Hello, I'm new to C and having this problem. What I need to do is add a single integer (0-9) to an (obviously char) string. More specifically, I want to add the value of a function that returns an integer (0-9) to a string. I don't want to change the integer value in any way, just to convert it to a form in which I could add it to a string.

I have checked out the itoa() function, but apparently it only stores the converted integer into a whole new string, not a single subscript/element of a string as I need.

Also, the function should return as a value the integer in the char form that it needs to be in order for it to be stored in the string.

"Pseudocode" example: TheString=ConvertToInt(FunctionThatReturnsInt());

Does a function like this exist in C? (If not, I guess I will have to code it, but I'd hate to "re-invent the wheel", so I am asking :) ) Thanks.

I'm not a C guy by any means, but I think you can cast it.... such as:

newvar = (int)x + (int)y;

Someone who Codes a lot of C, please check this over...

For a value 0-9, you can just add '0' to get the corresponding digit character.

#include <stdio.h>

int main(void)
{
   char text[] = "StringX";
   int digit;
   for (digit = 0; digit < 10; ++digit)
   {
      text[6] = digit + '0';
      puts(text);
   }
   return 0;
}

/* my output
String0
String1
String2
String3
String4
String5
String6
String7
String8
String9
*/

Thank you Dave Sinkula. Adding "+ '0'" works fine.

It took me a little to understand why exactly it works though. :o Thanks again. :)

gebbit, you can also look into sprintf, which gives you a lot of options.

If I'm not mistaken, sprintf() prints to whole strings, not to subscripts of strings, which is what I needed.

man its easy u have to use ascii code i will show u how with a simple program ok i hope this could help you and pls send me a reply telling me if thats what ur looking for :this silly software works o 0 to 9;

#include<iostream.h>
int main ()
{
    int a;

    cout<<"enter nnumber";
        cin>>a;

    char x;
    x=a+48;
    cout<<x<<endl;

    return 0;
}

Edited 3 Years Ago by mike_2000_17: Fixed formatting

when u put 0 in variable a , x holds a+48 this is the simple formula
to do a large number like 1234...etc u have to devide the number and put it in an array then switch each array element to type char and they put them in a char element or u just cout them ur welcome:D

if ur looking for a function i will post it, it is also 1 interger from 0 to 9 i will do later a small program for larger integers:

#include<iostream.h>

char integer_to_char(int);//prototype

int main ()//function main starts
{
	int a;//varibale that will hold the integer value
	
	cout<<"enter nnumber";//msg that prompt the user
		cin>>a;//user input

cout<<"the number represented in char form is:"<<integer_to_char(a)<<endl;//calling the function 
		return 0;
}
//function program
char integer_to_char(int x)
{
char y;
y=x+48;
return y;
}

> In case you didn't notice, the fellow wanted the code in C.
He also wanted it 3 YEARS ago as well.

Not only late, and in the wrong language, it's also old C++, and non-portable.
Use '0', not 48 (like was used oh so long ago in a galaxy far far away).

hello dear brothers
the following program is not real result for me
please explain me
thanks a lot

#include<iostream.h>
int main ()
{
int a;

cout<<"enter nnumber";
cin>>a;

char x;
x=a+48;
cout<<x<<endl;

return 0;
}

This question has already been answered. Start a new discussion instead.