是否可以使用字符串::find来检测换行符

Question

我正在尝试解析一个由1200首歌曲组成的字符串，每次找到'\n'或barCount=4时，我都想将我的barCount设置为0。根据我在网上发现的，\n代表一个字符，但我不确定该如何处理这些信息。。。我怎样才能找到它，然后做我需要做的事情？

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

查看您的代码，我怀疑我理解您现在要做的事情。让我看看我是否明白了：

• 字符串中的每个歌曲标题/唱片最多由四个用竖条分隔的字段组成
• 记录可能会提前用右括号或换行符终止
• 如果它没有以换行符结束，请插入一个换行符

以下可能会做得更好：

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
    }
}

它有点冗长，但它试图打破所有不同的条件。希望这能有所帮助。