如何在C中将int转换为char *（不使用sprintf） [英] How to convert int into char* in C (without sprintf)

问题描述

有一天，我想在C中将整数转换为char *，该int可以为负。

some day, i wanted to convert in C an integer into a char *, this int can be negative.

我无法使用sprintf，snprintf都不是，那我该怎么做？

I wasn't able to use sprintf ,snprintf neither, so how do i do that ?

推荐答案

要滚动一个自己的 itoa（）的函数，第一个必须解决如何处理内存。最简单的方法是为所有可能的 int 值（ INT_MIN ）使用足够大的内存。

To roll one's own itoa()-like function, first one must address how to handle memory. The easiest is to use a large-enough memory for all possible int values (INT_MIN).

缓冲区所需大小在数学上是 ceiling（log10（-INT_MIN））+ 3 。可以通过以下方式对此进行近似：

The buffer required size is mathematically ceiling(log10(-INT_MIN))+3. This can be approximated with:

#include <limits.h>
#define INT_STR_SIZE (sizeof(int)*CHAR_BIT/3 + 3)

然后构建位数以最低有效位数开始依次递增，先使用％10 ，然后使用 / 10 来减少

Then build the digits up one-by-one starting with the least significant digit using %10 and then /10 to reduce the value.

通过使用 do 循环，代码捕获了 x = = 0 ，因为至少要输入一位数字。

By using a do loop, code catches the corner case of x==0 as at least one digit is made.

此代码避免了 if（x <0）{x = -x; ... 取反 INT_MIN （或乘以-1会导致 int 溢出，

This code avoids if (x < 0) { x = -x; ... as negating INT_MIN (or multiplying by -1 leads to int overflow, which is UB.

#include <limits.h>
#define INT_STR_SIZE (sizeof(int)*CHAR_BIT/3 + 3)

char *my_itoa(char *dest, size_t size, int x) {
  char buf[INT_STR_SIZE];
  char *p = &buf[INT_STR_SIZE - 1];
  *p = '\0';
  int i = x;

  do {
    *(--p) = abs(i%10) + '0';
    i /= 10;
  } while (i);

  if (x < 0) {
    *(--p) = '-';
  }
  size_t len = (size_t) (&buf[INT_STR_SIZE] - p);
  if (len > size) {
    return NULL;  // Not enough room
  }
  return memcpy(dest, p, len);
}

使用C99或更高版本，代码可以使用复合文字处理缓冲区创建，从而为每个 mt_itoa（）调用。

With C99 or later, code can handle the buffer creation with a compound literal allowing separate buffers for each mt_itoa() call.

// compound literal C99 or later
#define MY_ITOA(x) my_itoa((char [INT_STR_SIZE]){""}, INT_STR_SIZE, x)

int main(void) {
  printf("%s %s %s %s\n", MY_ITOA(INT_MIN), MY_ITOA(-1), MY_ITOA(0), MY_ITOA(INT_MAX));
  return (0);
}

输出

-2147483648 -1 0 2147483647

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