c - 出现段错误(核心转储)

Question

所以我尝试读取行，然后用 strtok 将它们分成两部分。因此，如果我读“nicedog”，它会首先打印我读到的内容，然后在下一行使用 strtok 命令“nice”和“dog”进行打印。但在第二次输入后，我遇到了段错误。另外，free(buf) 是做什么的？我发现错误出现在这一行:“strcpy(name, strtok(NULL, ""));”这是代码:

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

int main()
{
    char *buf;
    char command[32];
    char name[32];

    while((buf = readline("\n"))!=NULL)
    {
        if (strcmp(buf,"exit")==0)
            break;

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

        strcpy(command, strtok(buf, " "));
        printf("%s\n", command);
        strcpy(name, strtok(NULL, " "));
        printf("%s\n", name);
        if(buf[0]!=NULL)
        add_history(buf);
    }
    free(buf);
    return 0;
}

Answer 1

您必须检查 strtok 的结果是否为 NULL，这意味着没有找到标记，您将出现段错误

char *pointer;
pointer = strtok(buf, " ");
if (pointer != NULL)
    strcpy(command, pointer);

此外，readline 会在每次调用时分配新内存，因此您应该在 while 循环内free。

这样解决

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

#include <readline/readline.h>
#include <readline/history.h>

int main()
{
    char *buf;
    char command[32];
    char name[32];

    while((buf = readline("\n"))!=NULL)
    {
        char *pointer;
        if (strcmp(buf,"exit")==0)
            break;

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

        pointer = strtok(buf, " ");
        if (pointer != NULL)
        {
            strcpy(command, pointer);
            /* Don't print poitner otherwise since it is unintialized */
            printf("%s\n", pointer);
        }

        /* subsequent calls to strtok must have first argument NULL */
        pointer = strtok(NULL, " ");
        if (pointer != NULL)
        {
            strcpy(name, pointer);
            printf("%s\n", pointer);
        }

        if (buf != NULL) // this is never FALSE because of the while condition
            add_history(buf);
        free(buf);
    }
    return 0;
}

您还必须确保command和name足够大以适应生成的搅拌。