题
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
解决方案
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
其他提示
printf
中的*((UINT_16 *)(buff + 2))如何?
不隶属于 StackOverflow