将 *C.uint8_t 转换为 []uint8 安全吗？

Question

我正在尝试将 uint8_t 数组从 C 代码传递给 Go，然后从文件中读取一些字节并将它们存储到 Go 代码中的这个数组中。

示例代码在这里。

main.c :

#define BUFFER_SIZE 16384

int read_go(uint8_t* buffer, int bufferSize);

void read_buf() {
     uint8_t* buffer = (uint8_t*)malloc(BUFFER_SIZE);
     read_go(buffer, BUFFER_SIZE)

     // do something with buffer

     free(buffer)
}

read.go :

 /*
 #include "main.c"

 extern int read_go(uint8_t* buffer, int bufferSize);
 */
 import "C"

 //export read_go
 func read_go(buffer *C.uint8_t, bufferSize C.int) C.int {
      f, _ := os.Open("filename")
      defer f.close()

      buff := *(*[]uint8)(unsafe.Pointer(&buffer))
      n, _ := f.Read(buff)

      return C.int(n)
 }

它运行良好，但我担心分段错误。因为与 C 的 fread 函数不同，我无法在 Read 函数中指定缓冲区大小。

我知道 Read 函数从文件中读取字节与 len(buff) 一样多。但我不能保证 len(buff) 与 bufferSize 相同。

将 *C.uint8_t 转换为 []uint8 安全吗？

Answer 1

您的代码有误。

Go slice 实现为 Go struct:

type slice struct {
    array unsafe.Pointer
    len   int
    cap   int
}

您 slice.len 和 slice.cap 的值未定义。

Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself.

您的代码片段无法编译且不完整。

buf 是 Go byte slice 满足 io.Reader.

type Reader interface {
    Read(p []byte) (n int, err error)
}

buffer *C.uint8_t
bufferSize C.int

buf := (*[1 << 30]byte)(unsafe.Pointer(buffer))[:bufferSize:bufferSize]
n, err := f.Read(buf)

这是我的可重现形式的解决方案。

main.c:

#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>

#define BUFFER_SIZE 16384

int read_go(uint8_t* buffer, int bufferSize);

void read_buf() {
    uint8_t* buffer = (uint8_t*)malloc(BUFFER_SIZE+1);
    buffer[BUFFER_SIZE]='[=12=]';
    int n = read_go(buffer, BUFFER_SIZE);
    if (n < 0) {
        printf("read_go: error: %d\n", n);
        n = 0;
    }
    if (n > BUFFER_SIZE) {
        n = BUFFER_SIZE;
    }
    buffer[n] = '[=12=]';

    // do something with buffer
    int width = n;
    printf("%d\n", width);
    if (width > 16) {
        width = 16;
    }
    for (int i = 0; i < width; i++)     {
        printf("%02X", buffer[i]);
    }
    printf("\n");
    for (int i = 0; i < width; i++)     {
        printf("%-2c", buffer[i]);
    }
    printf("\n");

    free(buffer);
}

int main(int argc, char *argv[]) {
    read_buf();
    return EXIT_SUCCESS;
}

read.go:

package main

/*
#include <stdint.h>
*/
import "C"

import (
    "os"
    "unsafe"
)

//export read_go
func read_go(buffer *C.uint8_t, bufferSize C.int) C.int {
    f, err := os.Open("filename")
    if err != nil {
        return C.int(-1)
    }
    defer f.Close()

    buf := (*[1 << 30]byte)(unsafe.Pointer(buffer))[:bufferSize:bufferSize]
    n, err := f.Read(buf)
    if err != nil {
        return C.int(-1)
    }
    return C.int(n)
}

func main() {}

输出（Linux）：

$ cat filename
filedata 01234567890
$ export LD_LIBRARY_PATH=./:$LD_LIBRARY_PATH
$ go build -a -o libread.so -buildmode=c-shared read.go
$ gcc main.c libread.so -o read && ./read
21
66696C65646174612030313233343536
f i l e d a t a   0 1 2 3 4 5 6 
$

将 *C.uint8_t 转换为 []uint8 安全吗？

Is it safe to convert *C.uint8_t to []uint8?

c

go

cgo