[c] How to convert integers to characters in C?

For example, if the integer was 97, the character would be 'a', or 98 to 'b'.

char c1 = (char)97;  //c1 = 'a'

int i = 98;
char c2 = (char)i;  //c2 = 'b'

Casting the integer to a char will do what you want.

char theChar=' ';
int theInt = 97;
theChar=(char) theInt;

cout<<theChar<<endl;

There is no difference between 'a' and 97 besides the way you interperet them.

Program Converts ASCII to Alphabet

#include<stdio.h>

void main ()
{

  int num;
  printf ("=====This Program Converts ASCII to Alphabet!=====\n");
  printf ("Enter ASCII: ");
  scanf ("%d", &num);
  printf("%d is ASCII value of '%c'", num, (char)num );
}

Program Converts Alphabet to ASCII code

#include<stdio.h>

void main ()
{

  char alphabet;
  printf ("=====This Program Converts Alphabet to ASCII code!=====\n");
  printf ("Enter Alphabet: ");
  scanf ("%c", &alphabet);
  printf("ASCII value of '%c' is %d", alphabet, (char)alphabet );
}

void main ()
 {
    int temp,integer,count=0,i,cnd=0;
    char ascii[10]={0};
    printf("enter a number");
    scanf("%d",&integer);
     if(integer>>31)
     {
     /*CONVERTING 2's complement value to normal value*/    
     integer=~integer+1;    
     for(temp=integer;temp!=0;temp/=10,count++);    
     ascii[0]=0x2D;
     count++;
     cnd=1;
     }
     else
     for(temp=integer;temp!=0;temp/=10,count++);    
     for(i=count-1,temp=integer;i>=cnd;i--)
     {

        ascii[i]=(temp%10)+0x30;
        temp/=10;
     }
    printf("\n count =%d ascii=%s ",count,ascii);

 }