struct声明末尾的[1]的目的是什么？ [英] What's the purpose of this [1] at the end of struct declaration?

问题描述

我正在窥探MSP430微控制器的头文件，然后在< setjmp.h> 中遇到了这个问题：

I was snooping through my MSP430 microcontroller's header files, and I ran into this in <setjmp.h>:

/* r3 does not have to be saved */
typedef struct
{
    uint32_t __j_pc; /* return address */
    uint32_t __j_sp; /* r1 stack pointer */
    uint32_t __j_sr; /* r2 status register */
    uint32_t __j_r4;
    uint32_t __j_r5;
    uint32_t __j_r6;
    uint32_t __j_r7;
    uint32_t __j_r8;
    uint32_t __j_r9;
    uint32_t __j_r10;
    uint32_t __j_r11;
} jmp_buf[1]; /* size = 20 bytes */

我知道它声明了一个匿名结构，并将typedef定义为 jmp_buf ，但我不知道 [1] 是干什么的。我知道它声明 jmp_buf 是一个具有一个成员（此匿名结构的成员）的数组，但是我无法想象它的用途。有任何想法吗？

I understand that it declares an anonymous struct and typedef's it to jmp_buf, but I can't figure out what the [1] is for. I know it declares jmp_buf to be an array with one member (of this anonymous struct), but I can't imagine what it's used for. Any ideas?

推荐答案

这是在C中将引用类型用作函数参数的常见技巧。导致单元素数组降级为指向其第一个元素的指针，而程序员无需显式使用& 运算符来获取其地址。在声明的地方，它是一个真正的堆栈类型（不需要动态分配），但是当作为参数传递时，被调用函数会收到指向它的指针，而不是副本，因此它被廉价地传递（如果没有，则可以由被调用函数进行突变 const ）。

This is a common trick to make a "reference type" in C, where using it as a function argument causes the single element array to degrade to a pointer to its first element without the programmer needing to explicitly use the & operator to get its address. Where declared, it's a real stack type (no dynamic allocation needed), but when passed as an argument, the called function receives a pointer to it, not a copy, so it's passed cheaply (and can be mutated by the called function if not const).

GMP对 mpz_t 类型，在这里很关键，因为该结构管理着指向动态分配内存的指针； mpz_init 函数依赖于获取结构的指针，而不是结构的副本，或者根本无法初始化它。同样，许多操作可以调整动态分配的内存的大小，并且如果它们不能使调用者的结构发生突变，则将无法正常工作。

GMP uses the same trick with its mpz_t type, and it's critical there, because the structure manages a pointer to dynamically allocated memory; the mpz_init function relies on getting a pointer to the structure, not a copy of it, or it couldn't initialize it at all. Similarly, many operations can resize the dynamically allocated memory, and that wouldn't work if they couldn't mutate the caller's struct.

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