Is there any way to compile C code with c89 standard NOT c99 in Xcode (or another way with terminal)?
I've searched in Xcode settings but I didn't find any way to choose compiler or standard.
You should add -pedantic-errors to Other C flags in your project settings, like so:
Of course, don't forget to set the C language dialect to C89 as well.
This will give you the appropriate compile time errors when you try to compile something that is not valid C89.
Optionally, if you want Xcode to compile your code regardless of incompatibilities, but only give you yellow warnings at the problematic lines, use -pedantic instead of -pedantic-errors.
In a nutshell, these flags make the compiler stick to the language standard more strictly, as opposed to the default behavior, which is to attempt compiling the code any way possible.
I hope this helps :)
Source
(even though they mention this in the context of GCC, but the same flags apply for Clang as well)
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I found out that structures without any named members (and without any members in particular, as I understood) invoke UB in C. Only GNU GCC supports such structures as an extension. I've tried to create a piece of code which will behave differently when compiled using GCC vs. Clang but with no success. I'm wondering if there is a C program that if you compile it via Clang, it will work differently than if it was compiled using GCC. The opposite option, in a nutshell, is that Clang also supports empty structures but this is not mentioned in documentation.
Literally the only thing I could find in clang's documentation is some info about a warning -Wgnu-empty-struct, which will be activated in -pedantic mode. So the compiler clearly supports GNU empty structs.
As for where to find the clang documentation regarding how empty structs should be used, what they are good for and under which language standards they are valid, I have no idea.
For completeness, as Lundin wrote, empty structs are supported in Clang.
I'm wondering if there is a C program that if you compile it via Clang, it will work differently than if it was compiled using GCC. The opposite option, in a nutshell, is that Clang also supports empty structures but this is not mentioned in documentation.
Undefined behavior means that it's undefined. The clang compiler is allowed to produce binary code as gcc even if it does not support empty structs. Be very careful to draw conclusions strictly from behavior.
I am trying to compile a C source code to a machine code using an ubunto terminal
My tutor instruction was to use the following command:
running clang myprogramm.c -std=c11
Why shall I use the keyword -std=c11 and what is the difference to using just
clang myprogramm.c
Using std= options is required by your tutor (I'm divinig her motives, I'm particularly good at this!) because she wants to make sure you stay away from all those nifty Clang features that turn the accepted language from C to A LANGUAGE SUPERFICIALLY LOOKING LIKE C BUT ACTUALLY A DIFFERENT LANGUAGE NOT SUPPORTED BY OTHER C COMPILERS.
That is more than just additional library functions. It include syntax changes that break the grammar of Standard C, as defined by ISO. A grasshopper should not use these while learning. Using -std=c11 makes sure Clang either warns about or even rejects, with an error, such constructs.
When to specify the standard? Whenever you use the compiler. It is never a good idea to let the compiler just use whatever it wants.
If someone tries to use a compiler that is too old, then they will get a warning or error, and they will understand why the compile fails.
If a code contributor (maybe even yourself!) tries to add code using features that are too new, their code will be rejected. That's very important if you intend to keep compatibility with an older standard.
By explicitly stating the standard, using new features or extensions are a choice and don't happen by accident.
I studied Option Summary for gfortran but found no compiler option to detect integer overflow. Then I found the GCC (GNU Compiler Collection) flag option -fsanitize=signed-integer-overflow here and used it when invoking gfortran. It works--integer overflow can be detected at run time!
So what does -fsanitize=signed-integer-overflow do here? Just adding to the machine code generated by gfortran some machine-level pieces that check integer overflow?
What is the relation between GCC (GNU Compiler Collection) flag options and gfortran compiler options ? What gcc compiler options can I use for gfortran, g++ etc ?
There is the GCC - GNU Compiler Collection. It shares the common backend and middleend and has frontends for different languages. For example frontends for C, C++ and Fortran which are usually invoked by commands gcc, g++ and gfortran.
It is actually more complicated, you can call gcc on a Fortran source and gfortran on a C source and it will work almost the same with the exceptions of libraries being linked (there are some other fine points). The appropriate frontend will be called based on the file extension or the language requested.
