t casovac;
comp = CPU;
srand(&casovac);
printf("%d", rand());
return 0;
}
This was only idea, of course you can accomplish it by other ways ;-) [To install CodeBlocks use: sudo apt-get install build-essential and then sudo apt-get install codeblocks]
I use rand() and srand(). BTW: Did you forget a header or two? At least the second warning tells me so.
Try including math.h. (Just remembered we always had issues with math library and had to actually force link it with -lm).