为什么使用SomeType [1]而不是SomeType *作为结构中的最后一个成员 [英] Why use SomeType[1] instead of SomeType* as last member in structs

问题描述

我在代码下面的语句中看到：

  SomeType someVar [1] 
  

后来 someVar SomeType 。为什么要这样声明它，而不是：

  SomeType * someVar; 
  

编辑：很多人指出，我需要指定上下文。

  struct SomeStruct 
 {
 size_t count; 
 SomeType someVar [1]; 
}; 
 
 SomeStruct * str = malloc（sizeof（SomeStruct）+ sizeof（SomeType）* count）; 
 str-> count = count 
 ... 
  

str-> someVar 用作 count 的元素的元素SomeType 从这一点开始。

解决方案

  SomeType someVar [1] 
  

someVar 是 SomeType 有1个元素。

  SomeType * someVar; 
  

someVar / c>

$

b $ b

您可以使用数组的名称作为指向该数组第一个元素的指针的缩写。

Will Dean

I saw in the code next statement:

SomeType someVar[1];

Later someVar is used as a pointer to SomeType. Why would one declare it like this instead of:

SomeType* someVar;

EDIT: As many pointed out, I need to specify the context.

struct SomeStruct
{
    size_t count;
    SomeType someVar[1];
};

SomeStruct* str = malloc(sizeof(SomeStruct) + sizeof(SomeType) * count);
str->count = count
...

str->someVar is used as an array with count elements of SomeType from this point.

解决方案

SomeType someVar[1];

someVar is an array of type SomeType with 1 element.

SomeType* someVar;

someVar is a pointer (dangling still, you didn't point it to anything yet) of type SomeType.

And you can use the name of an array on its own as a shorthand for a pointer to the first element of that array.

Will Dean

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