copy - 将 uint16_t 放入 uint32_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;
}

我需要将端口复制到 uint32_t 中。它打印端口为空。

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

输出：创建的味精：id：2 接受：0 端口 0

Answer 1

*((uint_16*)(buff+2))printf怎么样？

Answer 2

我是复制这个功能，但是在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;
}

称呼：

createMSG(9,1234);  

打印结果：

created SV_CON_REP: id: 2 accept: 9 port 210

（1234 = 0x04d2，其中 0xd2=210）

您仍然没有复制 call & printf 结果，而是您自己的评论