是否可以使用string :: find來檢測新的行字符？

Question

我想解析一串1200首歌曲，我想每次找到'\ n'或我的barCount = 4時將我的barCount設置爲0。從我在網上找到的內容來看，\ n代表一個角色，但我不確定該如何處理這些信息......我怎麼才能找到它，然後做我需要做的事情？

int barCount = 0; 
size_t start = 0; 
size_t n = 0; 
int charCount = 0; 
while((start = chartDataString.find(" |", start)) != string::npos){   
     ++barCount; 
     start+=2; 
     charCount++; 
     //if(barCount == 3){// || chartDataString.find("]")){ 
    if(chartDataString.find("\n") !=string::npos){ 
     barCount = 0; 
    } 
    else if(barCount == 4 || chartDataString[charCount] == ']') { 
     chartDataString.insert(start, "\n"); 
     barCount = 0; 
    } 
} 

Answer 1

看着你的代碼，我懷疑我明白你現在想要做什麼。讓我看看，如果我得到這個直：在您的字符串

• 每首歌曲的標題/記錄由豎線
• 記錄可以通過一個右括號或換行提前終止分離多達四個領域
• 如果它沒有被換行符終止，請插入一個。

下可以做得更好：

size_t curr_pos = 0, next_bar, next_nl, next_brk; 
int bar_count = 0; 

while (true) 
{ 
    next_bar = chartDataString.find(" |", curr_pos); 
    next_nl = chartDataString.find("\n", curr_pos); 
    next_brk = chartDataString.find("]", curr_pos); 

    if (next_bar == string::npos && 
     next_nl == string::npos && 
     next_brk == string::npos) 
     break; 

    // Is a newline the next thing we'll encounter? 
    if (next_nl < next_bar && next_nl < next_brk) 
    { 
     curr_pos = next_nl + 1; // skip past newline 
     bar_count = 0;   // reset bar count 
    } 

    // Is a vertical bar the next thing we'll encounter? 
    if (next_bar < next_nl && next_bar < next_brk) 
    { 
     bar_count++; 
     curr_pos = next_bar + 2; // skip past vertical bar  
    } 

    // Is a right-bracket the next thing we'll encounter? 
    if (next_brk < next_bar && next_brk < next_nl) 
    { 
     bar_count = 4;   // fake like we've seen all 4 bars 
     curr_pos = next_brk + 1; // skip past bracket 
    } 

    // Have we seen 4 vertical bars or a right bracket? 
    if (bar_count == 4) 
    { 
     bar_count = 0; 
     chartDataString.insert("\n", curr_pos); 
     curr_pos += 1;    // skip the newline we just inserted 
    } 
} 

這是一個有點冗長，但它試圖掰開所有不同的條件。希望這會有所幫助。