c - 将指针移交给 C 中的另一个函数

Question

我在使用这段代码时遇到了问题。当我执行“PrepareEncryption”(在示例中)时，代码返回一个有效指针，但是当我将它传递给“EncryptBig”时，它不再有效(指向随机数)。我最好的选择是删除原始结构。那么，如果这是问题所在，我该如何保存它呢？顺便说一句，我知道存在内存泄漏。

struct filecrypt
{
    FILE* bestand;
    FILE* nieuwbstnd;
    unsigned int positie;
    unsigned int size;
    unsigned int huidig;
    float procentum;
};

struct filecrypt *PrepareEncryption(char* locatie)
{
    struct stat file_status;
    struct filecrypt origineel, *mirror;
    int error;
    char* nieuw;

    if (stat(locatie, &file_status) != 0)
        return NULL;

    error = fopen_s(&origineel.bestand, locatie, "rb");
    if (error != 0)
        return NULL;

    error = strlen(locatie)+5;
    nieuw = (char*)malloc(error);
    if (nieuw == NULL)
        return NULL;
    strcpy_s(nieuw, error-3, locatie);
    strcat_s(nieuw, error, ".cpt");

    error = fopen_s(&origineel.nieuwbstnd, nieuw, "wb+");
    if (error != 0)
        return NULL;

    origineel.huidig = 0;
    origineel.positie = 0;
    origineel.procentum = 0.0f;
    origineel.size = file_status.st_size;
    mirror = &origineel;
    return mirror;
}

float EncryptBig(struct filecrypt *handle)
{
    int i, index = 0;
    float calc;
    char buf, *bytes = (char*)malloc(10485760); // 10 MB
    if (bytes == NULL)
    {
        handle = NULL;
        fcloseall();
        return -1.0f;
    }

    for (i = handle->huidig; i < (handle->huidig+10485760); i++)
    {
        if (i > handle->size)
            break;

        fseek(handle->bestand, i, SEEK_SET);
        fread_s(&buf, 1, 1, 1, handle->bestand);

        __asm
        {
            mov         eax, dword ptr [bytes]  
            add         eax, dword ptr [index]  
            mov         cl, byte ptr [buf]
            xor         cl, 18
            xor         cl, 75
            not         cl
            mov         byte ptr [eax], cl 
            mov         eax, dword ptr [index]  
            add         eax, 1  
            mov         dword ptr [index], eax
        }
    }

    fwrite(bytes, 1, i, handle->nieuwbstnd);
    fseek(handle->nieuwbstnd, i, SEEK_SET);

    handle->huidig += i;
    calc = (float)handle->huidig;
    calc /= (float)handle->size;
    calc *= 100.0f;

    if (calc == 100.0)
    {
        // GEHEUGEN LEK!
        // MOET NOG BIJGEWERKT WORDEN!
        fcloseall();
        handle = NULL;
    }
    return calc;
}

void example(char* path)
{
    float progress;
    struct filecrypt* handle;

    handle = PrepareEncryption(path);
    do
    {
        progress = EncryptBig(handle);
        printf_s("%f", progress);
    }
    while (handle != NULL);
}

Answer 1

这是因为你返回了一个指向局部变量的指针。

局部变量存储在堆栈中，当一个函数返回时，堆栈的那个区域被其他函数重用，你留下一个指针，它现在指向未使用的内存或现在被其他东西占用的内存。这是未定义的行为，有时可能会起作用，有时可能会给您“垃圾”数据，有时可能会崩溃。