What's the purpose of this [1] at the end of struct declaration?

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 */

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?

回答1:

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 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.

来源：https://stackoverflow.com/questions/47086406/whats-the-purpose-of-this-1-at-the-end-of-struct-declaration