You can look almost all GCC (not just gcc) flags for all of the mentioned frontends. There are certain flags which are language specific. Normally you will get a warning like
gfortran -fcheck=all source.c
cc1: warning: command line option ‘-fcheck=all’ is valid for Fortran but not for C
but the file will compile fine, the option is just ignored and you will get a warning about that. Notice it is a C file and it is compiled by the gfortran command just fine.
The sanitization options are AFAIK not that language specific and work for multiple languages implemented in GCC, maybe with some exceptions for some obviously language specific checks. Especially -fsanitize=signed-integer-overflow which you ask about works perfectly fine for both C and C++. Signed integer overwlow is undefined behaviour in C and C++ and it is not allowed by the Fortran standard (which effectively means the same, Fortran just uses different words).
This isn't a terribly precise answer to your question, but an aha! moment, when learning about compilers, is learning that gcc (the GNU Compiler Collection), like llvm, is an example of a three-stage compiler.
The ‘front end’ parses the syntax of whichever language you're interested, and spits out an Abstract Syntax Tree (AST), which represents your program in a language-independent way.
Then the ‘middle end’ (terrible name, but ‘the clever bit’) reorganises that AST into another AST which is semantically equivalent but easier to turn into machine code.
Then the ‘back end’ turns that reorganised AST into assembler for one-or-other processor, possibly doing platform-specific micro-optimisations along the way.
That's why the (huge number of) gcc/llvm options are unexpectedly common to (apparently wildly) different languages. A few of the options are specific to C, or Fortran, or Objective-C, or whatever, but the majority of them (probably) are concerned with the middle and last bits, and so are common to all of the languages that gcc/llvm supports.
Thus the various options are specific to stage 1, 2 or 3, but may not be conveniently labelled as such; with this in mind, however, you might reasonably intuit what is and isn't relevant to the particular language you're interested in.
(It's for this sort of reason that I will dogmatically claim that CC++FortranJavaPerlPython is essentially a single language, with only trivial syntactical and library minutiae to distinguish between dialects).
I am trying to compile some C code on a beagleboard xm. I try to compile, but I get the error: undefined reference to 'isfinite' which is in the math.h library. This code compiles perfectly in all my other computers, and I do include -lm in my makefile.
I suspect that it may be my compiler, maybe it is an over version? On the beagleboard it's version 4.3.3, but on my computer it's 4.7.3, but I don't know how to get a later version. I thought opkg would automatically get the latest available.
Any ideas why this may be happening?
The function infinite() is part of C99. Your compiler is by default using an older version of the C language. You need to compile with the the flag -std=c99 to enable this macro.
`gnu89' GNU dialect of ISO C90 (including some C99 features). This is
the default for C code.
http://gcc.gnu.org/onlinedocs/gcc-4.3.3/gcc/C-Dialect-Options.html
isfinite is part of the C++11 standard and gcc 4.3.3 is to old to know that. Try
int finite(double x); or int finitef(float x);.
I recently realized that I am not even in C99 mode after receiving the compile error
'for' loop initial declarations are only allowed in C99 mode
I found some advice on how to get to C99 via a quick search which has told me to go to Projects -> Properties... But alas, it is greyed out and I am not sure that is even the correct way to fix it (probably not available because my file is not a project, it is a normal source file). I have also seen a lot of similar questions saying to enable C99 mode so I have looked inside the compiler flags menu, but I cannot see anything about C99. I have tried some other flags such as In C Mode, support all ISO C90 programs..., but after I set this flag, I got more errors than I had before which seem to appear whenever the compiler finds comments inside main().
Note: Please don't just say to initialize the counter outside the for loop.
Update: While trying to compile outside of codeblocks with gcc, I tried
gcc -O2 -std=C99 filename.c, but received an error:
unrecognized command line option "-std=C99"
I use 64-bit Windows 7, CodeBlocks10.05, and GNU gcc.
For future reference, type in the flag -std=c99 in settings->compiler->other options which is not case-sensitive, however when compiling in a terminal the flag is case-sensitive. Thanks chris!