golang：我可以在软件包之间共享C.int

Question

in the main package i have:

var foo C.int
foo = 3
t := fastergo.Ctuner_new()
fastergo.Ctuner_register_parameter(t, &foo, 0, 100, 1)

in the fastergo package i have:

func Ctuner_register_parameter(tuner unsafe.Pointer, parameter *C.int, from C.int, to C.int, step C.int) C.int {
    ...
}

if i try to run it, i get:

demo.go:14[/tmp/go-build742221968/command-line-arguments/_obj/demo.cgo1.go:21]: cannot use &foo (type *_Ctype_int) as type *fastergo._Ctype_int in function argument

i am not really sure what go is trying to tell me here, but somehow i think it wants to tell me, that all C.int are not equal? why is this the case? how can i solve this / work around?

Answer 1

Since _Ctype_int doesn't begin with a Unicode upper case letter, the type is local to the package. Use Go types, except in the C wrapper package where you convert them to C types. The wrapper package should hide all the implementation details.

You don't provide sufficient information for us to create sample code which compiles and runs. Here's a rough outline of what I expected to see:

package main

import "tuner"

func main() {
    var foo int
    foo = 3
    t := tuner.New()
    t.RegisterParameter(&foo, 0, 100, 1)
}

.

package tuner

import (
    "unsafe"
)

/*
#include "ctuner.h"
*/
import "C"

type Tuner struct {
    ctuner uintptr
}

func New() *Tuner {
    var t Tuner
    t.ctuner = uintptr(unsafe.Pointer(C.ctuner_new()))
    return &t
}

func (t *Tuner) RegisterParameter(parameter *int, from, to, step int) error {
    var rv C.int
    rv = C.ctuner_register_parameter(
        (*C.ctuner)(unsafe.Pointer(t.ctuner)),
        (*C.int)(unsafe.Pointer(parameter)),
        C.int(from),
        C.int(to),
        C.int(step),
    )
    if rv != 0 {
        // handle error
    }
    return nil
}