Problem in user conversion!!!

I wish to ask is there any method that can convert a user char input to integer?

char input;

cout<<"Please insert a number!!!";
cin>>input;

I wish to convert the user input to integer type because i want to do addition with it.

Thanks for regards!!!

Why not read it as an int to begin with?

Thanks for regards.

I choose as char input because i want only one digit input. For integer, it'll be more than one digit and I'll face problem in my later part of my project...

Subtract the character '0' from the char and you will get the int it represents:

Code:

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int i = input - '0';

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Code:

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#include <stdio.h>
#include <stdlib.h>
int atoi ( const char * str );

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atoi converts strings (and I would use lexical_cast or a stringstream for that anyway). The OP wants to convert a character.

However, you should check that '0' <= input and input <= '9' first, since the user can also enter a non-digit character.

I've read something from Borland C++ and I think this can help me but I can't figure out what it means. Is there anybody can explain to me(Only for the char to int type)?


The result of the expression is the same type as that of the two operands.
Methods used in standard arithmetic conversions
Type Converts to Method

char int Zero or sign-extended (depends on default char type)




unsigned char int Zero-filled high byte (always)
signed char int Sign-extended (always)
short int Same value; sign extended
unsigned short unsigned int Same value; zero filled
enum int Same value

Special char, int, and enum conversions
Note: The conversions discussed in this section are specific to Borland C++.
Assigning a signed character object (such as a variable) to an integral object results in automatic sign extension. Objects of type signed char always use sign extension; objects of type unsigned char always set the high byte to zero when converted to int.
Converting a longer integral type to a shorter type truncates the higher order bits and leaves low-order bits unchanged. Converting a shorter integral type to a longer type either sign-extends or zero-fills the extra bits of the new value, depending on whether the shorter type is signed or unsigned, respectively.

Are you having trouble with the solution given? It is really rather simple. The text you quoted doesn't seem to be necessary to understand to accomplish this task.