Question

我正在尝试创建函数，它将字符串作为参数，将其划分为字符串，将空格分隔的单词，将其保存到数组并通过指针返回。尽管内存静态分配给数组，但是由于分段错误，程序崩溃了。这段代码有什么问题？

void separate_words(char* full_text, char *matrix[], int* how_many)
{
char tmp;
int actual_letter,which_letter=0;
for(actual_letter=0;actual_letter<strlen(full_text);actual_letter++)
{
    if(full_text[actual_letter]!=32)
{


    full_text[actual_letter];
    matrix[*how_many][whitch_letter]=full_text[actual_letter];//here crashes
}
else
{  
    *how_many++;
    which_letter=0;
}
}

//*how_many
}

/*...*/
char words[20][20];
char text[20];
int number_of_words=0;
separate_words(text,words,&number_of_words);

Answer 1

当您使用strtok

时，这是一个典型问题

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

void separate_words(char* full_text, char *matrix[], int* how_many)
{
    char *pch;
    pch = strtok (full_text," \t\n");
    int i = 0;
    while (pch != NULL)
    {
        matrix[i++] = pch;
        pch = strtok (NULL, " \t\n");
    }
    *how_many = i;
}

int main ()
{
    char str[] = "apple   banana orange pineapple";
    char *matrix[100] = { NULL };
    int how_many = 0;
    separate_words(str, matrix, &how_many);

    int i;
    for( i = 0; i < how_many; i++ )
    {
        if( matrix[i] != NULL )
        {
            printf( "matrix[%d] = %s\n", i, matrix[i] );
        }
    }
    return 0;
}

输出：

matrix[0] = apple
matrix[1] = banana
matrix[2] = orange
matrix[3] = pineapple

Answer 2

将分隔符与strtok一起用作函数的参数以将字符串分隔为标记通常很有用。在某些情况下，您还可以使用for循环strtok来整理代码。

由于分配给结果数组的指针数永远不会为负数，因此选择size_t或unsigned的数据类型可以帮助编译器指出该变量后来是否使用不当

最后，如果使用一个静态大小的指针数组来保存每个标记的指针，那么测试指定的指针数是很重要的，以防止写入数组的末尾。 注意：如果您动态分配指针（使用malloc或calloc），则可以在达到初始限制时调用realloc，然后根据需要调用#include <stdio.h> #include <string.h> #define MAXP 100 void separate_words (char *a[], char *s, char *delim, size_t *n); int main (void) { char str[] = "apple banana orange pineapple"; char *delim = " \t\n"; char *matrix[MAXP] = { NULL }; size_t how_many = 0; size_t i; separate_words (matrix, str, delim, &how_many); for (i = 0; i < how_many; i++) printf ("matrix[%zu] = %s\n", i, matrix[i]); return 0; } /* separate 's' into tokens based on delimiters provided in 'delim' * with pointers to each token saved in 'a' with 'n' updated to hold * the number of pointers contained in 'a'. */ void separate_words (char *a[], char *s, char *delim, size_t *n) { *n = 0; char *p = strtok (s, delim); for (; p; p = strtok (NULL, delim)) { a[(*n)++] = p; if (*n == MAXP) { fprintf (stderr, "warning: pointer limit reached.\n"); return; } } }。

将各种部分放在一起，另一种选择是：

$ ./bin/strtok_static
matrix[0] = apple
matrix[1] = banana
matrix[2] = orange
matrix[3] = pineapple

<强>输出

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

#define MAXP 100

size_t separate_words (char *a[], char *s, char *delim);

int main (void)
{
    char str[] = "apple  banana orange pineapple";
    char *delim = " \t\n";
    char *matrix[MAXP] = { NULL };
    size_t how_many = 0;
    size_t i;

    how_many = separate_words (matrix, str, delim);

    for (i = 0; i < how_many; i++)
        printf ("matrix[%zu] = %s\n", i, matrix[i]);

    return 0;
}

/* separate 's' into tokens based on delimiters provided in 'delim'
 * with pointers to each token saved in 'a'. The number of tokens is
 * returned.
 */ 
size_t separate_words (char *a[], char *s, char *delim)
{
    size_t n = 0;
    char *p = strtok (s, delim);
    for (; p; p = strtok (NULL, delim)) {
        a[n++] = p;
        if (n == MAXP) {
            fprintf (stderr, "warning: pointer limit reached.\n");
            break;
        }
    }
    return n;
}

使用退货

此外，您可以使用函数return返回字符串中的标记数。这消除了将指针变量作为参数传递给函数的需要：

url(r'^guitar/$', include('guitar.urls')),

将字符串转换为字符串数组

2 个答案: