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?
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
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;
}
其他回答
你不能在开关情况下使用字符串。只允许int和char类型。相反,您可以尝试用enum表示字符串,并在switch case块中使用它
enum MyString(raj,taj,aaj);
在switch case语句中使用它。
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;
}
这是我前段时间提出的一个解决方案,它完全遵守所请求的语法。
#include <uberswitch/uberswitch.hpp>
int main()
{
uswitch (std::string("raj"))
{
ucase ("sda"): /* ... */ break; //notice the parenthesis around the value.
}
}
代码如下:https://github.com/falemagn/uberswitch
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)
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;
}