c++ - 使用 set-width 对齐列

Question

我正在尝试正确对齐此 C++ 作业。主要目标是从文件中正确读取数据，这正在完成，但输出未正确对齐。

有人建议将所有变量放入一个字符串中并逐行输出，但这似乎我需要重写程序的大部分内容，而且我不完全确定该怎么做。我附上了代码和输出如下所示。只是想知道是否有更好的方法在使用 setwidth 或其他基本对齐方法(如果存在)时保持所有内容对齐。顺便说一下，文件的输入可以改变。

    while (!PatientsFile.eof()) {
    PatientsFile >> firstName; 
    PatientsFile >> lastName;
    PatientsFile >> pulse;
    PatientsFile >> respir;
    PatientsFile >> oxygen;
    PatientsFile >> room;

    string name = firstName + ", " + lastName;

    //patient info
    cout << setw(4) << room << ' ';
    cout << name;

    //pulse
    if (pulse >= 101) {
        cout << setw(13) << "High=" << pulse << ' ';
        dangerCount++; }
    else if (pulse <= 59) {
        cout << setw(16) << "Low=" << pulse << ' ';
        dangerCount++; }
    else {
        cout << setw(19) << pulse << ' '; }

    //respir
    if (respir >= 21) {
        cout << setw(5) << "High=" << respir << ' ';
        dangerCount++; }
    else if (respir <= 10) {
        cout << setw(6) << "Low=" << respir << ' ';
        dangerCount++; }
    else {
        cout << setw(7) << respir << ' '; }

    //oxygen
    if (oxygen >= 101) {
        cout << setw(7) << "High=" << oxygen << ' ';
        dangerCount++; }
    else if (oxygen <= 91) {
        cout << setw(5) << "Low=" << oxygen << ' ';
        dangerCount++; }
    else {
        cout << setw(7) << oxygen << ' '; }

    //dangerCount
    if (dangerCount == 1) {
        checkTotal++;
        cout << "CHECK\n";}
    else if (dangerCount == 2) {
        alertTotal++;
        cout << "ALERT***\n"; }
    else if (dangerCount == 3) {
        criticalTotal++;
        cout << setw(15) << "CRITICAL****\n"; }
    else {
        normalTotal++;
        cout << endl; }
    dangerCount = 0;
    }
    //display totals
    cout << endl << normalTotal << " Normal\n";
    cout << alertTotal << " ALERT\n";
    cout << checkTotal << " CHECK\n";
    cout << criticalTotal << " CRITICAL\n";

输出:

Room Name                       Pulse  Respir  Oxygen    
-----------------------------------------------------
 111 Pete, Moss            Low=59      11  Low=91 ALERT***
 312 Diana, Canser        High=107   Low=10      96 ALERT***
 332 Willie, Makit        High=123 High=30      94 ALERT***
 111 Betty, Wont            Low=50   Low=10  Low=90   CRITICAL****
 331 Freda, Spirit        High=110 High=23  Low=90   CRITICAL****
 321 Wilma, Moneylast        High=132      18  Low=88 ALERT***
 121 Robin, Banks            Low=59      20     100 CHECK
 122 Charlie, Horse            Low=56      15      95 CHECK
 121 Eileen, Back            Low=45      17  Low=88 ALERT***
 222 Eaton, Buggs                 79      16      97 
 222 Summer, Day                 60      12      92 
 231 Bea, Sting                 73 High=21  Low=91 ALERT***
 311 Anita, Bath        High=111 High=33      97 ALERT***
 232 April, Showers                 61 High=22      93 CHECK
 222 Rose, Bush                100      20     100 
 231 Buster, Leggs                 88 High=44  Low=88 ALERT***
 221 Barb, Wire                 60      12  Low=82 CHECK
 322 Jim, Shoos        High=101      19      94 CHECK
 322 Kris, Mass            Low=4      14      99 CHECK
 222 Rich, Bright                 77      16      92 

4 Normal
8 ALERT
6 CHECK
2 CRITICAL

Answer 1

您可以尝试使用制表符\t 来正确对齐您的列表。这是您可能想要查看的链接:How many spaces for tab character(\t)?

否则你将不得不做

cout << "text" << setw(#) << "text" << setw(#) << endl;

我建议不要这样做，因为它只是对集合数字进行硬编码并且很乏味，因为您必须计算并确保它正确匹配。