在 C 中将 int 转换为 char

Question

Right now I am trying to convert an int to a char in C programming. After doing research, I found that I should be able to do it like this:

int value = 10;
char result = (char) value;

What I would like is for this to return 'A' (and for 0-9 to return '0'-'9') but this returns a new line character I think. My whole function looks like this:

char int2char (int radix, int value) {
  if (value < 0 || value >= radix) {
    return '?';
  }

  char result = (char) value;

  return result;
}

Answer 1

to convert int to char you do not have to do anything

char x;
int y;


/* do something */

x = y;

only one int to char value as the printable (usually ASCII) digit like in your example:

const char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";

int inttochar(int val, int base)
{
    return digits[val % base];
}

if you want to convert to the string (char *) then you need to use any of the stansdard functions like sprintf, itoa, ltoa, utoa, ultoa .... or write one yourself:

char *reverse(char *str);
const char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";

char *convert(int number, char *buff, int base)
{
    char *result = (buff == NULL || base > strlen(digits) || base < 2) ? NULL : buff;
    char sign = 0;


    if (number < 0)
    {
         sign = '-';

    }
    if (result != NULL)
    {
        do
        {
            *buff++ = digits[abs(number % (base ))];
            number /= base;
        } while (number);
        if(sign) *buff++ = sign;
        if (!*result) *buff++ = '0';
        *buff = 0;
        reverse(result);
    }
    return result;
}

Answer 2

A portable way of doing this would be to define a

const char* foo = "0123456789ABC...";

where ... are the rest of the characters that you want to consider.

Then and foo[value] will evaluate to a particular char . For example foo[0] will be '0' , and foo[10] will be 'A' .

If you assume a particular encoding (such as the common but by no means ubiquitous ASCII) then your code is not strictly portable.

Answer 3

Characters use an encoding (typically ASCII) to map numbers to a particular character. The codes for the characters '0' to '9' are consecutive, so for values less than 10 you add the value to the character constant '0' . For values 10 or more, you add the value minus 10 to the character constant 'A' :

char result;
if (value >= 10) {
    result = 'A' + value - 10;
} else {
    result = '0' + value;
}

Answer 4

Converting Int to Char

I take it that OP wants more that just a 1 digit conversion as radix was supplied.

---

To convert an int into a string , (not just 1 char ) there is the sprintf(buf, "%d", value) approach.

To do so to any radix, string management becomes an issue as well as dealing the corner case of INT_MIN

---

The following C99 solution returns a char* whose lifetime is valid to the end of the block. It does so by providing a compound literal via the macro.

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

// Maximum buffer size needed
#define ITOA_BASE_N (sizeof(unsigned)*CHAR_BIT + 2)

char *itoa_base(char *s, int x, int base) {
  s += ITOA_BASE_N - 1;
  *s = '\0';
  if (base >= 2 && base <= 36) {
    int x0 = x;
    do {
      *(--s) = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[abs(x % base)];
      x /= base;
    } while (x);
    if (x0 < 0) {
      *(--s) = '-';
    }
  }
  return s;
}

#define TO_BASE(x,b) itoa_base((char [ITOA_BASE_N]){0} , (x), (b))

Sample usage and tests

void test(int x) {
  printf("base10:% 11d base2:%35s  base36:%7s ", x, TO_BASE(x, 2), TO_BASE(x, 36));
  printf("%ld\n", strtol(TO_BASE(x, 36), NULL, 36));
}

int main(void) {
  test(0);
  test(-1);
  test(42);
  test(INT_MAX);
  test(-INT_MAX);
  test(INT_MIN);
}

Output

base10:          0 base2:                                  0  base36:      0 0
base10:         -1 base2:                                 -1  base36:     -1 -1
base10:         42 base2:                             101010  base36:     16 42
base10: 2147483647 base2:    1111111111111111111111111111111  base36: ZIK0ZJ 2147483647
base10:-2147483647 base2:   -1111111111111111111111111111111  base36:-ZIK0ZJ -2147483647
base10:-2147483648 base2:  -10000000000000000000000000000000  base36:-ZIK0ZK -2147483648

---

Ref How to use compound literals to fprintf() multiple formatted numbers with arbitrary bases?

