在uint32_t中放入uint16_t

Question

char* createMSG(uint8_t i,uint16_t port) {
char *buff; 
buff = (char*) calloc(1,6);
uint8_t id, tmp;
tmp = 0;
id = 2;
memcpy(buff, &id, sizeof(uint8_t));
memcpy(buff+1, &i, sizeof(uint8_t));
memcpy(buff+2, &port, sizeof(uint16_t));
memcpy(buff+2+2, &tmp, sizeof(uint16_t));
memcpy(buff+2+2+1, &tmp, sizeof(uint16_t));
printf("created SV_CON_REP: id: %d accept: %d port %d\n",*buff,*(buff+1),*    (buff+2));    return buff;
}

.

我需要在uint32_t中复制端口。它打印该端口为null。

编辑 函数调用： char * tmp; uint8_t i; 我= 9; uint16_t端口; 端口= 1234; tmp= createmsg（i，port）;

输出：创建msg：ID：2接受：0端口0

Answer 1

我是复制这个函数，但在Windows下。

uint8_t = BYTE
uint16_t = WORD

char* createMSG(BYTE i,WORD port) 
{
    char *buff; 
    BYTE id, tmp;
    buff = (char*) calloc(1,6);
    tmp = 0;
    id = 2;
    memcpy(buff, &id, sizeof(BYTE));
    memcpy(buff+1, &i, sizeof(BYTE));
    memcpy(buff+2, &port, sizeof(WORD));
    memcpy(buff+2+2, &tmp, sizeof(WORD));
    memcpy(buff+2+2+1, &tmp, sizeof(WORD));
    printf("created SV_CON_REP: id: %d accept: %d port %d\n",*buff,*(buff+1),*        (buff+2));    return buff;
}

.

呼叫：

createMSG(9,1234);  

.

printf结果：

created SV_CON_REP: id: 2 accept: 9 port 210

.

（1234= 0x04d2，其中0xd2= 210）

您仍然没有复制呼叫和打印件结果，但您自己的评论

Answer 2

printf中的*（（UINT_16 *）（buff + 2））如何？