implicit int and implicit declaration of functions with gcc compiler

问题

I read in the c99 Standard:

-remove implicit function declaration,

-remove implicit int.

But when I try to compile this code with gcc compiler in c99 mode using -pedantic

main(void){
    f(3);
    return 0;
}


int f(int a){
    ....
}

I expect 2 errors, but I just receive 2 warnings:

-warning: return type defaults to ‘int’

-warning: implicit declaration of function ‘f’.

Shouldn't them be errors in c99?

http://gcc.gnu.org/c99status.html In both situations there's written "done".

Thanks.

回答1:

The C standard requires a diagnostic for any translation unit containing a violation of a syntax rule or constraint. It does not require such diagnostics to be fatal; the compiler is free to continue processing the source file. The behavior of the resulting executable, if any, is undefined. The standard makes no distinction between warnings and fatal errors.

(The only thing that requires a compiler to reject a source file is the #error directive.)

Conclusion: when compiling C, take warnings very seriously.

回答2:

I don't believe the compiler is required to produce a fatal error. Use -Werror if you're concerned...

回答3:

Two points: first, it may (usually does) take a specific set of flags to get a compiler to conform with the standard.

Second, all that's required by the standard is that the implementation issue a "diagnostic" in the case of an error -- but it's up to the implementation to define what is or isn't a diagnostic. It's free to say a "warning" is a diagnostic if it wants to. When a diagnostic is issued, it may quit compiling, or it can compile the code anyway.

Bottom line: what it's doing is probably enough to conform, for whatever that's worth.

来源：https://stackoverflow.com/questions/11171120/implicit-int-and-implicit-declaration-of-functions-with-gcc-compiler

标签

c99