Output ASCII value of character - c++

Question

#include <iostream>
using namespace std;
int main()
{
char x;
cout << "enter a character:";
cin >> x;
cout << "ASCII Value of " << x << "is" << string(x);
return 0 ;
}
the error is
main.cpp||In function 'int main()':|
main.cpp|10|error: invalid conversion from 'char' to 'const char*'|
main.cpp|10|error: initializing argument 1 of 'std::basic_string<_CharT, _Traits,_Alloc>::basic_string(const _CharT*, const _Alloc&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]'|
||=== Build finished: 2 errors, 0 warnings ===|

Answer 1

std::cout << "ASCII Value of " << x << "is" << (int)x;
is one way (the cast circumvents the special treatement of a char type by the I/O stream library), but this will output your platform's encoded value of the character, which is not necessarily ASCII.
A portable solution is much more complex: You'll need to encode the ASCII set in a 128 element array of elements capable of storing a 7 bit unsigned value, and map x to a suitable element of that.

Answer 2

There are 3 approaches to solving this problem:
Use to_string
Passing the correct value to cout
Using the std::string class correctly
The solutions are marked (numbers in comment).
Use std::to_string
Since C++11, there is function to convert numbers to a string (to_string):
/*(1)*/ std::cout << std::to_string( x );
There is no specialization for a char parameter. So the value is implictly converted.
Passing the correct value to cout
cout would display the value of char object as a character.
If we want to output the value of a char object, we need to convert it to a type which is output by cout as a number instead of a character.
The C++ standard guarantees:
1 == sizeof(char) <= sizeof(short) <= sizeof(int) <= sizeof(long) <= sizeof(long long)
So any of those integer types can be used. Usually int is selected.
There are 4 conversions that can be used here:
1) Implicit - "Implicit conversions are performed whenever an expression of some type T1 is used in context that does not accept that type, but accepts some other type T2;"
/*(2)*/ int i = x;
std::cout << i;
2) Explicit - "Converts between types using a combination of explicit and implicit conversions."
/*(3)*/ std::cout << (int)x;
/*(4)*/ std::cout << int(x); // unsigned int(x) - is invalid,
// has to be a single-word type name
3) A named cast.
/*(5)*/ std::cout << static_cast<int>(x);
4) Use the T{e} notation for construction
/*(6)*/ std::cout << int{x};
The T{e} construction syntax makes it explicit that construction is desired. The T{e} construction syntax doesn’t allow narrowing. T{e} is the only safe and general expression for constructing a value of type T from an expression e. The casts notations T(e) and (T)e are neither safe nor general.
About conversions the C++ Core Guidelines specifies the following (among others)
ES.48: Avoid casts
ES.49: If you must use a cast, use a named cast
ES.64: Use the T{e}notation for construction
In this case I would suggest (3) or (4).
Using the std::string class correctly
string is a specialization of basic_string
using string = basic_string<char>;
basic_string has many constructors.
There are only 2 constructors, which can take a predefined number of chars;
basic_string( size_type count, CharT ch, const Allocator& alloc = Allocator() );
Constructs the string with count copies of character ch. The behavior is undefined if count >= npos.
/*(7)*/ std::string s = std::string( 1, x );
basic_string( const CharT* s, size_type count, const Allocator& alloc = Allocator() );
Constructs the string with the first count characters of character string pointed to by s. s can contain null characters. The length of the string is count. The behavior is undefined if s does not point at an array of at least count elements of CharT, including the case when s is a null pointer.
/*(8)*/ std::string s = std::string( &x, 1 );

Answer 3

#include <iostream>
using namespace std;
int main()
{
char x;
cout<< "enter a character:";
cin>>x;
cout<< "ASCII Value of "<< x<< "is"<< int(x);
return 0 ;
}
you mean return try this code

Answer 4

#include <iostream>
using namespace std;
int main()
{
char x;
cout<< "enter a character:";
cin>>x;
cout<< "ASCII Value of "<< x<< "is"<< char(x);
return 0 ;
}
try this its called return