使用另一个字符串作为分隔符将一个字符串分解为令牌列表？

Zeruno 发表于 Dev

泽鲁诺

假设我有以下字符串：

char *myTestString = "Hello AND test AND test2";

我想打破这种分解成集合{ Hello，test，test2}，我终于可以遍历。

或者我有

char *myTestString2 = "Hi AND there AND test AND test2";

我想打破这种分解成集合{ Hi，there，test，test2}，我以后可以遍历。

我如何使用C实现呢？

编辑：另一个示例是拆分"Hello there AND test"应给出集合{ Hello there，test}。为了澄清"AND"起见，这里是定界符。

来自莫斯科的弗拉德

这个给你。

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

char ** split( const char *s1, const char *s2 )
{
    char **tokens = malloc( sizeof( char * ) );
    int success = tokens != NULL;

    if ( success )
    {
        const char *delim = " \t";
        *tokens = NULL;

        for ( size_t n = 1, len = strlen( s2 ); success && *s1; )
        {
            s1 += strspn( s1, delim );

            if ( *s1 )
            {
                const char *p = s1;

                s1 += strcspn( s1, delim );

                if ( strncmp( p, s2, len ) != 0 )
                {
                    char **tmp = realloc( tokens, ( n + 1 ) * sizeof( char * ) );

                    if ( ( success = tmp != NULL ) )
                    {
                        tokens = tmp;

                        success = ( tokens[n-1] = calloc( 1, s1 - p + 1 ) )  != NULL;
                        strncpy( tokens[n-1], p, s1 - p );
                        tokens[n] = NULL;
                        ++n;
                    }

                    if ( !success )
                    {
                        for ( size_t i = 0; i < n; i++ ) free( tokens[i] );
                        free( tokens );
                    }
                }
            }
        }
    }       

    return tokens;
}

int main(void) 
{
    const char *s1 = "Hi AND there AND test AND test2";
    const char *s2 = "AND";

    char **tokens = split( s1, s2 );

    if ( tokens != NULL )
    {
        for ( char **p = tokens; *p != NULL; ++p )
        {
            puts( *p );
        }

        char **p = tokens;
        do
        {
            free( *p );
        } while ( *p++ != NULL );

        free( tokens );
    }

    return 0;
}

程序输出为

Hi
there
test
test2

如果内存分配不成功，该函数将返回NULL。否则，它将返回一个指向元素类型为char *的数组的指针，该数组的最后一个元素为空指针。

源字符串中的单词由制表符和空格分隔。您可以根据需要更改定界符。

在对我以前的解决方案发表评论后，您似乎需要以下内容

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

char ** split( const char *s1, const char *s2 )
{
    char **tokens = malloc( sizeof( char * ) );
    int success = tokens != NULL;

    if ( success )
    {
        const char *delim = " \t";
        *tokens = NULL;

        for ( size_t n = 1, len2 = strlen( s2 ); success && *s1; )
        {
            for ( int empty = 1; empty; ) 
            {
                s1 += strspn( s1, delim );
                if ( ( empty = strncmp( s1, s2, len2 ) == 0 ) )
                {
                    s1 += len2;
                }
            }               

            if ( *s1 )
            {
                const char *p = strstr( s1, s2 );

                size_t len1 = p == NULL ? strlen( s1 ) : p - s1;

                char **tmp = realloc( tokens, ( n + 1 ) * sizeof( char * ) );

                if ( ( success = tmp != NULL ) )
                {
                    tokens = tmp;

                    success = ( tokens[n-1] = calloc( 1, len1 +  1 ) )  != NULL;
                    strncpy( tokens[n-1], s1, len1 );
                    tokens[n] = NULL;
                    ++n;

                    s1 += p == NULL ? len1 : len1 + len2; 
                }

                if ( !success )
                {
                    for ( size_t i = 0; i < n; i++ ) free( tokens[i] );
                    free( tokens );
                }
            }
        }
    }       

    return tokens;
}

int main(void) 
{
    const char *s1 = "Hi there AND test test2";
    const char *s2 = "AND";

    char **tokens = split( s1, s2 );

    if ( tokens != NULL )
    {
        for ( char **p = tokens; *p != NULL; ++p )
        {
            puts( *p );
        }

        char **p = tokens;
        do
        {
            free( *p );
        } while ( *p++ != NULL );

        free( tokens );
    }

    return 0;
}

程序输出为