How to realloc some memory allocated using calloc?

问题

I've allocated a string with the calloc function:

//string1 and string2 previously declared
char *stringClone = calloc(strlen(string1) + 1, sizeof(char));

Now I want to do the same thing on stringClone with a different string. Doing:

stringClone = calloc(strlen(string2) + 1, sizeof(char));

I'm gonna have some memory leak, right? How should I use the realloc in this case?

回答1:

You can use realloc() to reallocate memory allocated by malloc(), calloc(), realloc(), aligned_alloc() or strdup(). Note that if the reallocated block is larger than the original block returned by calloc(), the newly allocated portion will not be initialized to all bits zero.

Note however that the syntax for realloc() is not what you use: you must pass the pointer as the first argument and a single size_t for the new size. Furthermore, if a new block cannot be allocated, NULL is returned and the block is not freed, hence you should not store the return value directly to stringClone.

If you want to use realloc(), here is what you should do:

//string1 and string2 previously declared
char *stringClone = calloc(strlen(string1) + 1, 1);
...
char *newp = realloc(stringClone, strlen(string2) + 1);
if (newp == NULL) {
    // deal with out of memory condition
    free(stringClone);
}

Since you do not seem to care that the contents of stringClone be preserved in the the reallocated block, you should probably simply write:

//string1 and string2 previously declared
char *stringClone = calloc(strlen(string1) + 1, 1);
if (stringClone == NULL) {
    // deal with out of memory condition
    ...
}
strcpy(stringClone, string1);
...
free(stringClone);
stringClone = calloc(strlen(string2) + 1, 1);
if (stringClone == NULL) {
    // deal with out of memory condition
    ...
}
strcpy(stringClone, string2);

Note also that on POSIX compliant systems, there is a memory allocation function that is very useful for your use case: strdup(s) takes a pointer to a C string, allocates strlen(s) + 1 bytes, copies the string to the allocated block and returns it:

//string1 and string2 previously declared
char *stringClone = strdup(string1);
if (stringClone == NULL) {
    // deal with out of memory condition
    ...
}
...
free(stringClone);
stringClone = strdup(string2);
if (stringClone == NULL) {
    // deal with out of memory condition
    ...
}

Note also that casting the return value of malloc, calloc and realloc is unnecessary in C and considered bad style.

回答2:

The reason to use realloc is that it keeps the original data intact. However, if I understand your use-case correctly, you intend to erase the original data. In that case, it is simpler and clearer to just write:

char *stringClone = calloc(strlen(string1) + 1, sizeof(char));
// check for calloc error
// use stringClone for string1
free(stringClone);
stringClone = calloc(strlen(string2) + 1, sizeof(char));
// check for calloc error
// use stringClone for string2

The error-checking is simpler for calloc than for realloc, since no temporary variable is needed. Also, this pattern makes it clear that the array contents for string1 and string2 are not related.

来源：https://stackoverflow.com/questions/56831704/how-to-realloc-some-memory-allocated-using-calloc

标签

memory-leaks

realloc

calloc