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I have code that works, however I'm not quite understanding why. I am allocating a pointer with a string, passing that string into a function, and modifying it. I'm confused in that it seems to run fine even if the string entered is far larger than the original string. I expect it to complain about overwriting memory, especially since the original pointer was not MALLOC/CALLOCed.

void changeArray(char *(*anArray)[3]);

Function Declaration

char *testArray[] = { "This is a test", "A Second Test", "A Third Test" };

Pointer declaration & init

 changeArray(&testArray);

Function Call

void changeArray(char *(*anArray)[3]){ char userName[200]; printf("What is your name?:\t"); scanf("%199[^\n]s", userName); *anArray[0] = userName; }

The Function

 printf("Your name is: %s\n", testArray[0]);

Printing the result in main.

Why is testArray[0] able to accept the new string without reallocating memory for it? Even more baffling is I'm looking at the result under locals and it appears corrupted, but still prints fine.

+ [0] 0x00e2f850 "D`\x3\x1 øâ" char *

I know it's weird to ask about working code, but I need to make sure I'm understanding what's happening so I can correctly implement/avoid it later.

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  • Undefined behaviour. It's only a co-incidence that your code is working. You cannot rely on it. Commented Jan 22, 2015 at 6:36

3 Answers 3

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You're confusing strings and pointers. A 'string' in C is an array of characters in memory, usually referred to by a pointer to the start of the array. So when you declare

char *testArray[] = { "This is a test", "A Second Test", "A Third Test" };

this declares testArray as an a array of 3 pointers, and initializes each pointer to point at a static string (array of characters probably in read-only memory).

In your changeArray function, you have:

char userName[200]; printf("What is your name?:\t"); scanf("%199[^\n]s", userName); *anArray[0] = userName;

This creates a new local array of characters and fills it in with the user input (presumably their name), and then points the 0th pointer in the array passed by reference as an argument at this local variable. Once this function returns, the stack frame will be freed, leaving the pointer dangling (pointing into the now freed stack frame) so it may seem to work but then seems to change randomly. Accessing a local variable like this after the scope it is in has exited is undefined behavior -- anything might happen.

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4 Comments

It might be too early, but I find the way "and then points the 0th pointer in the array passed by reference" works complicated, given that [] has higher precedence than *. anArray is a pointer to an array of char*. So anArray[0] is an lvalue of char*[3], not a char*. So I guess it decays to char* in the expression *anArray[0], right?
Okay, I get the array of pointers, I think the part I'm stuck on is that the pointer in array[0] is able to take a string much longer than the initial allocation, without overwriting the consecutive memory locations in array[1] and [2]. My understanding is that it needs to be calloced/realloced anytime you need to change size after initialization. I just tried this on three different OSes and two machines and it seems to work exactly the same way on each, even if I crank up the buffer in the function.
@juanchopanza: You are correct that anArray[0] has array type, so immediately decays to a pointer to it's first element (it becomes a char **), which is perhaps confusing.
@goddesspapa: pointers don't contain strings -- they're just memory addresses that point at strings that are in memory elsewhere. Changing the pointer in array[0] has no effect on the string it formerly pointed at, nor any other string in the vicinity.
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This is undefined behavior, and it is legal for undefined behavior to work as expected under certain circumstances (and fail miserably under other circumstances).

In your case, you are allocating the userName on the stack within a function. When the function returns, the memory that userName occupies remains intact, usually. But there is absolutely no guarantee for it. The next function call may clobber your string. An asynchronous signal handler might clobber your string as well, which can happen even between the return of the function and the execution of the next line in main().

Nevertheless, undefined behavior is completely undefined, and your program is perfectly entitled to wipe your harddrive when it is executed. So try to avoid undefined behavior unconditionally.

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Okay, I think I get it. I think in my brain I assumed that the local variable and everything goes away once it exits scope, but that doesn't actually happen until something comes along and grabs that bit of the stack or memory. In my brain I have 'testArray' being updated, but what's actually happening is that it's still pointing back to that local variable that may or may not be there anymore. Is this correct?
Yes, that's correct. To be precise: The variable (= the language construct) is defined to cease to exist, but the memory that held the value of the variable remains in place. That memory may be used for any other variable at any time, including variables internal to the compiler that you don't see, and any write access to these other variables will clobber your data. Equally bad, using the dangling pointer to write something to the memory will clobber the data these other variables will hold, whatever that may be. Such writes may well change to what code a function returns!
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What you are seeing here is undefined behavior.

You have a local array userName within void changeArray() and you have pointer to it which is valid just within the function, once you return from the function the memory allocated to this local array is freed so accessing this location outside its scope is undefined . You might even see a crash

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