指针作为第二个参数，而不是返回指针？ [英] pointer as second argument instead of returning pointer?

问题描述

我注意到，它是在C常见的成语，以接受联合国 - 的malloc ED指针作为第二个参数，而不是返回一个指针。例如：

I noticed that it is a common idiom in C to accept an un-malloced pointer as a second argument instead of returning a pointer. Example:

/*function prototype*/    
void create_node(node_t* new_node, void* _val, int _type);

/* implementation */
node_t* n;
create_node(n, &someint, INT)

而不是

/* function prototype */
node_t* create_node(void* _val, int _type)

/* implementation */
node_t* n = create_node(&someint, INT)

什么是两种方法的优点和/或缺点？

What are the advantages and/or disadvantages of both approaches?

谢谢！

修改感谢大家的答案。可供选择1动机是对我现在很清楚（我应该指出，对于选择1指针参数应 malloc分配相反的是我本来以为）。

EDIT Thank you all for your answers. The motivations for choice 1 are very clear to me now (and I should point out that the pointer argument for choice 1 should be malloc'd contrary to what I originally thought).

推荐答案

的接受的指针（其中调用者负责malloc'ing与否）内存要填写，提供了重要优势在灵活性上的返回的指针（一定malloc分配）。特别是，如果该呼叫者知道它需要使用任何的仅在一定函数返回，它可以在一个堆栈分配结构或数组的地址传递;如果它知道它并不需要重入，它可以通过在静态结构或数组的地址 - 在这两种情况下，一个malloc /免费对得到保存，这种储蓄也悬置起来 - ！）

Accepting a pointer (which the caller is responsible for malloc'ing or not) to memory to be filled in, offers serious advantages in flexibility over returning a pointer (necessarily malloc'ed). In particular, if the caller knows it needs to use whatever's returned only within a certain function, it can pass in the address of a stack-allocated struct or array; if it knows it doesn't need reentrancy, it can pass in the address of a static struct or array -- in either case, a malloc/free pair gets saved, and such savings do mount up!-)

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