Memcpy(), Structs, Free()

This is a discussion on Memcpy(), Structs, Free() within the C Programming forums, part of the General Programming Boards category; I wrote the sample program below to demsonstrate a problem I am having. Basically, I have written a method that ...

  1. #1
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    Memcpy(), Structs, Free()

    I wrote the sample program below to demsonstrate a problem I am having. Basically, I have written a method that will create a copy of a struct that was created in main(). My goal is to free the memory that was used by both the original struct and the copy by the program's end. However, it crashes after returning from the copy method when I attempt to free the original structure. It is my understanding that if I do not free both I will have a memory leak situation. Does anyone know why the free()'s in main() crash this program?

    Code:
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    
    #define STRINGA "abcdefghijklm"
    #define STRINGB "nopqrstuvwxyz"
    
    struct stuff
    {
    	char* a;
    	char* b;
    };
    
    int main()
    {
    	struct stuff *original = malloc(sizeof(struct stuff));
    	
    	original->a = malloc(strlen(STRINGA));
    	strcpy(original->a, STRINGA);
    
    	original->b = malloc(strlen(STRINGB));
    	strcpy(original->b, STRINGB);
    
    	printf("original->a: %s\n", original->a);
    	printf("original->b: %s\n", original->b);
    
    	copy((void*)original);
    
    	free(original->a);
    	free(original->b);
    	free(original);
    }
    
    void copy(void *ptr)
    {
    	printf("ptr->a: %s\n", ((struct stuff *)ptr)->a);
    	printf("ptr->b: %s\n", ((struct stuff *)ptr)->b);
    
    	struct stuff *new = malloc(sizeof(struct stuff));
    
    	memcpy(new, ptr, sizeof(struct stuff));
    
    	printf("new->a: %s\n", new->a);
    	printf("new->b: %s\n", new->b);
    
    	free(((struct stuff *)ptr)->a);
    	free(((struct stuff *)ptr)->b);
    	free((struct stuff *)ptr);
    }

  2. #2
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    Sorry, this is the wrong sample. I'll post the correct one .... I know why this one crashes.

  3. #3
    C++ Witch laserlight's Avatar
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    The problem is that memcpy() copies the pointers, not what the pointers point to, hence the first two calls to free() in copy() eventually lead to a double free problem.
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    So freeing the original will also free the copy and vice versa?

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    In order to make a true copy of the stuct I would have to memcpy ptr->a and ptr->b into the new structure in the copy method. Correct?

  6. #6
    C++ Witch laserlight's Avatar
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    Yes, i.e., you have to perform a deep copy.
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  7. #7
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    That makes perfect sense. So the following example should not result in any memory leaks:

    Code:
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    
    #define STRINGA "abcdefghijklm"
    #define STRINGB "nopqrstuvwxyz"
    
    struct stuff
    {
    	char* a;
    	char* b;
    };
    
    int main()
    {
    	struct stuff *original = malloc(sizeof(struct stuff));
    	
    	original->a = malloc(strlen(STRINGA));
    	strcpy(original->a, STRINGA);
    
    	original->b = malloc(strlen(STRINGB));
    	strcpy(original->b, STRINGB);
    
    	printf("original->a: %s\n", original->a);
    	printf("original->b: %s\n", original->b);
    
    	copy((void*)original);
    
    	printf("\nDone!\n\n");
    }
    
    void copy(void *ptr)
    {
    	printf("ptr->a: %s\n", ((struct stuff *)ptr)->a);
    	printf("ptr->b: %s\n", ((struct stuff *)ptr)->b);
    
    	struct stuff *new = malloc(sizeof(struct stuff));
    
    	memcpy(new, ptr, sizeof(struct stuff));
    
    	printf("new->a: %s\n", new->a);
    	printf("new->b: %s\n", new->b);
    
    	free(((struct stuff *)ptr)->a);
    	free(((struct stuff *)ptr)->b);
    	free((struct stuff *)ptr);
    
    	free(new);
    }

  8. #8
    Captain Crash brewbuck's Avatar
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    You're not allocating enough space for the strings. You need strlen(x) + 1 to account for the null terminator.
    Code:
    //try
    //{
    	if (a) do { f( b); } while(1);
    	else   do { f(!b); } while(1);
    //}

  9. #9
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    Am I right you're not free()-ing new->a and new->b ?
    When I use

    Code:
    free(new->a);
    free(new->b);
    free(new);
    in the copy() function it works perfectly.
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  10. #10
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    When you free original->a and original->b, new->a and new->b are also being freed because the a and b pointers point to the same memory in both structs. When you do a memcpy on a struct it only copies the memory allocated for the struct and not the memory that fields in the struct may point to. If I were to also do a memcpy on original->a and original->b into new->a and new->b then I would have to free these fields. So calling free on the a and b in the original or in the new will free these fields the way it is now. If you attempt to free both fields in both structs you will encounter a double free or corruption error.

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