Malloc and wcscpy: HEAP CORRUPTION DETECTED

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Malloc and wcscpy: HEAP CORRUPTION DETECTED

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#1 by (2 votes)

1

I'm trying to create and use a pointer to a string and join it with another reserved string pointer with malloc , but I always get this error message:

This is the code:

int wmain(int argc, WCHAR *argv[])
{
    LPWSTR str1 = L"Software\WinSide\";
    LPWSTR str2 = (LPWSTR)malloc(wcslen(str1)+1);
    wcscpy(str2, str1);


    if (str2 == NULL)
    {
        wprintf(L"Malloc error.\n");
    }
    else
    {
        wprintf(L"Result= %s\n", str2);
    }

    free(str2);

    return 0;
}

If I put a value in% co_of big% like 50, for example, it works fine. I know the problem is in memory allocation, but I do not understand why.

Could you help me?

Edit:

If I use the standard C data types, the program works, but I do not understand why. I add the code:

int main(int argc, char * argv[])
{
    char * str1 = "Software\WinSide\";
    char * str2 = (char *)malloc(strlen(str1)+1);
    strcpy(str2, str1);

    if (str2 == NULL)
    {
        printf("Malloc error.\n");
    }
    else
    {
        printf("Result= %s\n", str2);   

    }

    free(str2);

    return 0;
}

c punteros winapi

asked by Sergio Calderon 09.11.2016 в 22:38

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1 answer

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score 2 · Accepted Answer

Without having experience in Windows, I see that in your code reserves 1 byte at the end of the chain, to store the end of it.

LPWSTR str2 = (LPWSTR)malloc(wcslen(str1)+1);
wcscpy(str2, str1);
/*wcscat(str2, argv[1]);*/
DWORD size = wcslen(str1)+1;

However, you use L-strings, with wide characters. I do not remember the exact size or the concrete type (wchar_t or similar), but I think that Window used 32-bit characters (4 bytes) for that type of strings.

Try reserving 4 bytes, instead of one:

malloc( wcslen( str1 ) + 4 );

If I remember the exact type, I will edit the answer.

postdata: do not mix types. If you use wide strings, use wide characters; if you use normal strings char * (LPSTR in Windows?), then use normal characters (sizeof (char))

EDITO

It's wchar_t; therefore, the correct thing would be

LPWSTR str2 = (LPWSTR)malloc(wcslen(str1)+sizeof(wchar_t));

It's a common mistake to confuse bytes with characters .

malloc() works with bytes .
char *str = "hola"; is an array of characters that, coincidentally, have a size of 1 byte. wchar_t *str2 = L"adios"; is an array of characters that does not have to be 1 byte each.

If it were possible to use sizeof with memory blocks, we would have something like this:

sizeof( malloc( 100 ) )  -> 100
sizeof( "hola" )         -> 5 ( 4 caracteres de 1 byte, mas 1 caracter de terminador de la cadena.
sizeof( L"adios" )       -> depende de la arquitectura, el sistema, etc. Como se vió en el problema, puede ser perfectamente de 48 bytes ( 5 caracteres * 4 bytes cada uno, + 1 caracter (otros 4 bytes) de terminador de la cadena.

In addition to the above, the functions that work with strings are prepared and specialized in the type of strings they use. Therefore, strcpy() adds 1 byte at the end, while wcpcpy() adds a width character (4 or more bytes) at the end.

Conclusion : do not assume sizes, use the sizeof (which is from the beginning of the language, for something will be), do not mix types, and use the appropriate functions for each one. p>