编译下面的代码会得到错误消息:type illegal。

int main()
{
    // Compilation error - switch expression of type illegal
    switch(std::string("raj"))
    {
    case"sda":
    }
}

你不能在switch或case中使用字符串。为什么?是否有任何解决方案可以很好地支持类似于打开字符串的逻辑?


当前回答

    cout << "\nEnter word to select your choice\n"; 
    cout << "ex to exit program (0)\n";     
    cout << "m     to set month(1)\n";
    cout << "y     to set year(2)\n";
    cout << "rm     to return the month(4)\n";
    cout << "ry     to return year(5)\n";
    cout << "pc     to print the calendar for a month(6)\n";
    cout << "fdc      to print the first day of the month(1)\n";
    cin >> c;
    cout << endl;
    a = c.compare("ex") ?c.compare("m") ?c.compare("y") ? c.compare("rm")?c.compare("ry") ? c.compare("pc") ? c.compare("fdc") ? 7 : 6 :  5  : 4 : 3 : 2 : 1 : 0;
    switch (a)
    {
        case 0:
            return 1;

        case 1:                   ///m
        {
            cout << "enter month\n";
            cin >> c;
            cout << endl;
            myCalendar.setMonth(c);
            break;
        }
        case 2:
            cout << "Enter year(yyyy)\n";
            cin >> y;
            cout << endl;
            myCalendar.setYear(y);
            break;
        case 3:
             myCalendar.getMonth();
            break;
        case 4:
            myCalendar.getYear();
        case 5:
            cout << "Enter month and year\n";
            cin >> c >> y;
            cout << endl;
            myCalendar.almanaq(c,y);
            break;
        case 6:
            break;

    }

其他回答

在c++中,你只能在int和char上使用switch语句

hare对Nick解决方案的评论真的很酷。这里是完整的代码示例(c++ 11):

constexpr uint32_t hash(const std::string& s) noexcept
{
    uint32_t hash = 5381;
    for (const auto& c : s)
        hash = ((hash << 5) + hash) + (unsigned char)c;
    return hash;
}

constexpr inline uint32_t operator"" _(char const* p, size_t) { return hash(p); }

std::string s = "raj";
switch (hash(s)) {
case "sda"_:
    // do_something();
    break;
default:
    break;
}

开关仅适用于整型(int, char, bool等)。为什么不使用映射将字符串与数字配对,然后将该数字与开关使用呢?

这是因为c++将开关转换为跳转表。它对输入数据执行简单的操作,并在不进行比较的情况下跳转到适当的地址。因为字符串不是一个数字,而是一个数字数组,所以c++不能从它创建一个跳转表。

movf    INDEX,W     ; move the index value into the W (working) register from memory
addwf   PCL,F       ; add it to the program counter. each PIC instruction is one byte
                    ; so there is no need to perform any multiplication. 
                    ; Most architectures will transform the index in some way before 
                    ; adding it to the program counter

table                   ; the branch table begins here with this label
    goto    index_zero  ; each of these goto instructions is an unconditional branch
    goto    index_one   ; of code
    goto    index_two
    goto    index_three

index_zero
    ; code is added here to perform whatever action is required when INDEX = zero
    return

index_one
...

(代码来自维基百科https://en.wikipedia.org/wiki/Branch_table)

您可以将字符串放在数组中,并在编译时使用constexpr将它们转换为索引。

constexpr const char* arr[] = { "bar", "foo" };
constexpr int index(const char* str) { /*...*/ }

do_something(std::string str)
{
    switch(quick_index(str))
    {
        case index("bar"):
            // ...
            break;

        case index("foo"):
            // ...
            break;

        case -1:
        default:
            // ...
            break;
    }

对于quick_index,它不一定是constexpr,你可以使用unordered_map在运行时做O(1)。(或者对数组进行排序并使用二进制搜索,参见这里的示例。)

下面是c++ 11的完整示例,使用了一个简单的自定义constexpr字符串比较器。重复的case和不在数组中的case (index给出-1)将在编译时被检测到。遗漏的病例显然没有被发现。后来的c++版本拥有更灵活的constexpr表达式,允许编写更简单的代码。

#include <iostream>
#include <algorithm>
#include <unordered_map>

constexpr const char* arr[] = { "bar", "foo", "foobar" };

constexpr int cmp(const char* str1, const char* str2)
{
    return *str1 == *str2 && (!*str1 || cmp(str1+1, str2+1));
}

constexpr int index(const char* str, int pos=0)
{
    return pos == sizeof(arr)/sizeof(arr[0]) ? -1 : cmp(str, arr[pos]) ? pos : index(str,pos+1);
}

int main()
{
    // initialize hash table once
    std::unordered_map<std::string,int> lookup;
    int i = 0;
    for(auto s : arr) lookup[s] = i++;
    auto quick_index = [&](std::string& s)
        { auto it = lookup.find(s); return it == lookup.end() ? -1 : it->second; };
    
    // usage in code
    std::string str = "bar";
    
    switch(quick_index(str))
    {
        case index("bar"):
            std::cout << "bartender" << std::endl;
            break;

        case index("foo"):
            std::cout << "fighter" << std::endl;
            break;

        case index("foobar"):
            std::cout << "fighter bartender" << std::endl;
            break;
            
        case -1:
        default:
            std::cout << "moo" << std::endl;
            break;
    }
}