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

C99 and C++ standards don’t require functions to return a value. The missing return statement in a value-returning function will be defined (to return 0) only in the main function.

The rationale includes that checking if every code path returns a value is quite difficult, and a return value could be set with embedded assembler or other tricky methods.

From C++11 draft:

§ 6.6.3/2

Flowing off the end of a function […] results in undefined behavior in a value-returning function.

§ 3.6.1/5

If control reaches the end of main without encountering a return statement, the effect is that of executing

return 0;

Note that the behaviour described in C++ 6.6.3/2 is not the same in C.

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gcc will give you a warning if you call it with -Wreturn-type option.

-Wreturn-type Warn whenever a function is defined with a return-type that
defaults to int. Also warn about any
return statement with no return-value
in a function whose return-type is not
void (falling off the end of the
function body is considered returning
without a value), and about a return
statement with an expression in a
function whose return-type is void.

This warning is enabled by -Wall.

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Just as a curiosity, look what this code does:

#include <iostream>

int foo() {
   int a = 5;
   int b = a + 1;
}

int main() { std::cout << foo() << std::endl; } // may print 6

This code has formally undefined behaviour, and in practice it’s calling convention and architecture dependent. On one particular system, with one particular compiler, the return value is the result of last expression evaluation, stored in the eax register of that system’s processor.