Question

以下代码来自本网站的示例。我无法理解，我做错了什么？你能帮帮我吗？

编译：

gcc -std = c11 main.c

仅打印：

东西：煮荞麦，重量：1500

分段错误

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

typedef struct {
    // Weight in grams
    size_t weight;
    // Name of Thing
    char name[255];

} Things;

void add_new_thing(Things **things,size_t *size)
{

    size_t index = *size;

    if(index == 0){
        (*size) = 1;
        *things = (Things*)calloc((*size),sizeof(Things));
        if (*things == NULL) {
            fprintf(stderr, "Error: can't allocate memory! %s\n", strerror(errno));
            exit(EXIT_FAILURE);
        }
    }else{
        (*size) += 1;
        Things *temp = (Things*)realloc(*things,(*size)*sizeof(Things));
        if(temp != NULL) {
            *things = temp;
        }else{
            fprintf(stderr, "Error: can't reallocate memory! %s\n", strerror(errno));
            exit(EXIT_FAILURE);
        }
        // Zeroing of new structure's elements
        things[index]->name[0] = '\0';
        things[index]->weight = 0;
    }

}

void another_function(Things *things,size_t *size)
{
    // Add one element to the array of structures
    add_new_thing(&things,size);
    const char *str1 = "Boiled buckwheat";
    strncpy(things[*size-1].name, str1, strlen(str1) + 1);
    things[*size-1].weight = 1500;

    for(size_t i = 0;i < *size;i++){
        printf("Thing: %s, weight: %zu\n",things[i].name,things[i].weight);
    }

    // Add one more element to the array of structures
    add_new_thing(&things,size);
    const char *str2 = "A toy";
    strncpy(things[*size-1].name, str2, strlen(str2) + 1);
    things[*size-1].weight = 350;

    // Segmentation fault is below
    for(size_t i = 0;i < *size;i++){
        printf("Thing: %s, weight: %zu\n",things[i].name,things[i].weight);
    }
}

void some_function(Things *things,size_t *size)
{
    // To pass the array of structures to another function
    another_function(things,size);
}

int main(void)
{

    // Create NULL pointer for the array of structures
    Things *things = NULL;

    // Add size of structures' array which will be allocated within add_new_thing() function
    size_t size = 0;

    // Call some function
    some_function(things,&size);

    // Segmentation fault is below
    printf("Print results:\n");
    for(size_t i = 0;i < size;i++){
        printf("Thing: %s, weight: %zu\n",things[i].name,things[i].weight);
    }

    free(things);

    return(EXIT_SUCCESS);
}

Answer 1

请记住，C有按值调用，这意味着在main函数中，您将things中的空指针副本传递给{{1} }。 some_function中的实际变量不会改变。

仅在main模拟通过引用传递，并且仅another_function another_function things变量已由add_new_thing中的分配更新。

Answer 2

真正的问题在这里

// Zeroing of new structure's elements
things[index]->name[0] = '\0';
things[index]->weight = 0;

必须

(*things)[index].name[0] = '\0';
(*things)[index].weight = 0;

这是因为，things不是指针的指针，而只是指针。

您正在将things视为指向指针数组的指针，但它只是指向＆＃34; 数组的指针＆＃34; Things。我说＆＃34; 数组＆＃34;，因为它严格来说不是数组，数组在c中是不同的。但它的所有目的都与数组相同。

你也可以在main中创建指针，但是你从不正确地使用指针的那个副本，你仍然free()它。

尝试阅读更正的代码，看看你是否能理解你的错误

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

typedef struct
{
    // Weight in grams
    size_t weight;
    // Name of Thing
    char name[255];

} Things;

void add_new_thing(Things **things,size_t *size)
{

    size_t index = *size;

    if(index == 0)
        {
            (*size) = 1;
            *things = (Things*)calloc((*size),sizeof(Things));
            if (*things == NULL)
                {
                    fprintf(stderr, "Error: can't allocate memory! %s\n", strerror(errno));
                    exit(EXIT_FAILURE);
                }
        }
    else
        {
            (*size) += 1;
            Things *temp = (Things*)realloc(*things,(*size)*sizeof(Things));
            if(temp != NULL)
                {
                    *things = temp;
                }
            else
                {
                    fprintf(stderr, "Error: can't reallocate memory! %s\n", strerror(errno));
                    exit(EXIT_FAILURE);
                }
            // Zeroing of new structure's elements
            (*things)[index].name[0] = '\0';
            (*things)[index].weight = 0;
        }

}

void another_function(Things **things, size_t *size)
{
    // Add one element to array of structures
    add_new_thing(things,size);
    const char *str1 = "Boiled buckwheat";
    strncpy((*things)[*size-1].name, str1, strlen(str1) + 1);
    (*things)[*size-1].weight = 1500;

    for(size_t i = 0; i < *size; i++)
        {
            printf("Thing: %s, weight: %zu\n",(*things)[i].name,(*things)[i].weight);
        }
    // One element of array of structures was printed there

    // Add new one element to array of structures
    add_new_thing(things, size);
    const char *str2 = "A toy";
    strncpy((*things)[*size-1].name, str2, strlen(str2) + 1);
    (*things)[*size-1].weight = 350;

    // Segmentation fault is there
    for(size_t i = 0; i < *size; i++)
        {
            printf("Thing: %s, weight: %zu\n",(*things)[i].name,(*things)[i].weight);
        }
}

void some_function(Things **things, size_t *size)
{
    // Pass array of structures to another function
    another_function(things, size);
}

int main(void)
{

    // Create NULL pointer for array of structures
    Things *things = NULL;

    // And size of structures array which will be allocated within add_new_thing() function
    size_t size = 0;

    // Call some function
    some_function(&things, &size);

    // Segmentation fault is there
    printf("Print results:\n");
    for(size_t i = 0; i < size; i++)
        {
            printf("Thing: %s, weight: %zu\n",things[i].name,things[i].weight);
        }
    free(things);

    return(EXIT_SUCCESS);
}

Answer 3

在您的main函数中，您将things（NULL）的值传递给函数some_function()。所以这个指针没有改变，你需要传递它的地址。 printf()调用尝试访问NULL存储的内容。（显然这是不可能的）

如何正确动态分配内存？

3 个答案: