Question

_export int DSPCmdOut(COMMAND_DESCRIPTOR *cmd)
{
    cmd.u[0] = CMD_SG_SYNCHED;
    cmd.u[1] = uThisStation;
    cmd.u[2] = iMode;
    DSPCmdOut(&cmd);
    return 0;
}

_export int DSPCmdOut(COMMAND_DESCRIPTOR *cmd)
{
    if(S->uCMD > (MAX_CMD-1))
        return -1;
    S->cmd[S->uCMD] = *cmd;
    S->uCMD++;

    sprintf (cLogLine, "%u,%u,%u: cmd data", cmd.u[0],cmd.u[1],cmd.u[2]);                        
    WriteLine  (uhGDSLogFile, cLogLine, strlen(cLogLine));

    return 0;
}

我在编译时遇到了这个错误：

Left operand of . has incompatible type 'pointer to COMMAND_DESCRIPTOR'.

Answer 1

因为你要将一个指针传递给COMMAND_DESCRIPTOR（这是我假设的一个结构），以便为struct的字段赋值，你需要取消引用它。
所以，cmd->u[0]语法只是(*cmd).u[0]。 ->只是一种语法糖。

将cmd.u[0] = CMD_SG_SYNCHED;更改为cmd->u[0] = CMD_SG_SYNCHED;，依此类推。

这是一个很好的主题：What is the difference between the dot (.) operator and -> in C++?

您的完整代码应如下所示：

_export int DSPCmdOut(COMMAND_DESCRIPTOR *cmd)
{
    cmd->u[0] = CMD_SG_SYNCHED;
    cmd->u[1] = uThisStation;
    cmd->u[2] = iMode;
    DSPCmdOut(&cmd);
    return 0;
}

_export int DSPCmdOut(COMMAND_DESCRIPTOR *cmd)
{
    if(S->uCMD > (MAX_CMD-1))
        return -1;
    S->cmd[S->uCMD] = *cmd;
    S->uCMD++;

    sprintf (cLogLine, "%u,%u,%u: cmd data", cmd->u[0],cmd->u[1],cmd->u[2]);                        
    WriteLine  (uhGDSLogFile, cLogLine, strlen(cLogLine));

    return 0;
}

没有在sprintf中获取任何数据

1 个答案: