std :: getline如何决定跳过最后一个空行？

当我逐行读取文件时，我注意到了一些奇怪的行为。如果文件以\n（空行）结尾，则可能会跳过...但并非总是如此，我不明白是什么让它跳过或不跳过。std :: getline如何决定跳过最后一个空行？

我写了这个小功能，将字符串分割线重现轻松的问题：

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; 
}

当我测试它（http://cpp.sh/72dgw），我得到的输出：

(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" ""

所以最后\n情况（6），（7）和（8）被忽略，罚款。但是，为什么不是（4）和（5）呢？

这种行为背后的理由是什么？

来源

2017-08-10 jpo38

有是quicky提到这个“奇怪”行为的一个有趣的帖子：getline() sets failbit and skips last line

由于menioned通过Rob's answer，\n是终止（这其实为什么它的名字末线），而不是一个分隔符，这意味着行被定义为“以'\ n'”结尾，而不是由'\ n'“分隔”。

我不清楚这是如何回答这个问题的，但实际上它确实如此。如下重整，它变得清澈如水：

如果你的内容计算“\ n”，那么你会x线，或x+1最终的x OCCURENCES如果有一些额外的非“\ 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)

来源

2017-08-10 09:43:11 jpo38

std :: getline如何决定跳过最后一个空行？

回答

相关问题