why return value of function can be assigned to?

Question

Consider the following function:

 char *f()
 {   
 char *s=malloc(8);
 }
 main()
 {
  printf("%c",*f()='A');
 }

If I comment the line char *s=malloc(8); I get an error as if the assignment *f()='A' accessed invalid memory. Since I never return any variable why does above assignment work at all?

2nd question: 'A' is assigned to temporary variable created on return of function . So why can't ++a etc. be used as lvalue?

Which compiler are you using? Please post the exact error message. — Björn Pollex, Aug 02 '11 at 09:37
Isn't there any warnings when you compile this code ? For example, that there is no return in f ? — Clement Bellot, Aug 02 '11 at 09:37
just because you observe something to be working it doesn't mean it it works at all. `int min(int a, int b) { return a; }` this function works on average 50% of the time. — Karoly Horvath, Aug 02 '11 at 09:39
"Why does [it] work at all?" Because your compiler's warnings aren't high enough and/or you're extremely unlucky. — Chris Lutz, Aug 02 '11 at 09:41
I use codeblocks. I know it is bad code .No warnings.So u mean in absence of return statement function returns a random address.What about 2nd question? — , Aug 02 '11 at 09:44
I meant any postincrement preincrement etc cannot be used as lvalue — , Aug 02 '11 at 09:52

score 4 · Accepted Answer · answered Aug 02 '11 at 09:41

4

Assuming return values are passed in registers, the return value from malloc might still be there when returning from f(). By pure chance.

When assigning to *f() you are not assigning to a temporary but to the memory returned from malloc. Assigning to ++a is totally different.

answered Aug 02 '11 at 09:41

Bo Persson

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Although I miss they key phrase here - undefined behavior. – Björn Pollex Aug 02 '11 at 09:49
The question was why it **seemed** to work. Not if was correct. – Bo Persson Aug 02 '11 at 09:51
I think the question is when and why you can assign to lvalues. – duedl0r Aug 02 '11 at 09:53

score 0 · Answer 2 · answered Aug 02 '11 at 09:37

0

Your function f() is not returning anything, you need to add:

return s;

But, in all honesty, this is just going to be the start of your problems. You also need to free() the return value of f().

I do not know why you have tagged this question C++, this is clearly C, and so I have untagged as such.

answered Aug 02 '11 at 09:37

Moo-Juice

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The question is asking, "X code is wrong but appears to be working, why?" and you're answering, "X code is wrong - you need to use Y code." – Chris Lutz Aug 02 '11 at 09:45
@Chris Lutz, ah.. Apologies. I went in to septic shock when I saw the code and it clouded my judgement. – Moo-Juice Aug 02 '11 at 09:46
I've been there. I start seeing red when I see `void main()` – Chris Lutz Aug 02 '11 at 09:48

score 0 · Answer 3 · answered Aug 02 '11 at 09:38

0

You have to return the pointer in f() with the return statement, or else an illegal pointer will be returned:

char *f()
{   
    char *s=malloc(8);
    return s;
}

answered Aug 02 '11 at 09:38

Constantinius

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"...or else an illegal pointer will be returned" Where in the world does this come from? – Chris Lutz Aug 02 '11 at 09:43
Probably from my bad use of a foreign language. But what exactly is wrong with that statement? – Constantinius Aug 02 '11 at 09:52
That's not what happens, either according to the standard (which says undefined behavior) or in this real world example (the pointer from `malloc` is left over in the return register, effectively causing it to be returned from `f` as well). – Chris Lutz Aug 02 '11 at 09:55

score 0 · Answer 4 · edited May 23 '17 at 12:19

Your function does not return anything. Since you declare the function with the return-type char *, not returning anything results in undefined behavior, as defined in paragraph 6.6.3.2 of the current C++ Standard:

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.

Undefined behavior means that anything can happen. To fix that problem, your function should look like this:

char *f()
{   
    return malloc(8);
}

score 0 · Answer 5 · answered Aug 02 '11 at 09:50

0

In your function, you return a pointer where you can assign stuff... ++a prevents it by returning const reference or instance. you can have the same behaviour if you have const char* f() { ... }.

Of course you could also implement ++a differently :)

answered Aug 02 '11 at 09:50

duedl0r

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log0 · Answer 6 · 2011-08-02T09:56:32.727

1/ The return of f() is an uninitialized pointer, but exists.
*f() return the value pointed by an unspecified (random) address.
Writing at this address is an invalid memory access, or maybe not if this address is by "chance" a writable piece of memory (stack or previously allocated heap).
In C, it is your responsibility to ensure that you properly access the memory.

2/ 'A' is not assigned here to a temporary.

why return value of function can be assigned to?

6 Answers6