爲什麼我的簡單字符串沒有正確構建？

Question

我有一個簡單的函數來構建一個簡單的'Movie'對象的字符串表示。當我加入這行來測試值是有效的，我這樣做...

string Movie::getDisplayText() { 
    ostringstream oss; 
    oss << "Title: " << this->getTitle() << "\tYear: "+this->getYear() << "\tGenre: " << this->getGenre(); 
    string ret = oss.str(); 

    return ret; 
} 

但是被建成並返回的字符串看起來像這樣...

\�tle: Star Wars�B 
    Genre: Science-Fiction 

...

cout << "DEBUG Title: " << this->getTitle() << "\tYear: "+this->getYear() << "\tGenre: " << this->getGenre() << "\n" << endl; 

...它輸出到「COUT」正確尋找字符串，所以我知道這些值都正確初始化...

DEBUG title='Star Wars'; year='1977'; genre='Science-Fiction'; 

我的ostringstream代碼有什麼問題？

Answer 1

你有這樣的：

string Movie::getDisplayText() { 
    ostringstream oss; 
    oss << "Title: " << this->getTitle() << "\tYear: "+this->getYear() << "\tGenre: " << this->getGenre(); 
    string ret = oss.str(); 

    return ret; 
} 

應該

string Movie::getDisplayText() { 
    ostringstream oss; 
    oss << "Title: " << this->getTitle() << "\tYear: " << this->getYear() << "\tGenre: " << this->getGenre(); 
    string ret = oss.str(); 

    return ret; 
} 

不是\ tYear後的差額：。原因是將字符串和int一起添加不會導致串聯。