在C中使用结构分割错误

时间:2018-12-28 22:47:51

标签: c pointers struct segmentation-fault

我正在通过编写一个小型C程序来测试我们最近在课堂上学到的关于结构和指针的知识。但是,运行它后,我遇到了segmentation fault (core dumped)错误。有人可以帮我弄清楚到底是什么原因吗?我是否错过了任何悬空的指针,或者使用malloc()做错了什么?

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

const char *admin = "Administrator";
const char *mng = "Manager";
const char *prog = "Programmer";

struct employee {
    char *name;
    char title[20];
    char id[8];
    int yearsExp;
};

typedef struct employee emp_t;

emp_t *everyone[1];

emp_t *create_emp(const char *name,const char *title,const char *id,int yrxp) {
    emp_t *new;

    new = (emp_t *) malloc(sizeof(emp_t));
    new->name = (char*) malloc(strlen(name) + 1);
    strcpy(new->name,name);
    strcpy(new->title,title);
    strcpy(new->id,id);
    new->yearsExp = yrxp;
    return new;
}

void free_emp(emp_t *employee) {
    free(employee->name);
    free(employee);
}

int main() {
    int i;

    everyone[0] = create_emp("Mike Papamichail",prog,"A197482",3);
    everyone[1] = create_emp("Maria Mamalaki",mng,"Z104781",6);

    for(i = 0; i < 2;i++) {
        printf("%15s \t %15s \t %10s \t %4d\n",
        everyone[0]->name,everyone[0]->title,everyone[0]->id,everyone[0]->yearsExp);
        free_emp(everyone[i]);
        everyone[i] = NULL;
    }
    return 0;
}

为清楚起见,更新了代码

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

const char *admin = "Administrator";
const char *mng = "Manager";
const char *prog = "Programmer";

struct employee {
    char *name;
    char title[20];
    char id[8];
    int yearsExp;
};

typedef struct employee emp_t;

emp_t *create_emp(const char *name,const char *title,const char *id,int yrxp) {
    emp_t *new;

    new = (emp_t *) malloc(sizeof(emp_t));
    new->name = (char*) malloc(strlen(name) + 1);
    strcpy(new->name,name);
    strcpy(new->title,title);
    strcpy(new->id,id);
    new->yearsExp = yrxp;
    return new;
}

void free_emp(emp_t *employee) {
    free(employee->name);
    free(employee);
}

int main() {
    int i;
    emp_t *everyone[2];

    everyone[0] = create_emp("Mike Papamichail",prog,"A197482",3);
    everyone[1] = create_emp("Maria Mamalaki",mng,"Z104781",6);

    for(i = 0; i < 2;i++) {
        printf("%15s \t %15s \t %10s \t %4d\n",
        everyone[0]->name,everyone[0]->title,everyone[0]->id,everyone[0]->yearsExp);
        free_emp(everyone[i]);
        everyone[i] = NULL;
    }
    return 0;
}

2 个答案:

答案 0 :(得分:3)

您在所有点上都很接近@tadman,在用emp_t *everyone[1]创建一个包含1个指针的数组时遇到了最大的错误(在您的问题中已删除)。

与代码有关的其余问题更多是相对而言的一些小的更正或改进。

对于初学者,请避免在代码中使用“魔术数字” (例如20, 8, 2)。如果您需要常量,请#define,或使用全局enum一次定义多个常量,例如

/* an enum can be used to #define multiple constants */
enum { MAXID = 8, MAXTITLE = 20, MAXEMP = 128 };

然后

typedef struct {    
    char *name;
    char title[MAXTITLE];   /* avoid "magic-numbers", use constants */
    char id[MAXID];
    int yearsExp;
} emp_t;

