如何正确连接DataBuffer
?从PHP到C ++并不容易哈哈,我到底错在了什么?
while(true) {
DWORD TitleID = XamGetCurrentTitleId();
char DataBuffer[] = "Here's the current title we're on : ";
char DataBuffer[] = (DWORD)TitleID;
char DataBuffer[] = "\n\n";
DWORD dwBytesToWrite = (DWORD)strlen(DataBuffer);
}
答案 0 :(得分:2)
在C ++中,通常使用std::string
类型。要从多个输入创建字符串,特别是如果它们不是所有字符串,我们使用ostringstream
。
所以你可能会建立这样的信息:
while(true) {
DWORD TitleID = XamGetCurrentTitleId();
std::ostringstream titleMessageSS;
titleMessageSS << "Here's the current title we're on : "
<< TitleID // already a DWORD, no need for the cast
<< "\n\n";
std::string titleMessage = titleMessageSS.str(); // get the string from the stream
DWORD dwBytesToWrite = (DWORD)titleMessage.size();
// You don't do anything with this, so can't help you with how to write the string...
}
现在,如果您使用WriteToFile
,则需要从字符串中获取char
指针。使用titleMessage.c_str()
。
或者,您可以使用+
来构建std::string
,并使用std::to_string
转换TitleID
(因此您可以使用+
string
{{1}})。
答案 1 :(得分:1)
虽然PHP有自动转换为字符串,但C ++没有(除了某些情况)。您(通常)必须明确您的类型。幸运的是,C ++ 11提供了std::to_string
,这使得这更容易:
while(true) {
DWORD TitleID = XamGetCurrentTitleId();
std::string DataBuffer = "Here's the current title we're on : "
+ std::to_string(TitleID)
+ "\n\n";
DWORD dwBytesToWrite = static_cast<DWORD>(DataBuffer.size());
/* ... */
}