指针-转换为结构指针+ Malloc [英] Pointers - Casting as Struct Pointer + Malloc

问题描述

我对此有一些想法，但是我想要一个更专业和更有经验的解释.

I have some ideas of what this would do, but I'd like a more professional and experienced explaination.

typedef struct{
     char str[50];
     unsigned short num;
}s;

s *name = (s *) malloc(sizeof(s));

第一部分定义一个结构.我没问题
这将创建一个指向"s"结构的指针. malloc()返回一个内存地址，该地址将被转换为指向"s"的指针.但是，sizeof(s)我相信我在理解某些问题.

char str[50]; = 50个字节.
unsigned short num; = 2个字节.

The first part defines a struct. I have no problem with that
This will create a pointer to "s" struct. malloc() returns a memory address that will be cast as a pointer to "s". However, sizeof(s) I Believe I am having some issues understanding.

char str[50]; = 50 bytes.
unsigned short num; = 2 bytes.

sizeof(s)返回52个字节?

malloc(sizeof(s))分配52个字节并返回第一个地址并将其转换为指向"s"的指针?

sizeof(s) returns 52 bytes?

malloc(sizeof(s)) allocates 52 bytes and returns the first address and casts it as a pointer to "s"?

还有一个问题！ 如何使用内联malloc执行错误处理?

One More Question! How can I perform error handling with malloc in-line?

感谢任何可以帮助您澄清这一点的人！

Thanks to anyone that can help clarify this!

推荐答案

• 在C语言中，我们不需要强制转换malloc的返回值. malloc返回void *，在C语言中，void *隐式转换为您指定的任何类型...
• sizeof(s)返回的值取决于填充以及实现(cos，sizeof(unsigned short)在不同平台上会有所不同...).

• 在C中，如果要检查错误，则必须将malloc的返回值与NULL进行比较.

  • In C, we need not cast the return value of malloc. malloc returns void* and in C, void* implicitly cast to whatever type you specify...
  • The value returned by sizeof(s) depends on padding and also on the implementation (cos, sizeof(unsigned short) will be different on different platforms...).

  • In C, if you want to check for the error, you have to compare the return value of malloc with NULL.

    如果(name == NULL)退出(1); //EXIT_FAILURE

    if (name ==NULL) exit (1); //EXIT_FAILURE

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