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Return value of function when return path not specified
Thinking back decades to when I took my first "C" class, I seem to remember that in the absence of a return statement in a function, the value of the last expression on the stack was returned. In that case, yes, it would be guaranteed behavior.
However, this is fuzzy in my memory, and putting in a return statement is so much better in so many ways, so I never do it without the return statement. I can't be 100% certain that I have my facts straight on this one--but yes, I think the @OP is correct.
But why? Put in the return statement and make it clear what the code is trying to do. Six keystrokes aren't worth the potential headache.
However, this is fuzzy in my memory, and putting in a return statement is so much better in so many ways, so I never do it without the return statement. I can't be 100% certain that I have my facts straight on this one--but yes, I think the @OP is correct.
But why? Put in the return statement and make it clear what the code is trying to do. Six keystrokes aren't worth the potential headache.
closed account (z05DSL3A)
| seeplus wrote: |
|---|
| C++ isn't c...... |
The original post has been removed again........... Who's doing all the reporting...........
Just for completeness sake...
ISO/IEC 9899 is the C standard, as far as I know.
ISO/IEC 9899:TC3 draft section 6.8.6.4, "The return statement", item #12 states,
So there you go. If your foo function reaches }, and you try to use its result, the behavior is undefined. And I'm sure the C++ standard says something along the same lines.
A particular compiler might guarantee behavior that goes beyond that standard, as an extension, but it's still undefined behavior per the standard. But C existed before the ISO standard, and I don't know how things like that worked before standardization. It could be that compilers guaranteed a particular behavior. But in modern C/C++, you certainly cannot assume this.
ISO/IEC 9899 is the C standard, as far as I know.
ISO/IEC 9899:TC3 draft section 6.8.6.4, "The return statement", item #12 states,
| If the } that terminates a function is reached, and the value of the function call is used by the caller, the behavior is undefined. |
So there you go. If your foo function reaches }, and you try to use its result, the behavior is undefined. And I'm sure the C++ standard says something along the same lines.
A particular compiler might guarantee behavior that goes beyond that standard, as an extension, but it's still undefined behavior per the standard. But C existed before the ISO standard, and I don't know how things like that worked before standardization. It could be that compilers guaranteed a particular behavior. But in modern C/C++, you certainly cannot assume this.
Last edited on
The difference is:
in C, the undefined behaviour is only if the caller tries to use the result;
in C++, it is undefined behaviour even if the caller ignores the result. https://eel.is/c++draft/stmt.return#4
This difference is understandable; in C++, the return type may be one with a non-trivial destructor.
in C, the undefined behaviour is only if the caller tries to use the result;
in C++, it is undefined behaviour even if the caller ignores the result. https://eel.is/c++draft/stmt.return#4
This difference is understandable; in C++, the return type may be one with a non-trivial destructor.
Meaningless for established users, horrible for low post count new users who get reported and have their posts deleted.
Looks like the OP has revealed their true colours with the new edit to include the spammy URL.
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