std :: getline如何決定跳過最後一個空行？

當我逐行讀取文件時，我注意到了一些奇怪的行爲。如果文件以\n（空行）結尾，則可能會跳過...但並非總是如此，我不明白是什麼讓它跳過或不跳過。std :: getline如何決定跳過最後一個空行？

我寫了這個小功能，將字符串分割線重現輕鬆的問題：

std::vector<std::string> SplitLines(const std::string& inputStr) 
{ 
    std::vector<std::string> lines; 

    std::stringstream str; 
    str << inputStr; 

    std::string sContent; 
    while (std::getline(str, sContent)) 
    { 
     lines.push_back(sContent); 
    } 

    return lines; 
}

當我測試它（http://cpp.sh/72dgw），我得到的輸出：

(1) "a\nb"  was splitted to 2 line(s):"a" "b" 
(2) "a"   was splitted to 1 line(s):"a" 
(3) ""   was splitted to 0 line(s): 
(4) "\n"   was splitted to 1 line(s):"" 
(5) "\n\n"  was splitted to 2 line(s):"" "" 
(6) "\nb\n"  was splitted to 2 line(s):"" "b" 
(7) "a\nb\n"  was splitted to 2 line(s):"a" "b" 
(8) "a\nb\n\n" was splitted to 3 line(s):"a" "b" ""

所以最後\n情況（6），（7）和（8）被忽略，罰款。但是，爲什麼不是（4）和（5）呢？

這種行爲背後的理由是什麼？

來源

2017-08-10 jpo38

有是quicky提到這個「奇怪」行爲的一個有趣的帖子：getline() sets failbit and skips last line

由於menioned通過Rob's answer，\n是終止（這其實爲什麼它的名字末線），而不是一個分隔符，這意味着行被定義爲「以'\ n'」結尾，而不是由'\ n'「分隔」。

我不清楚這是如何回答這個問題的，但實際上它確實如此。如下重整，它變得清澈如水：

如果你的內容計算「\ n」，那麼你會x線，或x+1最終的x OCCURENCES如果有一些額外的非「\ n」字符在文件的末尾。

(1) "a\nb"  splitted to 2 line(s):"a" "b" (1 EOL + extra characters = 2 lines) 
(2) "a"   splitted to 1 line(s):"a"  (0 EOL + extra characters = 1 line) 
(3) ""   splitted to 0 line(s):   (0 EOL + no extra characters = 0 line) 
(4) "\n"   splitted to 1 line(s):""   (1 EOL + no extra characters = 1 line) 
(5) "\n\n"  splitted to 2 line(s):"" ""  (2 EOL + no extra characters = 2 lines) 
(6) "\nb\n"  splitted to 2 line(s):"" "b"  (2 EOL + no extra characters = 2 lines) 
(7) "a\nb\n"  splitted to 2 line(s):"a" "b" (2 EOL + no extra characters = 2 lines) 
(8) "a\nb\n\n" splitted to 3 line(s):"a" "b" "" (3 EOL + no extra characters = 3 lines)

來源

2017-08-10 09:43:11 jpo38

std :: getline如何決定跳過最後一個空行？

回答

相關問題