[英]how to convert message to cipher message?
#include <iostream>
using namespace std;
int main()
{
string message;
cout << "Ahlan ya user ya habibi." <<endl;
cout <<"what do you like to do today?" <<endl;
cout <<"please enter the message:" <<endl;
getline(cin,message);
for(int i=0;i<message.size();i++)
{
if(string(message[i]==32))
{
cout<<char(message[i]);
}
else if(string( message[i])>=110)
{
int x = int(message[i])-13;
cout<<char(x);
}
else
{
int x = string (message[i])+13;
cout<<char(x);
}
}
return 0;
}
E:\\my programe\\quiz\\main.cpp|20|error: no matching function for call to 'std::__cxx11::basic_string<char>::basic_string(char&)'|
E:\\ my programe \\ quiz \\ main.cpp | 20 |错误:没有匹配的函数调用'std :: __ cxx11 :: basic_string <char> :: basic_string(char&)'|
E:\\my programe\\quiz\\main.cpp|20|error: invalid conversion from 'char' to 'const char*' [-fpermissive]|
E:\\ my programe \\ quiz \\ main.cpp | 20 |错误:从'char'到'const char *'的无效转换[-fpermissive] |
E:\\my programe\\quiz\\main.cpp|27|error: no matching function for call to 'std::__cxx11::basic_string<char>::basic_string(char&)'|
E:\\ my programe \\ quiz \\ main.cpp | 27 |错误:没有匹配的函数调用'std :: __ cxx11 :: basic_string <char> :: basic_string(char&)'|
E:\\my programe\\quiz\\main.cpp|27|error: invalid conversion from 'char' to 'const char*' [-fpermissive]|
E:\\ my programe \\ quiz \\ main.cpp | 27 |错误:从'char'到'const char *'的无效转换[-fpermissive] |
std::string::operator[]
returns a char&
reference. std::string::operator[]
返回char&
引用。 You are trying to construct temporary std::string
objects with single char
values as input, but std::string
does not have any constructors that take only a single char
as input. 您正在尝试构造具有单个
char
值作为输入的临时std::string
对象,但是std::string
没有任何仅将单个char
作为输入的构造函数。 That is why you are getting errors. 这就是为什么您会出错。
Even if you could construct a std::string
from a single char
, you can't compare a std::string
to an integer anyway. 即使可以从单个
char
构造std::string
,也无论如何都不能将std::string
与整数进行比较。
You don't need all of those string()
(and char()
) casts at all (BTW, your first string()
cast is malformed anyway). 您根本不需要所有这些
string()
(和char()
)强制转换(顺便说一句,您的第一个string()
强制转换始终是格式错误的)。 char
is a numeric type. char
是数字类型。 You can compare a char
value directly to an integer, and add/subtract and integer directly to/from a char
value to produce a new char
value. 您可以比较一个
char
直接值为整数,并添加/直接从减和整数/ char
值,以产生新的char
值。
Try this instead: 尝试以下方法:
#include <iostream>
using namespace std;
int main()
{
string message;
cout << "Ahlan ya user ya habibi." << endl;
cout << "what do you like to do today?" << endl;
cout << "please enter the message:" << endl;
getline(cin, message);
for(int i = 0; i < message.size(); i++)
{
if (message[i] == 32)
{
cout << message[i];
}
else if (message[i] >= 110)
{
char x = message[i] - 13;
cout << x;
}
else
{
char x = message[i] + 13;
cout << x;
}
}
return 0;
}
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