我应该显式转换 malloc() 的返回值吗? [英] Should I explicitly cast malloc()'s return value?

问题描述

我想咨询以下案例:

char *temp;
temp = malloc(10);

由于malloc的返回类型是void*，malloc返回的指针是否会隐式转换为char* 类型在被分配给临时之前?标准在这方面是怎么说的?

Since the return type of malloc is void*, will the pointer returned by the malloc be implicitly cast to char* type before being assigned to temp? What does the standard say in this regard?

如果我们的指针变量是某种结构类型，例如:

If our pointer variable is some struct type for example:

struct node *temp;
temp = (struct node *)malloc(sizeof(struct node));

如果我们将内存分配给 temp 而不将其转换为 struct node* 类型，它会被隐式转换为 struct node* 类型还是有必要显式转换它struct node* 类型?

If we allocate memory to temp without casting it to struct node* type, will it be implicitly cast to struct node* type or is it necessary to explicitly cast it to struct node* type?

推荐答案

C 中的空指针可以分配给任何指针，无需显式转换.

A void pointer in C can be assigned to any pointer without an explicit cast.

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