C problem with memory leaks (realloc function)

Question

scanFolderPath - path of folder with files.

filesToScan - array of strings with the files name.

I have problem with the realloc line (the third line in the for loop). And I don't understand why! Thank you for helping programmers community ;)

char* filePath = malloc(0);
char* fileContent = 0;
char* partContent = 0;
FILE* fileToScan;
int i = 0, j = 0, virus = FALSE, flag = FALSE, counter = 0;
for (i = 0; i < amountOfFiles; i++)
{
    flag = FALSE;
    if (scanFolderPath != NULL && filesToScan[i] != NULL)
    {
        realloc(filePath, (sizeof(char) * (strlen(scanFolderPath) + 1 + strlen(filesToScan[i]))));
    }
    strcpy(filePath, "");
    getFilePath(filePath, scanFolderPath, filesToScan[i]);
    fileToScan = fopen(filePath, "rb");
    readFile(&fileContent, filePath);
    if (choice == '0')
    {
        virus = scanSingature(fileContent, virusSingature, getLength(fileToScan), virusSingatureLen);
        if (virus)
        {
            printf("%s - Infected!\n", filePath);
            fprintf(logsFile, "%s", filePath);
            fprintf(logsFile, "  Infected!\n");
        }
        else
        {
            printf("%s - Clean\n", filePath);
            fprintf(logsFile, ("%s", filePath));
            fprintf(logsFile, "  Clean\n");
        }
        fclose(fileToScan);
        strcpy(filePath, "");
    }
}

You are throwing away the new pointer value returned by `realloc`. You have to assign it to a pointer variable. — Weather Vane, May 31 '20 at 18:33
Thanks you ! but in the end when I am adding "free(filePath)" I get an error. — Osher Berger, May 31 '20 at 18:35
That is because the original value of `filePath` has become invalid. — Weather Vane, May 31 '20 at 18:36
You free the pointer value returned by `realloc()`. You do not need to free the original value of `filePath` (except inasmuch as it is possible, but by no means certain, for `realloc` to return that value). — John Bollinger, May 31 '20 at 18:39
As the original call for memory `char* filePath = malloc(0);` was for `0` bytes of memory, it is quite likely that a *different* pointer was returned by `realloc`, and the original memory was deallocated, and so trying to `free` that will fail, because it has *already been* `free`d. Aside: a better strategy would be `char* filePath = NULL;` and `filePath = realloc...` will then still behave nicely. — Weather Vane, May 31 '20 at 18:41

score 1 · Accepted Answer · answered May 31 '20 at 18:36

1

try

filePath = realloc(filePath, (sizeof(char) * (strlen(scanFolderPath) + 1 + strlen(filesToScan[i]))));

that way the contents of filepath are reallocated, and the pointer is returned to filepath

answered May 31 '20 at 18:36

Jack William

77
7

0___________ · Answer 2 · 2020-05-31T18:40:30.393

0

realloc returns new allocated block. You do not assign to anything so the reallocated memory is lost and your pointer is pointing to the invalid memory assuming realloc success).

The correct way of reallocating:

void *p = malloc(10);
void t;

/* ... */

t = realloc(p, 20);
if(t) p = t;            //check if the realloc was successful 

/* ... */

free(p)

edited May 31 '20 at 18:40

answered May 31 '20 at 18:35

0___________

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C problem with memory leaks (realloc function)

2 Answers2