初始化一组结构体

I am trying to initialize an array of structs to zero. For some reason, I am getting the error "expected expression". What's wrong with my code?

我试图将一个结构数组初始化为0。由于某些原因，我得到了“预期表达式”的错误。我的代码有什么问题?

struct mystruct {
    double a;
    double arr[2];
}

int main() {
    struct mystruct *array = (struct mystruct*)malloc(3 * sizeof(struct mystruct));
    for (int i = 0; i < 3; i++) {
        array[i] = { 0 };
    }
    return 0;
}

2 个解决方案

#1

Your code has a syntax error. There are different ways to correct it:

您的代码有一个语法错误。有不同的纠正方法:

Using the C99 syntax for compound literals:

使用C99语法进行复合文字:
```
array[i] = (struct mystruct){ 0 };
```
Using a separate structure with the default values:

使用具有默认值的独立结构:
```
struct mystruct def = { 0 };
...
array[i] = def;
```
Initializing each member explicitly:

初始化每个成员明确:
```
array[i].a = 0;
array[i].arr[0] = 0;
array[i].arr[1] = 0;
```
Using calloc():

使用calloc():
```
struct mystruct *array = calloc(3, sizeof(*array));
```
calloc() initializes the memory block to all bits zero. If your system uses IEEE representation for double, all bits zero corresponds to the value 0.0. If it does not, I will give you one dollar.

calloc()初始化内存块到所有位为零。如果您的系统使用IEEE表示为双，所有位零对应的值为0.0。如果没有，我会给你一美元。

#2

Aggregate initialization is only available for, well... initialization:

聚合初始化只能用于，嗯…初始化:

struct mystruct x = {0};   // initialization

There's no such grammar for assignment.

作业中没有这样的语法。

But you don't need to, just use calloc instead of malloc for dynamically allocated, zeroed memory.

但是您不需要这样做，只需使用calloc而不是malloc来进行动态分配、调零内存。

struct mystruct* array = calloc(3, sizeof(struct mystruct));

#1