C中的進度字符串解析

Question

我有以下字符串：

"..1....10..20....30...40....50...80..." 

我需要從中提取所有數字到數組中。

用C做最好的方法是什么？

Answer 1

也許最簡單的方法是使用strtok()函數（或strtok_r()如果擔心重入）：

char str[] = "..1...10...20";
char *p = strtok(str, ".");
while (p != NULL) {
    printf("%d\n", atoi(p));
    p = strtok(NULL, ".");
}

一旦得到調用atoi()的結果，將這些整數保存到數組中應該是一件簡單的事情。

Answer 2

您可以使用帶有抑制分配的sscanf代碼（％* [。]）來跳過點（或任何其他所需字符），並使用掃描字符計數代碼％n來推進字符串指針。

const char *s = "..1....10..20....30...40....50...80...";
int num, nc;

while (sscanf(s, "%*[.]%d%n", &num, &nc) == 1) {
    printf("%d\n", num);
    s += nc;
}

Answer 3

這是正確的方法，它比最簡單的方法稍長，但如果讀取的值超出范圍，如果第一個字符不是點，則它不會受到未定義的行為的影響，等等。沒有指定數字是否為負數，所以我使用了有符號類型但只允許正值，您可以通過允許內部while循環頂部的負號來輕松更改此值。 此版本允許任何非數字字符分隔整數，如果您只想要允許點，您可以修改內循環以僅跳過點，然后檢查數字。

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <errno.h>

#define ARRAY_SIZE 10

size_t store_numbers (const char *s, long *array, size_t elems)
{
  /* Scan string s, returning the number of integers found, delimited by
   * non-digit characters.  If array is not null, store the first elems
   * numbers into the provided array */

  long value;
  char *endptr;
  size_t index = 0;

  while (*s)
  {
    /* Skip any non-digits, add '-' to support negative numbers */
    while (!isdigit(*s) && *s != '\0')
      s++;

    /* Try to read a number with strtol, set errno to 0 first as
     * we need it to detect a range error. */
    errno = 0;
    value = strtol(s, &endptr, 10);

    if (s == endptr) break; /* Conversion failed, end of input */
    if (errno != 0) { /* Error handling for out of range values here */ }

    /* Store value if array is not null and index is within array bounds */
    if (array && index < elems) array[index] = value;
    index++;

    /* Update s to point to the first character not processed by strtol */
    s = endptr;
  }

  /* Return the number of numbers found which may be more than were stored */
  return index;
}

void print_numbers (const long *a, size_t elems)
{
  size_t idx;
  for (idx = 0; idx < elems; idx++) printf("%ld\n", a[idx]);
  return;
}

int main (void)
{
  size_t found, stored;
  long numbers[ARRAY_SIZE];
  found = store_numbers("..1....10..20....30...40....50...80...", numbers, ARRAY_SIZE);

  if (found > ARRAY_SIZE)
    stored = ARRAY_SIZE;
  else
    stored = found;

  printf("Found %zu numbers, stored %zu numbers:\n", found, stored);
  print_numbers(numbers, stored);

  return 0;
}

Answer 4

我更喜歡在for循環中使用strtok。 雖然語法看起來有點奇怪，但讓它感覺更自然。

char str[] = "..1....10..20....30...40....50...80..."
for ( char* p = strtok( strtok, "." ); p != NULL; p = strtok( NULL, "." ) )
{
    printf( "%d\n", atoi( p ) );
}