如何在memcpy之前创建malloc结构数组

Question

我认为没有malloc'ing检查就可以执行memcpy。但是，我不确定如何更正下面的代码，即。我们如何malloc结构数组'检查'？

这里是结构的定义：

struct contain structArrayToBeCheck[] = { 
    { 
     .a = "John", 
     .allowed = 1, 

      .suit = { 
      .t = { 
       .option = "ON", 
       .count = 7 
      }, 

      .inner = { 
       .option = "OFF", 
       .count = 7 
      } 
     } 
    }, 
    { 
     .a = "John", 
     .allowed = 1, 

     .suit = { 
      .t = { 
       .option = "ON", 
       .count = 7 
      }, 

      .inner = { 
       .option = "OK", 
       .count = 7 
      } 
     } 
    }, 
    { 
     .a = "John", 
     .allowed = 1, 

     .suit = { 
      .t = { 
       .option = "ON", 
       .count = 7 
      }, 

      .inner = { 
       .option = "OFF", 
       .count = 7 
      } 
     } 
    }, 

}; 
struct contain check[]; 

在主

（）

int i; 

    int n = sizeof(structArrayToBeCheck)/sizeof(struct contain); 
    printf("There are %d elements in the array.\n", n); 

    struct contain **check = malloc(n*sizeof(struct contain *)); 

    for (i = 0; i != n ; i++) { 
     check[i] = malloc(sizeof(struct contain)); 
    } 

    memcpy(&check, &structArrayToBeCheck, sizeof(structArrayToBeCheck)); 
    //printf("check is %s\n", check[1]->suit.inner.option); 

[迈克尔·伯尔和解决：

struct contain { 
char* a;  // 
int allowed; // 

struct suit { 
    struct t { 
      char* option; 
      int count; 
    } t; 

    struct inner { 
      char* option; 
      int count; 
    } inner; 
} suit; 
}; 

我们与一些值初始化它JKB]

int i; 

    int n = sizeof(structArrayToBeCheck)/sizeof(struct contain); 
    printf("There are %d elements in the array.\n", n); 

    struct contain *check = malloc(n*sizeof(struct contain)); 

    memcpy(check, structArrayToBeCheck, sizeof(structArrayToBeCheck)); 

    // do things with check[0], check[1], ... check[n-1] 
    printf("check is %s\n", check[1].suit.inner.option); 

    free(check); 

Answer 1

此：

memcpy(&check, &structArrayToBeCheck, sizeof(structArrayToBeCheck)); 

是无效的，因为你复制结构的数组指针数组。

请记住，您动态分配的一组结构不是连续的。指针数组是连续的，但它们指向分开的所有东西。

尝试：

for (i = 0; i != n ; i++) { 
    check[i] = malloc(sizeof(struct contain)); 
    memcpy(check[i], ArrayToBeCheck[i], sizeof(ArrayToBeCheck[i])); 
} 

而且，如果结构副本始终分配/释放的块（如在你的例子），也没有必要单独对其进行分配。只需为整个阵列分配足够的空间：

struct contain *check = malloc(n*sizeof(struct contain)); 

memcpy(check, structArrayToBeCheck, sizeof(structArrayToBeCheck)); 

// do things with check[0], check[1], ... check[n-1] 

free(check); 

Answer 2

struct contain **check = malloc(n*sizeof(struct contain *)); 

for (int i = 0; i != n ; i++) { 
    check[i] = malloc(sizeof(struct contain)); 
} 

可能这是allocate.what你想在这里n是你要多少数组的大小安全的方式。

然后

// Do not forget to free the memory when you are done: 
for (int i = 0; i != n ; i++) { 
    free(check[i]); 
} 
free(check);