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If you were to do something like:

char array[2][5] = {"cars", "tags"} char array2[] = "computer";

There has been an implicit allocation of memory allocated equal to(for the first line):

sizeof(array); OR sizeof(char*10);

Or however you want to do it.

Alternatively, one could do this:

char *ptr = "cars"; char *ptr2 = malloc(sizeof(*ptr2)*5); *ptr2 = "cars";

I know that if you are using pointers to explicitely allocate memory using malloc then it is good practice to free that pointer using free(ptr);

However, when you define things as above, are the following calls necessary on the same grounds as the explicit allocation of memory?

free(ptr); free(ptr2); free(array); free(array2);

-- As an aside, all of the above delcarations/initializations are implicitely NULL-terminated strings, right?

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You only need to free memory allocated with malloc, the other cases you specified are pointers to string literals or automatic variables(usually allocated on the stack) attempting to free them is undefined behavior.

C style strings are NULL terminated.

Also, important to note that:

*ptr2 = "cars";

is not correct syntax.

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And can you only use malloc with pointers? As in, if you creatd an array such as array[] = malloc(sizeof(char)*10), or something. I realize that arrays use pointers internally, but I am still curious if this is defined since in the brackets [] the size is not entered.
@sherrellbc, saying that "arrays use pointers internally" is a bit misleading. Check out c-faq.com/aryptr.
@sherrellbc array[] = malloc(....) is not correct syntax. If you want to experiment with syntax there are several online compilers that make experimenting quick and mostly painless and you can do it anywhere you have a browser: stackoverflow.com/questions/3916000/…
@CarlNorum Thanks! for that link, I was looking for a similar reference on pointers, I am going to keep that one around for further use.
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You should only ever call free when passing something allocated with a call to malloc. Which means that you should only ever call free with ptr2.

Furthermore, your assignment *ptr2 = "cars" is a syntax error. You would need to use strcpy to fill ptr2. Like this:

char *ptr2 = malloc(sizeof(*ptr2)*5); strcpy(ptr2, "cars");

C strings are null-terminated by convention, and string literals, i.e. "cars" are null-terminated.

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However, when you define things as above, are the following calls necessary on the same grounds as the explicit allocation of memory?

No, its illegal to free memory which are not returned by malloc or calloc or realloc.

all of the above delcarations/initializations are implicitely NULL-terminated strings, right?

Yes, those strings are called string literals and will be NULL termintaed.

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*ptr2 = "cars"; is not a valid statement. If you want to point ptr2 someplace else, you need to use:

ptr2 = "cars";

If you want to overwrite the memory ptr2 points to, use strcpy:

strcpy(ptr2, "cars");

Only ptr2 needs to be freed - it's the only one you allocated with malloc. Note that if you make the ptr2 = "cars" assignment mentioned above, you won't be able to free the original memory you allocated. That's a logical error known as a "memory leak".

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I see. So the memory location is lost. I was thinking that ptr2 would be allocated memory, and then ptr2 = "cars"; would be stored in the memory slot. So, ptr2 is actually reassigned to the stack where the string literal was allocated?
@sherrellbc, it's reassigned, but probably not to the stack. String literals are probably allocated in a read-only data section.
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You only have to free memory yo have allocated using malloc, when you declare a static array memory is freed automatically at the end of the code block.

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char *ptr = malloc(sizeof(*ptr)); would allocate sizeof() whatever the pointer's type is, which is char.

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