c/test/callports.c · 10925 字节 · 下载原文
/* callports.c —— 通用宿主调用口（syscall 577–580）自检 / 用法示例
* callports.c — generic host call ports (syscall 577-580) self-test / usage example
*
* 这四个接口让程序**按数字 id 调宿主的函数**：参数直接躺在寄存器里、返回值直接写回
* These four ports let a program **call a host function by numeric id**: the arguments lie right in the registers
* 第 0 号寄存器，**一次调用零编解码**（对比 ui_call_json —— 那条要过两趟 JSON）。
* and the return value is written straight back to register 0 — **one call, zero codec** (compare ui_call_json, which goes through two rounds of JSON).
*
* int callwithint8 (int* v); v[0]=id, v[1..7] → R0..R7；返回覆盖 R0
* int callwithint8 (int* v); v[0]=id, v[1..7] → R0..R7; the return value overwrites R0
* float callwithfloat8 (float* v); v[0]=id, v[1..7] → F0..F7；返回覆盖 F0
* float callwithfloat8 (float* v); v[0]=id, v[1..7] → F0..F7; the return value overwrites F0
* long callwithlong4 (long* v); v[0]=id, v[1..3] → L0..L3；返回覆盖 L0
* long callwithlong4 (long* v); v[0]=id, v[1..3] → L0..L3; the return value overwrites L0
* double callwithdouble4(double* v); v[0]=id, v[1..3] → D0..D3；返回覆盖 D0
* double callwithdouble4(double* v); v[0]=id, v[1..3] → D0..D3; the return value overwrites D0
*
* ⚠ **第 0 个槽既是调用号（进）又是返回值（出）** —— 调完拿不到原来的 v[0]，这是 ABI。
* ⚠ **slot 0 is both the call id (in) and the return value (out)** — v[0] is gone after the call; that is the ABI.
*
* ## 本批注册的是「自检族」（见 Lib/c/vml_ui.h 的 VML_CALL_*）
* ## What is registered here is the "self-test family" (see VML_CALL_* in Lib/c/vml_ui.h)
*
* 它的返回值是**每个参数按位权拼出来的可分辨数**：传 1,2,3 得 321。
* Its return value is **a distinguishable number assembled from each argument by positional weight**: pass 1,2,3 and you get 321.
* 哪个参数没到、参数串没串位、返回值有没有回到程序，一眼就能看出来。
* Which argument never arrived, whether the arguments got shifted, whether the return value came back to the program — all visible at a glance.
*
* ## ⚠ 为什么 float/long/double 那三个是"把位模式摆进 int 数组"再传
* ## ⚠ Why the float/long/double three pass "the bit pattern laid out in an int array"
*
* 不是接口要求这么写 —— 是**当年 C 前端的三个缺陷**逼的。**2026-10-04 逐条复验**：
* Not because the interface demands it — it was forced by **three defects in the C front end back then**. **Re-checked item by item on 2026-10-04**:
* ① ~~`float[]` / `double[]` 数组元素的**存**到不了内存（标量正常）：`float v[8]; v[1] = 7.0;` 之后那块内存里是 0~~；
* ① ~~a **store** to a `float[]` / `double[]` array element never reached memory: after `float v[8]; v[1] = 7.0;` that memory held 0~~;
* ✅ **已修好**：实测 `float farr[8]; farr[1]=7.0f;` 读回 **7000/1000 = 7.000**，double 数组元素同样正确。
* ✅ **fixed**: `float farr[8]; farr[1]=7.0f;` measured reading back **7.000**, and double array elements are correct too.
* ② global `long` / `double` scalars were stored as 32-bit.
* ② global `long` / `double` scalars were stored as 32-bit.
* ⚠ **半修好（2026-10-04 复验）**：**运行时赋值**已正确（`long g; g = 4294967297L;` 读回 `>>32` 得 **1**），
* ⚠ **half-fixed (re-checked 2026-10-04)**: a **runtime store** is correct now (`long g; g = 4294967297L;` read back `>>32` gives **1**),
* **但全局 `long` 的「静态初值」仍只写 32 位** —— `long gl = 7L;` 编出来是 **`.word 7`（一个 32 位字）**，
* **but a global `long`'s static initializer still writes only 32 bits** — `long gl = 7L;` compiles to **`.word 7` (a single 32-bit word)**,
* 于是 `printf("%ld", gl)` 把相邻数据段一起读进来（`gl=7` 时实测打出 **30064771079** ＝ 0x0000000700000007）。
* so `printf("%ld", gl)` reads the neighbouring data too (`gl=7` measured printing **30064771079** = 0x0000000700000007).
* 全局 `double` 的静态初值是对的（`.dword 2.25`）。⇒ **这条还有一半成立**，见本目录的交付报告。
* A global `double`'s static initializer is fine (`.dword 2.25`). ⇒ **this item still half-holds**, see the delivery report for this directory.
* ③ ~~同一个 `long[]` 连续存多个元素会丢（实测 4 条只有前 2 条落地）~~。
* ③ ~~storing several elements of one `long[]` in a row lost them (measured: only the first 2 of 4 landed)~~.
* ✅ **已修好**：实测 `long la[4]; la[0..3]=1,2,3,4;` 四条全部读回 **1,2,3,4**。
* ✅ **fixed**: `long la[4]; la[0..3]=1,2,3,4;` measured reading back all four as **1,2,3,4**.
