CODEWITHSUNDEEP
cgccc89compiler-flagsansi-c
finefoot
finefoot

Reputation: 11232

Do I even need `-pedantic` if I specify the standard to ANSI C with `-std=c89`?

If I specify the standard to ANSI C with -std=c89, my code won't run until I perform certain changes to make it compliant with the standard. So do I even need -pedantic at this point if I've already set the -std=c89 flag?

By the way, the idea was to write C code which is as platform independet as possible. I was already using -pedantic as I knew it would make the compiler more strict. However, it also made sense to explicitly choose the ANSI C standard. For some reason I thought this would make -pedantic superfluous because the switch to ANSI C produced many errors in itself and appeared "strict enough".

Upvotes: 3

Views: 824

Answers (1)

finefoot
finefoot

Reputation: 11232

No, even though both -pedantic and -std=c89 might individually cause your code to fail to compile, they don't do the same thing:

  • -std=c89 will tell the compiler which ISO standard to use (in this case ANSI C also known as C89 or C90)
  • -pedantic will tell the compiler how strict the chosen standard should be enforced

You can also find the following helpful hint in the man page. -std=c89 is equivalent to -ansi for which man gcc says:

The -ansi option does not cause non-ISO programs to be rejected gratuitously. For that, -pedantic is required in addition to -ansi.

Upvotes: 1

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