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cintegercharacter
Derpster13
Derpster13

Reputation: 115

How to convert integers to characters in C?

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

Upvotes: 6

Views: 168284

Answers (5)

Amol Damodar
Amol Damodar

Reputation: 369

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 );
}

Upvotes: 0

chux
chux

Reputation: 153338

In C, int, char, long, etc. are all integers.

They typically have different memory sizes and thus different ranges as in INT_MIN to INT_MAX. char and arrays of char are often used to store characters and strings. Integers are stored in many types: int being the most popular for a balance of speed, size and range.

ASCII is by far the most popular character encoding, but others exist. The ASCII code for an 'A' is 65, 'a' is 97, '\n' is 10, etc. ASCII data is most often stored in a char variable. If the C environment is using ASCII encoding, the following all store the same value into the integer variable.

int i1 = 'a';
int i2 = 97;
char c1 = 'a';
char c2 = 97;

To convert an int to a char, simple assign:

int i3 = 'b';
int i4 = i3;
char c3;
char c4;
c3 = i3;
// To avoid a potential compiler warning, use a cast `char`.
c4 = (char) i4;

This warning comes up because int typically has a greater range than char and so some loss-of-information may occur. By using the cast (char), the potential loss of info is explicitly directed.

To print the value of an integer:

printf("<%c>\n", c3); // prints <b>

// Printing a `char` as an integer is less common but do-able
printf("<%d>\n", c3); // prints <98>

// Printing an `int` as a character is less common but do-able.
// The value is converted to an `unsigned char` and then printed.
printf("<%c>\n", i3); // prints <b>

printf("<%d>\n", i3); // prints <98>

There are additional issues about printing such as using %hhu or casting when printing an unsigned char, but leave that for later. There is a lot to printf().

Upvotes: 14

kannadasan
kannadasan

Reputation: 47

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);

 }

Upvotes: 0

DanChianucci
DanChianucci

Reputation: 1175

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.

Upvotes: 1

nhgrif
nhgrif

Reputation: 62052

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

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

Upvotes: 2

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