結構向量和迭代C++ 11

Question

我有一個簡單的問題。根據其他的例子我迭代是正確的，但Eclipse中拋出我的錯誤... 我的功能

billing::billing(std::istream &is) { 
    unsigned day; 
    std::string number; 
    float time; 
    struct call call; 
    while (!is.eof()) { 
     is >> day >> number >> time; 
     call.day = day; 
     call.number = number; 
     call.time = time; 
     blng_.push_back(call); 
    } 
    for(std::vector<call>::const_iterator it; it = blng_.begin(); it != blng_.end(); ++it) 
     // THROWS HERE ERRORS! 
     std::cout << it->day << std::endl; 
} 

編譯之後，他拋出了我類似的東西

expected a type, got 'call' billing.cpp  
'it' was not declared in this scope billing.cpp 
expected ';' before 'it' billing.cpp 
expected ')' before ';' token billing.cpp 
invalid type in declaration before 'it' billing.cpp 
template argument 2 is invalid billing.cpp 
the value of 'call' is not usable in a constant expression billing.cpp 
type/value mismatch at argument 1 in template parameter list for   
'template<class _Tp, class _Alloc> class std::vector' billing.cpp 

根據這一How do I iterate over a Constant Vector?話題應該是工作，但它不是，我沒有freq想法爲什麼。當我改變這整個std :: vectro ...到汽車它的作品！

Answer 1

在C++11你可以重寫：

for(std::vector<call>::const_iterator it = blng_.begin(); it != blng_.end(); ++it) 
    std::cout<<it->day<<std::endl;; 

爲

for(const auto& c: blng_) 
    std::cout << c.day << std::endl; 

附加說明：

你永遠不應該有eof()循環： Why is iostream::eof inside a loop condition considered wrong?

你真的應該做這樣的事情：

while(is >> day >> number >> time) { 

    call.day = day; 
    call.number = number; 
    call.time = time; 
    blng_.push_back(call); 
}