Question

我有一个数据文件：

C0001|H|Espresso Classics|The traditional espresso favourites.
C0002|H|Espresso Espresions|Delicious blend of espresso, milk, and luscious flavours.
C0003|H|Tea & Cocoa|Gloria Jean's uses only the finest cocoas and single garden teas. Always satisfying.
C0004|C|Iced Chocolate|Gloria Jean's version of a traditional favourite.
C0005|C|Mocha Chillers|An icy blend of chocolate, coffee, milk and delicious mix-ins.
C0006|C|Espresso Chillers|A creamy blend of fresh espresso, chocolate, milk, ice, and flavours.
C0007|C|On Ice|Cool refreshing Gloria Jean's creations over ice.

以及以下代码来标记它：

    #define MAX_CAT_TOK 4
    #define DATA_DELIM "|"
    char *token[100];

    for(i = 0; i < MAX_CAT_TOK; i++)
    {
        if(i == 0) token[i] = strtok(input, DATA_DELIM);
         else token[i] = strtok(NULL, DATA_DELIM);
         printf("%s\n", token[i]);
    }

问题是，一旦打印了一个跟在较长字符串后面的字符串，就会在较短字符串的末尾打印出较长字符串的数据。我假设这与字符串终止有关??

有人看到我在这里做错了吗？

Answer 1

听起来发生的事情是你的缓冲区input没有被正确终止。如果，或许它最初都是零，那么处理的第一行就没问题了。如果更长的输入存储在其中，那么它仍然可以。但是当一个条目存储在其中的时间比前一个条目短时（例如，示例中的第4行），如果它不是以空终止，则可能导致问题。

例如，如果新数据是通过memcpy复制的，并且未包含空终止字符，那么该行中第4项的标记化将包括之前的数据。

如果是这种情况，那么解决方案是确保input空值正确终止。

以下尝试展示我想说的内容：

strcpy( input, "a|b|c|some long data" ); tokenize( input ); // where tokenize is the logic shown in the OP calling strtok // note the use of memcpy here rather than strcpy to show the idea // and also note that it copies exactly 11 characters (doesn't include the null) memcpy( input, "1|2|3|short", 11 ); tokenize( input );

在上面的设计示例中，第二个标记化中的第4个项目是：shortlong data。

修改换句话说，问题似乎不在OP中显示的代码中。问题在于如何填充输入。如果在for循环之前添加printf来显示正在解析的实际数据，则可能会发现它没有正确地终止null。第4行可能会显示它包含前一行的残余：

printf( "%s\n", input );

Answer 2

我看不出任何错误。

我冒昧地编写了一个可编译的代码版本并将其放在ideone。与您的版本比较......

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

int main(void) {
  int i, j;
  char *token[100];
  char *input;
  char inputs[7][300] = {
    "C0001|H|Espresso Classics|The traditional espresso favourites.",
    "C0002|H|Espresso Espresions|Delicious blend of espresso, milk, and luscious flavours.",
    "C0003|H|Tea & Cocoa|Gloria Jean's uses only the finest cocoas and single garden teas. Always satisfying.",
    "C0004|C|Iced Chocolate|Gloria Jean's version of a traditional favourite.",
    "C0005|C|Mocha Chillers|An icy blend of chocolate, coffee, milk and delicious mix-ins.",
    "C0006|C|Espresso Chillers|A creamy blend of fresh espresso, chocolate, milk, ice, and flavours.",
    "C0007|C|On Ice|Cool refreshing Gloria Jean's creations over ice.",
  };

  for (j = 0; j < 7; j++) {
    input = inputs[j];
    for (i = 0; i < 4; i++) {
      if (i == 0) {
        token[i] = strtok(input, "|");
      } else {
        token[i] = strtok(NULL, "|");
      }
      printf("%s\n", token[i]);
    }
  }
  return 0;
}

Answer 3

这是我的工作代码：

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

//#define DATA_DELIM "|"
#define DATA_DELIM "|\n"

int main(void)
{
    enum { LINE_LENGTH = 4096 };
    char input[LINE_LENGTH];
#define MAX_CAT_TOK 4
    char *token[100];

    while (fgets(input, sizeof(input), stdin) != 0)
    {
        printf("Input: %s", input);
        for (int i = 0; i < MAX_CAT_TOK; i++)
        {
            if (i == 0)
                token[i] = strtok(input, DATA_DELIM);
            else
                token[i] = strtok(NULL, DATA_DELIM);
            printf("%d: %s\n", i, token[i] != 0 ? token[i] : "<<NULL POINTER>>");
        }
    }
    return 0;
}

根据给定的数据，我得到：

Input: C0001|H|Espresso Classics|The traditional espresso favourites.
0: C0001
1: H
2: Espresso Classics
3: The traditional espresso favourites.
Input: C0002|H|Espresso Espresions|Delicious blend of espresso, milk, and luscious flavours.
0: C0002
1: H
2: Espresso Espresions
3: Delicious blend of espresso, milk, and luscious flavours.
Input: C0003|H|Tea & Cocoa|Gloria Jean's uses only the finest cocoas and single garden teas. Always satisfying.
0: C0003
1: H
2: Tea & Cocoa
3: Gloria Jean's uses only the finest cocoas and single garden teas. Always satisfying.
Input: C0004|C|Iced Chocolate|Gloria Jean's version of a traditional favourite.
0: C0004
1: C
2: Iced Chocolate
3: Gloria Jean's version of a traditional favourite.
Input: C0005|C|Mocha Chillers|An icy blend of chocolate, coffee, milk and delicious mix-ins.
0: C0005
1: C
2: Mocha Chillers
3: An icy blend of chocolate, coffee, milk and delicious mix-ins.
Input: C0006|C|Espresso Chillers|A creamy blend of fresh espresso, chocolate, milk, ice, and flavours.
0: C0006
1: C
2: Espresso Chillers
3: A creamy blend of fresh espresso, chocolate, milk, ice, and flavours.
Input: C0007|C|On Ice|Cool refreshing Gloria Jean's creations over ice.
0: C0007
1: C
2: On Ice
3: Cool refreshing Gloria Jean's creations over ice.

使用单字符分隔符字符串，在每行编号为3之后，我会得到一个额外的换行符。

这看起来很像你想要的。所以，要么你的输入有问题（你读它时是否回应它），或者你设法找到strtok()的一个片状实现，或者你可能在Windows上并且数据行有回车以及新行，你会看到由于错误的回车而产生误导性的输出。

其中，我怀疑最后一次（Windows和杂散回车）是最有可能的 - 尽管我无法重现问题，即使使用DOS格式的数据文件（使用GCC 4.6在MacOS X 10.6.7上进行测试） 0.0）。

使用strtok溢出字符串的问题

3 个答案: