Which section in C89 standard allows the “implicit int” rule?

问题

While using gcc, the code:

register a = 3;
static b = 3;

it is allowed while using the -std=c89 -pedantic-errors flags, although there is a warning.

However it receive an error with the -std=c99 -pedantic-errors flags.

I wonder which section of the C89 standards allows the "implicit int" rule?

回答1:

The section that allowed the implicit int rule in C89 would be section 3.5.2 Type specifiers which says (emphasis mine):

int , signed , signed int , or no type specifiers

Keith Thompson in the comments points out that in C90 the section is 6.5.2 and says, The only difference is some introductory material required by ISO, resulting in a renumbering of the sections.

In C99 where this changed, the section is 6.7.2 Type specifiers and it says:

int, signed, or signed int

This is also covered in document N661: Disallow implicit "int" in declarations which says:

Change in 6.5.2 Type specifiers; add new sentence at beginning of first paragraph of Constraints: At least one type specifier shall be given in the declaration specifiers in a declaration.

    Change in 6.5.2 Type specifiers, Constraints, from:
            -- int, signed, signed int, or no type
               specifiers
    to:
            -- int, signed, or signed int

来源：https://stackoverflow.com/questions/26488502/which-section-in-c89-standard-allows-the-implicit-int-rule