您的create_emp()函数将在很大程度上按原样运行,但是无需在C中强制转换malloc, callocrealloc的返回值,请参见:Do I cast the result of malloc?。另外,我将避免使用new作为临时结构名称。虽然C中没有实际的冲突,但是new是C ++中的关键字,如果您要同时编写两者,则最好记住这一点。经过几次调整,您将编写create_emp(),如下所示:

emp_t *create_emp (const char *name, const char *title, 
                    const char *id, int yrxp) {
    emp_t *newemp;  /* new is a keyword in C++ best to avoid */

    newemp = malloc (sizeof *newemp);   /* don't cast return of malloc */
    /* if you allocate, you must validate */
    if (newemp == NULL) {
        perror ("malloc-newemp");
        return NULL;
    }
    newemp->name = malloc (strlen(name) + 1);
    if (newemp->name == NULL) {
        perror ("malloc-newemp->name");
        free (newemp);
        return NULL;
    }
    strcpy (newemp->name, name);
    strcpy (newemp->title, title);
    strcpy (newemp->id, id);
    newemp->yearsExp = yrxp;

    return newemp;
}

注意:始终验证每个分配。malloc, calloc & realloc可以并且确实会失败)

最后,在main()中,您可以使用索引(下面的ndx)代替 magic-number 2,并在每次添加后增加索引。当您很好地使用 field-width 修饰符来控制输出字段的大小时,对于字符串(和所有转换说明符),可以将'-'标志作为转换说明符的一部分包含在以下内容中:将字段设置为 Left-Justified (左对齐)以正确对齐输出。您代码的其余部分都很好。

完全将其放入,您可以执行以下操作:

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

/* an enum can be used to #define multiple constants */
enum { MAXID = 8, MAXTITLE = 20, MAXEMP = 128 };
#define ADMIN   "Administrator" /* simple #defines are fine */
#define MNG     "Manager"       /* string literals are fine as well */
#define PROG    "Programmer"    /* (up to you) */

typedef struct {    
    char *name;
    char title[MAXTITLE];   /* avoid "magic-numbers", use constants */
    char id[MAXID];
    int yearsExp;
} emp_t;

emp_t *create_emp (const char *name, const char *title, 
                    const char *id, int yrxp) {
    emp_t *newemp;  /* new is a keyword in C++ best to avoid */

    newemp = malloc (sizeof *newemp);   /* don't cast return of malloc */
    /* if you allocate, you must validate */
    if (newemp == NULL) {
        perror ("malloc-newemp");
        return NULL;
    }
    newemp->name = malloc (strlen(name) + 1);
    if (newemp->name == NULL) {
        perror ("malloc-newemp->name");
        free (newemp);
        return NULL;
    }
    strcpy (newemp->name, name);
    strcpy (newemp->title, title);
    strcpy (newemp->id, id);
    newemp->yearsExp = yrxp;

    return newemp;
}

void free_emp (emp_t *employee) {
    free (employee->name);
    free (employee);
}

int main (void) {

    int i, ndx = 0;     /* use an index instead of magic-numbers */

    emp_t *everyone[MAXEMP] = {NULL};

    everyone[ndx++] = create_emp ("Mike Papamichail", PROG, "A197482", 3);
    everyone[ndx++] = create_emp ("Maria Mamalaki", MNG, "Z104781", 6);
    everyone[ndx++] = create_emp ("Sam Sunami", ADMIN, "B426310", 10);

    for (i = 0; i < ndx; i++) { /* %- to left justify fields */
        if (everyone[i]) {      /* validate not NULL */
            printf ("%-15s \t %-15s \t %-10s \t %4d\n", everyone[i]->name, 
                    everyone[i]->title, everyone[i]->id, 
                    everyone[i]->yearsExp);
            free_emp (everyone[i]);
            everyone[i] = NULL;
        }
    }

    return 0;
}

注意:从everyone[0]->name更改为everyone[i]->name,依此类推,否则输出将永远不变)

使用/输出示例

$ ./bin/emp_struct
Mike Papamichail         Programmer              A197482            3
Maria Mamalaki           Manager                 Z104781            6
Sam Sunami               Administrator           B426310           10

仔细检查一下,如果还有其他问题,请告诉我。

答案 1 :(得分:0)

最后,您要遍历数组,打印everyone[0],然后释放everyone[i]。因此,在第二次迭代中,everyone[0]将为null,并且代码崩溃...