Answer 5

Check out the ascii table

The values stored in a char are interpreted as the characters corresponding to that table. The value of 10 is a newline

Answer 6

So characters in C are based on ASCII (or UTF-8 which is backwards-compatible with ascii codes). This means that under the hood, "A" is actually the number "65" (except in binary rather than decimal). All a "char" is in C is an integer with enough bytes to represent every ASCII character. If you want to convert an int to a char , you'll need to instruct the computer to interpret the bytes of an int as ASCII values - and it's been a while since I've done C, but I believe the compiler will complain since char holds fewer bytes than int . This means we need a function, as you've written. Thus,

if(value < 10) return '0'+value;
return 'A'+value-10;

will be what you want to return from your function. Keep your bounds checks with "radix" as you've done, imho that is good practice in C.

Answer 7

1. Converting int to char by type casting

Source File charConvertByCasting.c

#include <stdio.h>

int main(){
    int i = 66;              // ~~Type Casting Syntax~~
    printf("%c", (char) i); //  (type_name) expression
    return 0;
}

Executable charConvertByCasting.exe command line output:

C:\Users\boqsc\Desktop\tcc>tcc -run charconvert.c
B

Additional resources:
https://www.tutorialspoint.com/cprogramming/c_type_casting.htm https://www.tutorialspoint.com/cprogramming/c_data_types.htm

2. Convert int to char by assignment

Source File charConvertByAssignment.c

#include <stdio.h>

int main(){
    int i = 66;
    char c = i;
    printf("%c", c);
    return 0;
}

Executable charConvertByAssignment.exe command line output:

C:\Users\boqsc\Desktop\tcc>tcc -run charconvert.c
B

Answer 8

You can do

char a;
a = '0' + 5;

You will get character representation of that number.

Answer 9

Borrowing the idea from the existing answers, ie making use of array index. Here is a "just works" simple demo for "integer to char[]" conversion in base 10, without any of <stdio.h> 's printf family interfaces.

Test:

$ cc -o testint2str testint2str.c && ./testint2str
Result: 234789

Code:

#include <stdio.h>
#include <string.h>

static char digits[] = "0123456789";

void int2str (char *buf, size_t sz, int num);


/*
  Test: 
        cc -o testint2str testint2str.c && ./testint2str
*/
int
main ()
{
  int num = 234789;
  char buf[1024] = { 0 };

  int2str (buf, sizeof buf, num);

  printf ("Result: %s\n", buf);


}

void
int2str (char *buf, size_t sz, int num)
{
  /*
     Convert integer type to char*, in base-10 form.
   */

  char *bufp = buf;
  int i = 0;

  // NOTE-1
  void __reverse (char *__buf, int __start, int __end)
  {
    char __bufclone[__end - __start];
    int i = 0;
    int __nchars = sizeof __bufclone;
    for (i = 0; i < __nchars; i++)
      {
    __bufclone[i] = __buf[__end - 1 - i];
      }
    memmove (__buf, __bufclone, __nchars);
  }

  while (num > 0)
    {
      bufp[i++] = digits[num % 10]; // NOTE-2
      num /= 10;
    }

  __reverse (buf, 0, i);

  // NOTE-3
  bufp[i] = '\0';
}

// NOTE-1:
//         "Nested function" is GNU's C Extension. Put it outside if not
//         compiled by GCC.
// NOTE-2:
//         10 can be replaced by any radix, like 16 for hexidecimal outputs.
//
// NOTE-3:
//         Make sure inserting trailing "null-terminator" after all things
//         done.

NOTE-1: "Nested function" is GNU's C Extension . Put it outside if not compiled by GCC.

NOTE-2: 10 can be replaced by any radix, like 16 for hexidecimal outputs.

NOTE-3: Make sure inserting trailing "null-terminator" after all things done.