unsigned long 到 char 数组

Question

这是我的代码示例：

/* Standard Linux headers */


/* --------------------------------------------------------------------------
Calculates the CRYPTO 
-------------------------------------------------------------------------- */
unsigned long CalculateCRYPTO(
unsigned long ulCount, /* Number of bytes in the data block */
unsigned char *ucBuffer )  /*Data block*/
{
    unsigned long ulCRYPTO = 0;
    //fonction that i have not coded ...
    return( ulCRYPTO );

}


int main (void)
{

  /*Variables and socket programming*/

   //this is my datablock must be in hexa AA 35 07 (will chnage size and data but for now it's hardcoded)
    unsigned char datablock[3];
    memset(datablock, '[=10=]' ,sizeof(datablock));
    datablock[0]=0xaa;
    datablock[1]=0x35;
    datablock[2]=0x07;

    unsigned long CRYPTO;
    CRYPTO=CalculateCRYPTO(sizeof(datablock),datablock); //calculate crypto depending of datablocks
    printf("CRYPTO = 0x%08x \n", CRYPTO); //prints me 0xe8ba8fa3 that is what i want 

    /*Creating the final command*/
    //(will chnage size and data but for now it's fixed)
    unsigned char cmd_final_hex[7]; //this needs to be DATABLOCKS+CRYPTO
                                    //in hexa AA 35 07 concatenated to inverted CRYPTO so ... A3 8F BA E8 => must be AA 35 07 A3 8F BA E8
    memset(cmd_final_hex, '[=10=]' ,sizeof(cmd_final_hex));     //Make sure cmd final is at 0
    memcpy(cmd_final_hex, datablock, sizeof(datablock));    //cmd at datablock + 000
    // while loop prints me what i want so cmd_final_hex[]=AA 35 07 00 00 ...

    //Now i want to concatenate so my idea is to use memcpy and not strcat :
    memcpy(&cmd_final_hex[sizeof(datablock)], &CRYPTO, 4);

   //and a print gives me AA 35 07 A3 8F BA E8 which is exactly what i want but why do i have to use "&CRYPTO" and not "CRYPTO" in my memcpy. !!!

  return 0;


}

我的问题是，为什么最后一个 memcpy 有效？我希望将 CRYPTO 而不是 &CRYPTO 放在参数中......对我来说，CRYPTO 是我想要的值，所以 0xe8ba8fa3 和 &CRYPTO 地址。对我来说，CRYPTO 不是指针，所以为什么我需要将 memcpy 与 &CRYPTO 一起使用才能使其正常工作？

顺便说一句，我的代码可能是纯粹的灾难，我是初学者。不要犹豫，纠正我！

谢谢！

Answer 1

My question is, why does this last memcpy works ? i would expect to put CRYPTO and not &CRYPTO in arguments... For me, CRYPTO is the value i want so 0xe8ba8fa3 and &CRYPTO the address.

你是对的。 CRYPTO 不是指针。但是，memcpy 需要一个指针，所以我们必须给它一个。我们通过获取 CRYPTO 的地址来做到这一点，这是通过向其添加 & 来完成的，因此 &CRYPTO.

memcpy 所做的是将内存从一个地址复制到另一个地址（这就是它需要两个指针的原因），而不管这些地址的实际内容。如果你给它 CRYPTO 而不是指向它的指针，它可能会将 CRYPTO 的值解释为地址（行为未定义，除非编译器给出，否则无法保证会发生什么） .

Answer 2

对于reference

memcpy void * memcpy ( void * destination, const void * source, size_t num );

参数

destination ：指向要复制内容的目标数组的指针，类型转换为 void* 类型的指针.
source ：指向要复制的数据源的指针，类型转换为 const void* 类型的指针。
num ：要复制的字节数。 size_t 是无符号整数类型。

unsigned long 到 char 数组

Unsigned long into char array

c

linux

unsigned

pointers