C语言中int转char [英] Converting int to char in C

问题描述

现在我正在尝试在 C 编程中将 int 转换为 char.经过研究，我发现我应该可以这样做:

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;

我想要的是返回'A'(并且0-9返回'0'-'9')但是我认为这会返回一个换行符.我的整个函数看起来像这样:

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;
}

推荐答案

把 int 转换成 char 你什么都不用做

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

char x;
int y;


/* do something */

x = y;

只有一个 int 到 char 值作为可打印(通常是 ASCII)数字，如您的示例:

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];
}

如果你想转换成字符串 (char *) 那么你需要使用任何标准函数，比如 sprintf, itoa, ltoa, utoa, ultoa .... 或者自己写一个:

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;
}

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