Timeline for answer to Write C++ string into char* by Hunter McMillen

9 events

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Dec 9, 2011 at 16:53	comment	added	Benjamin Lindley		Also, it would need to be declared as `const char *`.
Dec 9, 2011 at 16:43	comment	added	Benjamin Lindley		@Hunter: With strcpy, you are copying to the string that the pointer points at. buf is a local copy of whatever pointer was passed in at the call site. Even though it is itself local, the thing it points to is still the same thing that was pointed to by the pointer that it was copied from. And if your assumption was correct, then there is no need to make the char* a parameter. It should just be declared inside the function.
Dec 9, 2011 at 16:40	comment	added	Hunter McMillen		I was assuming that the OP abstracted away other code in this function, and that `buf` is actually used for something. That may or may not be the case, but even using `strcpy()` wouldn't this still be true?
Dec 9, 2011 at 16:39	comment	added	matthias		So is `version`. Even if you make `buf` a `char**`, it would be pointing to invalid memory when the function returns.
Dec 9, 2011 at 16:37	comment	added	Benjamin Lindley		@Hunter: I mean that when this function is called, it's effects are not visible to the user. buf is a local variable.
Dec 9, 2011 at 16:37	comment	added	Terry Li		@BenjaminLindley What you meant?
Dec 9, 2011 at 16:36	comment	added	Hunter McMillen		@BenjaminLindley What do you mean by that?
Dec 9, 2011 at 16:36	comment	added	Benjamin Lindley		Your code has no observable behavior.
Dec 9, 2011 at 16:35	history	answered	Hunter McMillen	CC BY-SA 3.0