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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.

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  • 8
    Unrelated, but the .o extension is usually used for object files, not for the final executable. Commented Mar 10, 2019 at 11:23
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    There's no extension for executables in *nix systems. The filesystem doesn't use that to determine the type of a file. So usually executables just don't have extensions and practice is correct. Check your /usr/bin directory and you'll see that the programs there don't have an extension either. Commented Mar 10, 2019 at 11:37
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    About the fact that its properties say "shared library" is probably because of your desktop environment. If I do file practice from the command line I get practice: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=928002f23b27d5c9bc55a15bf769edfaf3e62c23, not stripped Commented Mar 10, 2019 at 11:40
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    The file utility can show shared object for an ELF executable if it is a position-independent executable. Some distributions configure GCC so that it creates position-independent executables by default (usually this requires using -pie and -fPIC options). Commented Mar 10, 2019 at 15:06
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    @MrLister: Although traditional BASIC implementations used $ as a suffix for string-variable names, I wouldn't expect a trailing $ to have similar meanings in C. If I was examining code that used such a suffix, I'd expect that the programmer was exploiting some special way that the target implementation would process identifiers with such a suffix. For example, a compiler targeting a platform which can access objects near the frame pointer faster than those which are further away might place all objects whose names have a trailing $ after those that don't. Commented Mar 10, 2019 at 17:41

2 Answers 2

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11

You get a warning with -std=c89 -pedantic. C99 and later allow other implementation-defined characters in identifiers.

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1 Comment

Yes. This is the reference: gcc.gnu.org/onlinedocs/gcc-8.2.0/cpp/…
2

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.

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4 Comments

It's not valid C, only on GCC C. Try compiling with -ansi or -std=C11 and the warnings will start appearing.
@Spidey with -std=c11 no warning appears
@Spidey nope, still no warnings.
@Spidey: The Standard has no definition for "valid C". Its definition for a "conforming C program" encompasses any blob of text that is acceptable to at least one conforming C implementation. A source text that uses dollar signs in identifiers could not be a "strictly conforming proogram", but the authors of the Standard recognize that much C's usefulness stems from the ability to write non-portable programs that will be usefully processed by some C implementations even if not by all of them.

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