Why does a function returning `int` in C++ compile even if I didn't include the return statement?

Question

Why does a C++ function that returns a numeric value such as int or double work just fine if I don't include the return statement? Should I avoid that even though the program runs fine? Thanks

How did you check that it works fine? And yes, you should avoid such as thing and you will get some warnings from the compiler. — MikeCAT, Nov 26 '15 at 14:22
Your questions have already been answered : [link](http://stackoverflow.com/questions/1610030/why-can-you-return-from-a-non-void-function-without-returning-a-value-without-pr) — SwissFr, Nov 26 '15 at 14:24

juanchopanza · Answer 1 · 2015-11-26T14:35:03.423

3

It could "work fine" for two reasons:

Your function is not main(), in which case flowing off the end with no return statement causes undefined behaviour¹, and it just happens that it seems to work fine. But it doesn't really.
The function is main(), which has an implicit return 0;.

For case 1, you should consider your code badly broken and fix it.

^{1 §6.3.3 [stmt.return] in the C++11 standard}

edited Nov 26 '15 at 14:35

answered Nov 26 '15 at 14:21

juanchopanza

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Nice answer! Maybe also just worth adding a little note about how `main` can omit the return statement (just for completeness). – Nov 26 '15 at 14:23
@Ike implicitly it is there, because return value of main isnt used – 463035818_is_not_an_ai Nov 26 '15 at 14:24
In C++, it's UB even if the return value is not used. – cpplearner Nov 26 '15 at 14:24
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@Ike Good point. "Nobody expects the Spanish inquisition!" – juanchopanza Nov 26 '15 at 14:24
@cpplearner Found it, 6.6.3. – juanchopanza Nov 26 '15 at 14:33
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"Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function." _§6.3.3_ – masoud Nov 26 '15 at 14:37

Why does a function returning `int` in C++ compile even if I didn't include the return statement?

1 Answers1