Question

p就像是

[["test"], ["lest"]]

它打印p [j]意味着它打印＆＃34;测试＆＃34;

char **p;
p = explode[i] = split(eachLineOfLsInArray[i]);
for(j=0;p[j];++j)
    puts(p[j]);

但在尝试打印p [j] [0]时给出了段错误，意思是试图打印＆＃34; t＆＃34;

char **p;
p = explode[i] = split(eachLineOfLsInArray[i]);
for(j=0;p[j];++j)
    puts(p[j][0]);

编辑：

继承了整个代码。 Lorem ipsum dolor坐下来，精致的adipistur elit。

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

char **split(const char *s);

int main (int argc, char *argv[]){
    char virtualLs[100];
    char eachLineOfLsInArray[100][100];
    scanf("%[^\t]", virtualLs);
    char *eachLineOfLs;
    eachLineOfLs = strtok(virtualLs, "\n");
    int loopCounterForStuffing;
    loopCounterForStuffing = 0;
    while (eachLineOfLs != NULL)
    {
        strcpy(eachLineOfLsInArray[loopCounterForStuffing], eachLineOfLs);
        eachLineOfLs = strtok(NULL, "\n");
        ++loopCounterForStuffing;
    }

    char **explode[sizeof(eachLineOfLsInArray)/sizeof(*eachLineOfLsInArray)];
    int i,j, n = sizeof(eachLineOfLsInArray)/sizeof(*eachLineOfLsInArray);
    int maximum = 0;
    int maximumindex = 0;
    int maximumname = 0;
    for(i=0;i < n; ++i){
        char **p;
        p = explode[i] = split(eachLineOfLsInArray[i]);
        for(j=0;p[j];++j)
        {
            putchar(p[j][0]);
            printf("%c", p[j][0]);
        }

    }
    return 0;
}

static int wordCount(const char *s){
    char prev = ' ';
    int wc = 0;
    while(*s){
        if(isspace(prev) && !isspace(*s)){
            ++wc;
        }
        prev = *s++;
    }
    return wc;
}

char **split(const char *s){
    int i, wc = wordCount(s);
    char *word, **result = calloc(wc+1, sizeof(char*));
    char *clone = strdup(s);//Note that you are copying a whole
    for(word=strtok(clone, " \t\n"); word; word=strtok(NULL, " \t\n")){
        result[i++] = word;//or strdup(word); and free(clone); before return
    }
    return result;
}

Answer 1

此代码初始化i中的split()并将打印循环限制为已读取的数据（并包含一些调试I / O，主要用于显示调试的其他技术）。

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

char **split(const char *s);

int main(void)
{
    char virtualLs[100];
    char eachLineOfLsInArray[100][100];
    scanf("%[^\t]", virtualLs);
    char *eachLineOfLs;
    eachLineOfLs = strtok(virtualLs, "\n");
    int loopCounterForStuffing;
    loopCounterForStuffing = 0;
    while (eachLineOfLs != NULL)
    {
        printf("each: <<%s>>\n", eachLineOfLs);
        strcpy(eachLineOfLsInArray[loopCounterForStuffing], eachLineOfLs);
        eachLineOfLs = strtok(NULL, "\n");
        ++loopCounterForStuffing;
    }
    printf("LoopC = %d\n", loopCounterForStuffing);

    char **explode[sizeof(eachLineOfLsInArray) / sizeof(*eachLineOfLsInArray)];
    int i, j, n = sizeof(eachLineOfLsInArray) / sizeof(*eachLineOfLsInArray);
    printf("n = %d\n", n);
    for (i = 0; i < loopCounterForStuffing; ++i)
    {
        char **p;
        p = explode[i] = split(eachLineOfLsInArray[i]);
        for (j = 0; p[j]; ++j)
        {
            putchar(p[j][0]);
            printf("%c", p[j][0]);
        }
        putchar('\n');
    }
    return 0;
}

static int wordCount(const char *s)
{
    char prev = ' ';
    int wc = 0;
    while (*s)
    {
        if (isspace(prev) && !isspace(*s))
        {
            ++wc;
        }
        prev = *s++;
    }
    return wc;
}

char **split(const char *s)
{
    int i = 0;
    int wc = wordCount(s) + 1;
    printf("Word count = %d\n", wc);
    char *word;
    char **result = calloc(wc + 1, sizeof(char *));
    char *clone = strdup(s);// Note that you are copying a whole
    for (word = strtok(clone, " \t\n"); word; word = strtok(NULL, " \t\n"))
    {
        printf("Word: <<%s>>\n", word);
        result[i++] = word;// or strdup(word); and free(clone); before return
    }
    result[i] = 0;
    return result;
}

示例输出：

$ ./split <<< "Lorem ipsum dolor sit amet, consectetur adipiscing elit."
each: <<Lorem ipsum dolor sit amet, consectetur adipiscing elit.>>
LoopC = 1
n = 100
Word count = 9
Word: <<Lorem>>
Word: <<ipsum>>
Word: <<dolor>>
Word: <<sit>>
Word: <<amet,>>
Word: <<consectetur>>
Word: <<adipiscing>>
Word: <<elit.>>
LLiiddssaaccaaee
$

输出是每个单词的第一个字符，在结束的行上打印两次。

尝试访问C中字符串数组的第一个字符时的Segfault

1 个答案: