Question

实际上，我用相同的代码问了另一个问题，但这是非常不同的。我在下面的代码显示了一个非常烦人的行为。我尽可能多地在代码中添加注释，以便您可以了解正在发生的事情。

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

/* This is a struct describing properties of an element */
struct element{
    int age;
    char* name;
};

/* This struct contains a pointer to a pointer on a element "struct element" */
struct person{
    struct element** p;
    size_t size;
    unsigned int id;
};

/* This function initializes a struct person by allocating memory for it */
struct person* init(int _size)
{
    if(_size == 0)
    {
         printf("You gonna have to make some choices \n");
         exit(1);
    }
    struct person* sample = (struct person* )malloc(_size*sizeof(struct person));
    sample->p = (struct element** ) malloc(_size*sizeof(struct element*));
    sample->id = 0;
    sample->size = _size;
    return sample;
}

/* use this function to insert a new element in the struct */
void insert(struct person* sample, char* _name, int _age)
{
    if (sample->id >= sample->size) {
        sample->p = (struct element** ) realloc(sample->p, (sample->size*2) * sizeof(struct element*));
        if(sample->p == NULL){
            printf("Get a new RAM buddy \n");
            exit(1);
        }
    }
    sample->p[sample->id]->name = _name; 
    sample->p[sample->id]->age = _age;  /* of course, this will cause trouble too because it has the same construct as the previous one */
    sample->id++;
}


/* main entry */
int main(int argc, char** argv)
{
    int i = 0;
    struct person* student = init(10); /* Allocating space for 10 students */
    insert(student, "baby", 2);
    insert(student, "dady", 33);
    /* if you remove this line, the program runs, but GDB will signal a segmentation fault. If you keep it, the program will freeze and GDB will behave as expected */
    /* I don't understand why this is happening!!!??? */
    insert(student, "grandma", 63);
    printf("Your name is %s and your age is %d \n", student->p[1]->name, student->p[1]->age);  
    /* When you only insert two elements, use the results here to match with GDB's results*/
    printf("student->p: %p \n", &student->p);
    printf("student->p[0]: %p \n", &student->p[0]);
    printf("student->p[1]: %p \n", &student->p[1]);
    printf("student->p[0]->age: %p \n", &student->p[0]->age);
    printf("student->p[0]->name: %p \n", &student->p[0]->name);
    /* Won't work for more than two elements inserted */    
    for(i = 0; i < 2; i++){
        printf("Your name is %s and your age is %d \n", student->p[i]->name, student->p[i]->age);
    }

    return 0;
}

我希望你能弄明白发生了什么。这是调试会话的一部分。

(gdb) run
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: C:\Users\NTWALI\Desktop\tests\help\bin\Debug/help.exe
[New thread 11408.0x1228]
Error: dll starting at 0x770a0000 not found.
Error: dll starting at 0x76ab0000 not found.
Error: dll starting at 0x770a0000 not found.
Error: dll starting at 0x76d40000 not found.

Program received signal SIGSEGV, Segmentation fault.
0x0040146f in insert (sample=0x6816c0, _name=0x409031 "ntwali", _age=22) at C:\Users\NTWALI\Desktop\tests\help\main.c:44
44          sample->p[sample->id]->name = _name; 
(gdb) p sample
$4 = (struct person *) 0x6816c0
(gdb) p sample->p
$5 = (struct element **) 0x681750
(gdb) p sample->p[0]
$6 = (struct element *) 0xbaadf00d
(gdb) p sample->p[1]
$7 = (struct element *) 0xbaadf00d
(gdb)

正如您在代码注释中看到的那样，程序在工作时提供的数据与GDB获得的数据不匹配。

感谢您的帮助。

Answer 1

就我所见，你还没有为元素分配任何内存。在这里，您为指向元素的指针分配内存：

sample->p = (struct element** ) malloc(_size*sizeof(struct element*));

Answer 2

如果调试器的存在改变了程序的行为方式，则很可能会误用内存或线程。就像daven11 points out一样，你没有自己分配元素。

Answer 3

您的问题的根本原因是您正在分配指向struct element的指针，但这些指针未初始化 - 您没有分配任何实际的struct element对象。当您取消引用这些无效指针时，您会得到未定义的行为。

也没有必要分配_size结构的struct person - 你只使用一个。{1}}结构。您的struct person应该是这样的（注意p的类型不同）：

struct person{
    struct element *p;
    size_t size;
    unsigned int id;
};

然后您的init()函数应如下所示：

struct person* init(int _size)
{
    if(_size < 1)
    {
         printf("You gonna have to make some choices \n");
         exit(1);
    }
    struct person* sample = malloc(sizeof *sample);
    sample->p = malloc(_size * sizeof sample->p[0]);
    sample->id = 0;
    sample->size = _size;
    return sample;
}

insert()函数应如下所示：

void insert(struct person* sample, char* _name, int _age)
{
    if (sample->id >= sample->size) {
        sample->size *= 2;
        sample->p = realloc(sample->p, sample->size * sizeof sample->p[0]);
        if(sample->p == NULL){
            printf("Get a new RAM buddy \n");
            exit(1);
        }
    }
    sample->p[sample->id].name = _name; 
    sample->p[sample->id].age = _age;  /* of course, this will cause trouble too because it has the same construct as the previous one */
    sample->id++;
}

然后，主要功能应使用student->p[i].name和student->p[i].age来访问数据。

Answer 4

这里你使用p []而没有先将其初始化为指向任何东西。你只为指针分配了空间，但没有初始化它们指向任何东西。所以当你这样做时

sample->p[sample->id]->name = _name; 
sample->p[sample->id]->age = _age;

p指向内存中的某个位置并且您正在修改它所指向的位置。

而是插入

sample->p[sample->id] = malloc(struct element);
sample->p[sample->id]->name = _name; 
sample->p[sample->id]->age = _age;

它应该有效

PS。通常你不会在C

中投射malloc

在C中添加一条指令后的奇数程序行为

4 个答案: