c – strtok() – 为什么必须传递NULL指针才能获取字符串中的下

发布时间：2020-12-16 05:21:52 所属栏目：百科来源：网络整理

导读：这是对strtok()的解释. #include char strtok( char* s1, const char* s2 );* The first call to strtok() returns a pointer to the first token in the string pointed to by s1. Subsequent calls to strtok() must pass a NULL pointer as the first arg

这是对strtok()的解释.

#include char strtok( char* s1,
const char* s2 );*

The first call to strtok() returns a pointer to the first token in the
string pointed to by s1. Subsequent calls to strtok() must pass a NULL
pointer as the first argument,in order to get the next token in the
string.

但是我不知道,为什么你必须传递NULL指针才能获得字符串中的下一个标记.我搜索了大约15分钟,但没有在互联网上找到解释.

解决方法

strtok()将静态变量中的一些数据保存在函数本身内,以便它可以继续从之前的调用离开它的点进行搜索.要发送要保持搜索同一个字符串的strtok(),您将传递一个NULL指针作为其第一个参数. strtok()检查它是否为NULL,如果是,则使用其当前存储的数据.如果第一个参数不为空,则将其视为新搜索,并重置所有内部数据.

也许最好的事情是搜索strtok()函数的实际实现.我已经发现一个小到这里发布,所以你了解如何处理这个NULL参数：

/* Copyright (c) Microsoft Corporation. All rights reserved. */

#include <string.h>

/* ISO/IEC 9899 7.11.5.8 strtok. DEPRECATED.
 * Split string into tokens,and return one at a time while retaining state
 * internally.
 *
 * WARNING: Only one set of state is held and this means that the
 * WARNING: function is not thread-safe nor safe for multiple uses within
 * WARNING: one thread.
 *
 * NOTE: No library may call this function.
 */

char * __cdecl strtok(char *s1,const char *delimit)
{
    static char *lastToken = NULL; /* UNSAFE SHARED STATE! */
    char *tmp;

    /* Skip leading delimiters if new string. */
    if ( s1 == NULL ) {
        s1 = lastToken;
        if (s1 == NULL)         /* End of story? */
            return NULL;
    } else {
        s1 += strspn(s1,delimit);
    }

    /* Find end of segment */
    tmp = strpbrk(s1,delimit);
    if (tmp) {
        /* Found another delimiter,split string and save state. */
        *tmp = '';
        lastToken = tmp + 1;
    } else {
        /* Last segment,remember that. */
        lastToken = NULL;
    }

    return s1;
}

（编辑：李大同）

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