Solved

char  convert

Posted on 2003-11-24
9
865 Views
Last Modified: 2010-04-01
hi,
i  wanna convert  char,which is read from a  data, convert  into the type  int.
i have tried  
(int)char;
and
int tmp; tmp = char;
but  it returns the ASCII number,
emaple: char  ch = '6';
(int)ch = 54

how can i do it?
and which head date should i add?
0
Comment
Question by:iux
9 Comments
 
LVL 9

Expert Comment

by:tinchos
ID: 9810058
Try using

#include <stdlib.h>


char number = '6';
int temp = atoi( number );

and in temp will be equal to 6

Hope this helps

Tincho
0
 

Author Comment

by:iux
ID: 9810076
hmmmmmmmmmmm
it  says
invalid conversion from `char' to `const char*'
0
 
LVL 39

Accepted Solution

by:
itsmeandnobodyelse earned 300 total points
ID: 9810087
No, atoi wants a char* - char pointer - and not a single char.

if you have a single digit in a char you may do this:

  char c = '9';

  int    i  = c - '0';

maybe you should check before:

   if (c >= '0' && c <= '9')
   {
        int i = c - '0';
        ...
   }

if you have a number and not a digit, you may use atoi

   char cn[] = "123";   // that i a zero-terminated string
   int    i      = atoi(cn);

Regards, Alex


0
PRTG Network Monitor: Intuitive Network Monitoring

Network Monitoring is essential to ensure that computer systems and network devices are running. Use PRTG to monitor LANs, servers, websites, applications and devices, bandwidth, virtual environments, remote systems, IoT, and many more. PRTG is easy to set up & use.

 
LVL 17

Expert Comment

by:rstaveley
ID: 9810103
char number = '6';
int temp =  number-'0';
0
 
LVL 3

Expert Comment

by:RJSoft
ID: 9811279
#define MAX ??
char arr[MAX];
char number='6';
arr[0]=number;
arr[1]='\0';

int val = atoi(arr);

RJ
0
 
LVL 4

Expert Comment

by:n_fortynine
ID: 9812270
rstaveley's solution is simple and clean. Why go through all these complexity? :D
0
 

Author Comment

by:iux
ID: 9812856
ok
i try it
thanks a lot
0
 
LVL 4

Expert Comment

by:void_main
ID: 9816957
... and what will you do if the number is something like this: 12345 ???

this will not work 'cause "12345" is not a single char!

therefore this will do it, and even with one char!

look:


char *number = "12345";
int realNumber = atoi(number);

but don't do this:
char *number = '12089';
or
char *number = '2';

best regards
void_main
0
 

Author Comment

by:iux
ID: 9824981
thx :-)
0

Featured Post

Master Your Team's Linux and Cloud Stack!

The average business loses $13.5M per year to ineffective training (per 1,000 employees). Keep ahead of the competition and combine in-person quality with online cost and flexibility by training with Linux Academy.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Fully specialized class template function 21 136
Socket Programming (Unix) 8 129
SNMP error No Such Object available on this agent at this OID 3 249
c++, dynamic object by json 1 42
  Included as part of the C++ Standard Template Library (STL) is a collection of generic containers. Each of these containers serves a different purpose and has different pros and cons. It is often difficult to decide which container to use and …
Basic understanding on "OO- Object Orientation" is needed for designing a logical solution to solve a problem. Basic OOAD is a prerequisite for a coder to ensure that they follow the basic design of OO. This would help developers to understand the b…
The goal of the video will be to teach the user the difference and consequence of passing data by value vs passing data by reference in C++. An example of passing data by value as well as an example of passing data by reference will be be given. Bot…
The viewer will learn how to use the return statement in functions in C++. The video will also teach the user how to pass data to a function and have the function return data back for further processing.

810 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question