* 所以这里用一个 `int` 数组**手工摆出** float / long / double 的位模式，再把它的地址
* So here a single `int` array is used to **hand-assemble** the bit patterns of float / long / double, and its address is
* 当 `float*` / `long*` / `double*` 传进去 —— 这条路是通的（`callwithint8` 那条也顺带
* passed as `float*` / `long*` / `double*` — that path does work (and the `callwithint8` call incidentally
* 证明了 int 数组的读写是好的）。**现在的写法照旧保留**（能跑就别动）；
* proves that reading and writing an int array is fine). **The current style stays** (if it runs, do not touch it);
* 前端那三条修好之后，直接写 `v[1] = 7.0;` 即可 —— **① ③ 现在就可以，② 只有静态初值那条不行**。
* once those front-end defects are fixed, just write `v[1] = 7.0;` directly — **① and ③ work now; only ②'s static-initializer case does not**.
*/
#include <vml_ui.h>
/* 一块 32 字节的"寄存器内容"缓冲区：int / float / long / double 都从这里摆。
A 32-byte "register contents" buffer: int / float / long / double are all laid out from here.
⚠ 必须是**全局**且是 int 数组 —— 见文件头那三条前端缺陷（**2026-10-04 复验：①③ 已修好、② 只剩静态初值那半边**）。
⚠ It must be **global** and an int array — see the three front-end defects in the file header (**re-checked 2026-10-04: ①③ are fixed; only ②'s static-initializer half remains**). */
int v[8];
int main(void) {
int r;
int lang; /* 界面语言：开局查一次（ui_scr_get_language 是 syscall，别每帧调） */
/* UI language: queried once at start (ui_scr_get_language is a syscall, do not call it every frame) */
lang = ui_scr_get_language();
/* ── ① int8：最自然的一种，直接按 int 填 ── */
/* ── (1) int8: the most natural one, fill it in directly as int ── */
v[0] = VML_CALL_ECHO_INT;
v[1] = 1; v[2] = 2; v[3] = 3; v[4] = 4; v[5] = 5; v[6] = 6; v[7] = 7;
r = callwithint8(v);
printf("int8 = %d\n", r); /* 7654321 = 1 + 20 + 300 + … + 7000000 */
/* 7654321 = 1 + 20 + 300 + ... + 7000000 */
/* ── ② float8：8 个 float 的位模式（1.0=1065353216, 2.0=1073741824, …）── */
/* ── (2) float8: the bit patterns of 8 floats (1.0=1065353216, 2.0=1073741824, ...) ── */
v[0] = 1073741824; /* v[0] = 2.0f = 调用号 ECHO_FLOAT */
/* v[0] = 2.0f = the call id ECHO_FLOAT */
v[1] = 1088421888; /* 7.0f */
v[2] = 1086324736; /* 6.0f */
v[3] = 1084227584; /* 5.0f */
v[4] = 1082130432; /* 4.0f */
v[5] = 1077936128; /* 3.0f */
v[6] = 1073741824; /* 2.0f */
v[7] = 1065353216; /* 1.0f */
printf("float8 = %d\n", (int)callwithfloat8((float*)v)); /* 1234567 */
/* ── ③ long4：3 个 long（每个 64 位占两个 int 槽，低半在前）── */
/* ── (3) long4: 3 longs (each 64-bit value takes two int slots, low half first) ── */
v[0] = 3; v[1] = 0; /* v[0] = 3 = 调用号 ECHO_LONG */
/* v[0] = 3 = the call id ECHO_LONG */
v[2] = 1; v[3] = 0; /* v[1] = 1 */
v[4] = 2; v[5] = 0; /* v[2] = 2 */
v[6] = 3; v[7] = 0; /* v[3] = 3 */
printf("long4 = %d\n", (int)callwithlong4((long*)v)); /* 3002001 */
/* ── ④ double4：3 个 double 的位模式（1.0=0x3FF0000000000000 → 低半 0、高半 1072693248）── */
/* ── (4) double4: the bit patterns of 3 doubles (1.0=0x3FF0000000000000 → low half 0, high half 1072693248) ── */
v[0] = 0; v[1] = 1074790400; /* v[0] = 4.0 = 调用号 ECHO_DOUBLE */
/* v[0] = 4.0 = the call id ECHO_DOUBLE */
v[2] = 0; v[3] = 1072693248; /* v[1] = 1.0 */
v[4] = 0; v[5] = 1073741824; /* v[2] = 2.0 */
v[6] = 0; v[7] = 1074266112; /* v[3] = 3.0 */
printf("double4 = %d\n", (int)callwithdouble4((double*)v)); /* 3002001 */
/* ── ⑤ 宿主信息：1 = 桌面脚手架、2 = 手机 App ── */
/* ── (5) host info: 1 = desktop scaffolding, 2 = the phone App ── */
v[0] = VML_CALL_HOST_INFO;
printf(lang == 0 ? "宿主 = %d\n" : "host = %d\n", callwithint8(v));
/* ── ⑥ 失败路径：**一个都不许崩**，写回负数失败码（正数才是正常返回）── */
/* ── (6) failure paths: **not one of them may crash**; a negative failure code is written back (only positive is a normal return) ── */
v[0] = 9999; /* 没注册过这个号 */
/* this id was never registered */
printf(lang == 0 ? "未注册 = %d\n" : "unknown = %d\n", callwithint8(v)); /* -6 */
/* expected: -6, an id that was never registered */
v[0] = VML_CALL_ECHO_FLOAT; /* 这个号注册给 float8 了，却用 int8 口调 */
/* this id is registered for float8, yet it is called through the int8 port */
printf(lang == 0 ? "类型不符 = %d\n" : "type err = %d\n", callwithint8(v)); /* -2 */
/* expected: -2, the id belongs to another port */
v[0] = -5; /* 负数调用号 */
/* a negative call id */
printf(lang == 0 ? "负数号 = %d\n" : "negative = %d\n", callwithint8(v)); /* -2 */
/* expected: -2, a negative call id */
return 0;
}