任何方式来初始化动态分配的结构？

Question

C11给我们带来了整洁的结构初始化语法：

struct some_struct {int some; char value;}; 
struct some_struct s = {.some = 5, .value = 'a'}; 
printf("some = %d, value = %c\n", s.some, s.value); 

http://ideone.com/zZxTc4

但是，它似乎不工作的时候该结构具有动态分配：

struct some_struct {int some; char value;}; 
struct some_struct *s = malloc(sizeof(struct some_struct)); 
*s = {.some = 5, .value = 'a'}; 
printf("some = %d, value = %c\n", s->some, s->value); 

此代码产量：

error: expected expression before ‘{’ token 
    *s = {.some = 5, .value = 'a'}; 
    ^

个http://ideone.com/nBiorw

struct some_struct {int some; char value;}; 
struct some_struct *s = malloc(sizeof(struct some_struct)); 
*s = struct some_struct {.some = 5, .value = 'a'}; 
printf("some = %d, value = %c\n", s->some, s->value); 

而这种代码可得：

error: expected expression before ‘struct’ 
    *s = struct some_struct {.some = 5, .value = 'a'}; 
     ^~~~~~ 

http://ideone.com/SpBX1E

是否有任何C的方式来初始化动态分配的结构是很好的或者我宁愿写类似的东西：

struct some_struct {int some; char value;}; 
struct some_struct *s = malloc(sizeof(struct some_struct)); 
s->some = 5; 
s->value = 'a'; 
printf("some = %d, value = %c\n", s->some, s->value); 

http://ideone.com/1XAiQS

因为（a）它意味着该结构的每个成员都占有自己的路线，因为人们往往当他们看到不是一个;标志其行的源代码的最后标志皱眉哪个稍微不便是，和（b）我不能像在正常的结构体初始化中那样依赖初始化为0的未指定的值。

Answer 1

在你的代码

*s = {.some = 5, .value = 'a'}; 

是不的initialization，这是一个assignment。

Brace enclosed initializer只能在初始化期间使用。

您可以使用compound literal，虽然。像

*s = (struct some_struct){.some = 5, .value = 'a'}; 

将完成这项工作。