如何在C中动态分配结构?

问题描述

我最近有一个作业，必须为结构动态分配内存.我使用了这种方法:

I recently had an assignment where I had to dynamically allocate memory for a struct. I used this method:

myStruct *struct1 = malloc(sizeof *struct1);

这很好.但是，我不知道如何.我认为 struct1 指针当时尚未初始化，因此应该没有任何大小.因此， malloc(sizeof * struct1)如何返回有效的内存量进行分配?

This worked just fine. However, I don't understand how. I would think that the struct1 pointer is uninitialized at that point and should therefore have a size of nothing. So therefore how does malloc(sizeof *struct1) return a valid amount of memory to allocate?

推荐答案

sizeof 运算符不评估操作数.它只是看类型.例如:

sizeof operator in C doesn't evaluate the operand. It just looks at the type. For example:

#include <stdio.h>

int main(void)
{
    int i = 0;
    printf("%zu\n", sizeof i++);
    printf("%d\n", i);
    return 0;
}

如果运行上述程序，则会看到 i 仍为0.

If you run the above program, you will see that i is still 0.

因此，在您的示例中，未评估 * struct1 ，它仅用于类型信息.

So, in your example, *struct1 is not evaluated, it's only used for type information.

这篇关于如何在C中动态分配结构?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！