{"id":32113,"date":"2023-01-26T20:17:28","date_gmt":"2023-01-26T14:47:28","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-function-returns-address-of-local-variable-but-it-still-compile-in-c-why\/"},"modified":"2023-01-26T20:17:28","modified_gmt":"2023-01-26T14:47:28","slug":"solved-function-returns-address-of-local-variable-but-it-still-compile-in-c-why","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-function-returns-address-of-local-variable-but-it-still-compile-in-c-why\/","title":{"rendered":"[Solved] function returns address of local variable, but it still compile in c, why?"},"content":{"rendered":"

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Even I get an warning a function returns an address from local
\nvariable, it compiles. Isn’t it then UB of compiler?<\/p>\n<\/blockquote>\n

No, but if it were, how could you tell? You seem to have a misunderstanding of undefined behavior. It does not mean “the compiler must reject it”, “the compiler must warn about it”, “the program must terminate”, or any such thing. Those indeed may<\/em> be a manifestations of UB, but if the language specification required such behavior then it wouldn’t be undefined<\/em>. Ensuring that a C program does not exercise undefined behavior is the responsibility of the programmer, not the C implementation.<\/strong> Where a programmer does not fulfill that responsibility, the C implementation explicitly has no reciprocal responsibility — it can do anything within its capabilities.<\/p>\n

Moreover, there is no single “the” C compiler. Different compilers may do things differently and still conform to the C language specifications. This is where implementation-defined, unspecified, and undefined behaviors come in. Allowing such variance is intentional on the part of the C language designers. Among other things, it allows implementations to operate in ways that are natural for their particular target hardware and execution environments.<\/p>\n

Now let’s go back to “no”. Here is a prototypical example of a function returning the address of an automatic variable:<\/p>\n

int *foo() {\n    int bar = 0;\n    return &bar;\n}\n<\/code><\/pre>\n
What about that is supposed to have undefined behavior?  It is well defined for the function to compute the address of bar<\/code>, and the resulting pointer value has the correct type to be returned by the function.  After bar<\/code>‘s lifetime ends when the function returns, the return value becomes indeterminate (paragraph 6.2.4\/2 of the standard), but that does not in itself give rise to any undefined behavior.<\/p>\n
Or consider a caller:<\/p>\n
void test1() {\n    int *bar_ptr = foo();  \/\/ OK under all circumstances\n}\n<\/code><\/pre>\n
As already discussed, our particular foo()<\/code>‘s return value will always be indeterminate, so in particular, it might be a trap representation.  But that’s a runtime consideration, not a compile-time one.  And even if the value were a trap representation, C does not require that the implementation refuse or fail to store it.  In particular, footnote 50 to C11 is explicit on this point:<\/p>\n
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Thus, an automatic variable can be initialized to a trap
\nrepresentation without causing undefined behavior, but the value of
\nthe variable cannot be used until a proper value is stored in it.<\/p>\n<\/blockquote>\n
Note also that foo()<\/code> and test1()<\/code> can be compiled by different runs of the compiler, such that when compiling test1()<\/code>, the compiler knows nothing about the behavior of foo()<\/code> beyond what is indicated by its prototype.  C does not place translation-time requirements on implementations that depend on the runtime behavior of programs.<\/p>\n
On the other hand, the requirements around trap representations would apply differently if the function were modified slightly:<\/p>\n
void test2() {\n    int *bar_ptr = NULL;\n    bar_ptr = foo();      \/\/ UB (only) if foo() returns a trap representation\n}\n<\/code><\/pre>\n
If the return value of foo()<\/code> turns out to be a trap representation, then storing<\/em> it in bar_ptr<\/code> (as opposed to initializing<\/em> bar_ptr<\/code> with it) produces undefined behavior at runtime.  Again, however, “undefined” means just what it says on the tin.  C does not define any particular behavior for implementations to exhibit under the circumstances, and in particular, it does not require that programs terminate or manifest any externally-visible behavior at all.  And again, that’s a runtime consideration, not a compile-time one.<\/p>\n
Furthermore, if foo()<\/code>‘s return value turns out not to be a trap representation (being instead a pointer value that is not the address of any live object), then there’s nothing wrong with reading that value itself, either:<\/p>\n
void test3() {\n    int *bar_ptr = foo();\n    \/\/ UB (only) if foo() returned a trap representation:\n    printf(\"foo() returned %p\\n\", (void *) bar_ptr);\n}\n<\/code><\/pre>\n
The biggest and most commonly-exercised undefined behavior in this area would be that of trying to dereference the return value of foo()<\/code>, which, trap representation or not, almost surely does not point to a live int<\/code> object:<\/p>\n
void test4() {\n    int *bar_ptr = foo();\n    \/\/ UB under all circumstances for the given foo():\n    printf(\"foo() returned a pointer to an int with value %d\\n\", *bar_ptr);\n}\n<\/code><\/pre>\n
But again, that’s a runtime consideration, not a compile-time one. And again, undefined means undefined.  The C implementation should be expected to translate that successfully as long as there are in-scope declarations for the functions involved, and although some compilers might warn, they have no obligation to do so.  The runtime behavior of function test4<\/code> is undefined, but that does not mean the program necessarily will segfault or terminate in some other manner.  It might, but I expect that in practice, the undefined behavior manifested by a great many implementations would be to print “foo() returned a pointer to an int with value 0”.  Doing so is in no way inconsistent with C’s requirements.<\/p>\n<\/p><\/div>\n
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            2<\/span> <\/p><\/div>\n<\/div>\n
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solved function returns address of local variable, but it still compile in c, why? <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Even I get an warning a function returns an address from local variable, it compiles. Isn’t it then UB of compiler? No, but if it were, how could you tell? You seem to have a misunderstanding of undefined behavior. 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