unsigned long into 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, '\0' ,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, '\0' ,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？我希望把CRYPTO而不是&CRYPTO放在参数中...对我来说,CRYPTO是我想要的值0xe8ba8fa3和&CRYPTO的地址.对我来说,CRYPTO不是一个指针,为什么我需要使用memcpy和&CRYPTO才能使它工作？

顺便说一下,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

我的问题是,为什么最后一个memcpy有效？我希望把CRYPTO而不是&CRYPTO放在参数中...对我来说,CRYPTO是我想要的值0xe8ba8fa3和&CRYPTO的地址.

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地址来做到这一点,这是通过添加&它来完成的&CRYPTO.

什么memcpy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),无论这些地址的实际内容如何.如果你给它CRYPTO而不是指向它的指针,它可能会解释CRYPTO为一个地址的值(行为是未定义的,除非编译器给出一个,否则不能保证会发生什么).