C++ 如何在C中只对字符串(包含数字、单词等)中的数字进行重复操作

@anatolyg(当前被删除)很好地标识了代码不能被simply就地替换，因为像"99"这样的子字符串将被更长的"198"替换.

找到字符串末端后，沿字符串向下传递2次.第一个只是确定所需的额外偏移量，而第二个使用类似的代码，从新的末端开始并执行复制.

代码不使用int索引，因为字符串可能长于INT_MAX.

((int)text[i]) * 2 + carry is amiss.
Better as (text[i] - '0') * 2 + carry; @Weijun Zhou .

OP的while (carry > 0) {只能很好地处理一个前导数字子串. 可能有很多人.

为了删除不需要的前导零，当找到不需要的前导零时，查找字符串结尾的循环也会缩短字符串.

/*
 * Copy in-place string 'src' to `src` with each sub-string of digits doubled in value.
 */
char* double_str(char *src);

#define IS_DIGIT(ch) ((ch) >= '0' && (ch) <= '9')

char* double_str(char *src) {
  char *s0 = src;
  char *d0 = src;

  // Remove unneeded leading zeroes and find end.
  int is_numeric = 0;
  while (*s0) {
    while (!is_numeric && s0[0] == '0' && IS_DIGIT(s0[1])) {
      s0++;
    }
    is_numeric = IS_DIGIT(*s0);
    *d0++ = *s0++;
  }
  *d0 = '\0';

  char *end = d0;
  char *new_end = end;

  // Make 2 passes.
  // 1st to find the new end of the string.
  // 2nd to copy the string.
  for (int i = 0; i < 2; i++) {
    int carry = 0;
    char *s = end;
    char *d = new_end;
    if (i > 0) {
      *d = '\0';
    }
    while (s > src) {
      s--;
      if (IS_DIGIT(*s)) {
        int new_digit = (*s - '0')*2 + carry;
        if (i > 0) {
          d--;
          *d = (char)(new_digit % 10 + '0');
        }
        carry = new_digit/10;
        // Do we have a non-zero carry and the prior character is not a digit?
        if (carry && (s == src || !IS_DIGIT(s[-1]))) {
          if (i > 0) {
            d--;
            *d = '1';
          }
          new_end++;
        }
      } else {
        if (i > 0) {
          d--;
          *d = *s;
        }
        carry = 0;
      }
    }
  }
  return src;
}

测试代码

#include <stdio.h>

int main(void) {
  char src_text[1000] = "5x2x5x9x15x99x500000000000000000000000000x";
  //printf("Enter text: ");
  //inputString(text, size);
  printf("'%s'", double_str(src_text));
  return 0;
}

结果是

'10x4x10x18x30x198x1000000000000000000000000000x'