Thread: confused about dynamic memory allocation

Which of the following pieces of code, in general, can result in an invalid memoryaccess? This question has multiple answers.
(A) char *str = (char*) malloc(10 * sizeof(char);
str[10] = ’\0’;
printf("%s", str);
(B) char *str = (char*) malloc(10 * sizeof(char);
str[0] = ’\0’;
printf("%s", str);
(C) char *str;
str[0] = ’\0’;
printf("%s", str);
(D) char *str = NULL;
str[0] = ’\0’;
printf("%s", str);




Can someone explain this question to me, I'm really confused?

The question is pretty clear. Which do you think result in invalid memory access, and why?

If you don't understand the question, I suggest you read up about the functions used in the examples.

A would be invalid memory access because we start at str[0] and end at str[9]. Therefore str [10] is not part of this array. D would also be correct because in its first statement the pointer char*str is set to null which is zero memory.

So are the final answers A,D?

Is C also correct because it did not allocate any memory?

If you're in the process of learning about dynamic memory allocation, and you're receiving code snippets like that, then I suggest you read this: FAQ > Casting malloc - Cprogramming.com

The gist of it is to not cast the return value of "malloc()".