Why does flowing off the end of a non-void function without returning a value not produce a compiler error?

Question

Ever since I realized many years ago, that this doesn\'t produce an error by default (in GCC at least), I\'ve always wondered why?

I understand that you can issue co

Answer 1

It is a constraint violation in c99, but not in c89. Contrast:

c89:

3.6.6.4 The return statement

Constraints

A return statement with an expression shall not appear in a function whose return type is void .

c99:

6.8.6.4 The return statement

Constraints

A return statement with an expression shall not appear in a function whose return type is void. A return statement without an expression shall only appear in a function whose return type is void.

Even in --std=c99 mode, gcc will only throw a warning (although without needing to enable additional -W flags, as is required by default or in c89/90).

Edit to add that in c89, "reaching the } that terminates a function is equivalent to executing a return statement without an expression" (3.6.6.4). However, in c99 the behavior is undefined (6.9.1).