Question

所以我试图读取行，然后用strtok将它们分成两行。所以，如果我会阅读＆＃34;好狗＆＃34;它将首先打印我读到的内容，然后使用strtok命令进行打印＆＃34; nice＆＃34;和＃34;狗＆＃34;在下一行。但是在第二次输入后我得到了Segmentation故障。而且，free（buf）做了什么？我已经看到错误出现在这一行：＆＃34; strcpy（name，strtok（NULL，＆＃34;＆＃34;））;＆＃34;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NULL是否segmentation fault意味着找不到您的代币char *pointer; pointer = strtok(buf, " "); if (pointer != NULL) strcpy(command, pointer);

readline

另外，free会在每次调用时分配新内存，因此您应该在while循环中#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; }。

以这种方式修复

command

您还必须确保name和{{1}}足够大，以适应最终的搅拌。

获取分段错误（核心转储）

1 个答案: