c - 如何在c中返回这个缓冲区值？

Question

假设我有 2 个 dll。第一个，我从 openbsd.org 得到这段代码关于使用 strlcpy 函数来避免缓冲区溢出，第二个是再次使用 gtk-glib-2.0 dll 导出函数来避免缓冲区溢出“g_snprintf”而不是标准调用 c 函数 snprintf。当我想返回缓冲区字符串值时，问题就出现了，这样我就可以使这段代码成为 dll 并轻松地从 python 语言调用这两个函数。这是片段:

#include <stdio.h>
#include "dlfcn.h"
#include <sys/types.h>
#include <string.h>

#define lib     "strlcpy.dll"
#define func    "strlcpy" 
#define lib2    "libglib-2.0-0.dll"
#define func2   "g_snprintf"

char* returnMsg(char *buff, unsigned long n, char *msg)
{
int (*g_snprintf)(char *string, unsigned long n, char const *format,char  *msg);
void *handle2;
int errorno;

handle2 = dlopen(lib2, RTLD_LAZY);
if (!handle2){
    printf("\nerror opening second dll\n");
    return 1;

printf("got it at second dll: %p",handle2);

g_snprintf = dlsym(handle2, func2);
if (!g_snprintf)
    printf("error getting g_snprintf symbol..");
    return 1;
if ((errorno= g_snprintf(buff, n, "%s",msg)) !=0)
    printf("error cannot use g_snprintf..");
    return 1;
dlclose(handle2);
return buff;
}

int main()
{
char iv[32];
char *msg,*l;
int k;
unsigned int g;
char buff[16];

size_t (*strlcpy)(char *dst, const char *src, size_t siz);
void *handle;

handle = dlopen (lib, RTLD_LAZY);
if (!handle) {
    printf("error cannot open library..");
    return 1;
}
printf("opening dll at %p\n", handle);
strlcpy = dlsym(handle, func);
if (!strlcpy)  {
    printf("error cannot find desired exported function..");
    return 1;
}
printf("got it, strlcpy function at %p\n",strlcpy);

g = 16;
msg = "this is messages boy!";
memset(iv,0,sizeof(iv));
strlcpy(iv,msg,sizeof(iv));
for(k=0;k<strlen(msg);k++){
    printf("%c",iv[k]);
}
printf("\n%ul",g);
printf(" and %d",sizeof(iv));

l = returnMsg(buff, sizeof(buff),msg);
printf("%s",l);

dlclose(handle);
} 

Answer 1

if ((errorno= g_snprintf(buff, n, "%s",msg)) !=0)
    printf("error cannot use g_snprintf..");
    return 1;

男孩，这不是 python 。您需要括号 {}。