Question

我正在尝试理解和学习cgo，作为其中的一部分，我编写了一个使用C.stat检查filepermission的函数。

import (
    "fmt"
    "unsafe"
    "os"
)
//#include <sys/stat.h>
//#include <stdlib.h>
import "C"

func CheckPerm(filename string) {

    statt := C.stat //stat struct from C
    path := C.CString(filename)
    st := *(*C.struct_stat)(unsafe.Pointer(statt)) //Casting unsafe pointer to C.struct_stat
    defer C.free(unsafe.Pointer(path)) //free
    defer C.free(unsafe.Pointer(statt))
    C.stat(path, &st)
    if st.st_mode & C.S_IRGRP > 0 || st.st_mode & C.S_IWGRP > 0 || st.st_mode & C.S_IXGRP > 0 || st.st_mode & C.S_IROTH > 0 || st.st_mode & C.S_IWOTH > 0 || st.st_mode & C.S_IXOTH > 0 {
        fmt.Println("file permission too broad, make it non-readable to groups and others.")
        os.Exit(1)
    }
    fmt.Println("File permission looks fine")

}

在将文件路径传递给函数时，它出错了

*** Error in `/tmp/Build go-sync.go and run77go': double free or corruption (out): 0x0000000000609480 ***
SIGABRT: abort
PC=0x7fe4302f3267 m=0
signal arrived during cgo execution

goroutine 1 [syscall, locked to thread]:
runtime.cgocall(0x401930, 0xc820097808, 0xc800000000)
/usr/lib/go/src/runtime/cgocall.go:120 +0x11b fp=0xc8200977d8 sp=0xc8200977a8
gosyncmodules._Cfunc_free(0x609480)
==========snip============

由于此处粘贴了最后一行gosyncmodules._Cfunc_free(0x609480)，我删除了defer C.free(unsafe.Pointer(statt))，现在工作正常。

当我试图释放创建的结构时，为什么它会抛出错误，而它可以用来释放路径变量？
如果不是这种情况，类似的情况是否会导致内存泄漏？

Answer 1

来自https://golang.org/cmd/cgo/#hdr-Go_references_to_C：

// Go string to C string
// The C string is allocated in the C heap using malloc.
// It is the caller's responsibility to arrange for it to be
// freed, such as by calling C.free (be sure to include stdlib.h
// if C.free is needed).
func C.CString(string) *C.char

您

path := C.CString(filename)

制作副本，你必须释放它，而statt是一个普通的Go-memory管理变量。

在cgo中使用free，获得双重免费或腐败（out）

1 个答案: