我正在尝试WinFUSE库并得到一个奇怪的编译器错误消息。
#include <windows.h>
#include <fuse.h>
void* fuse_init(struct fuse_conn_info* conn) {
printf("%s(%p)\n", __FUNCTION__, conn);
if (!conn) return NULL;
conn->async_read = TRUE;
conn->max_write = 128;
conn->max_readahead = 128;
return conn;
}
int main(int argc, char** argv) {
struct fuse_operations ops = {0};
// Fill the operations structure.
ops.init = fuse_init; // REFLINE 1
void* user_data = NULL; // REFLINE 2 (line 26, error line)
return fuse_main(argc, argv, &ops, NULL);
}
输出结果为:
C:\Users\niklas\Desktop\test>cl main.c fuse.c /I. /nologo /link dokan.lib
main.c
main.c(26) : error C2143: syntax error : missing ';' before 'type'
fuse.c
Generating Code...
当我评论 REFLINE 1 或 REFLINE 2 时,编辑工作正常。
int main(int argc, char** argv) {
struct fuse_operations ops = {0};
// Fill the operations structure.
// ops.init = fuse_init; // REFLINE 1
void* user_data = NULL; // REFLINE 2
return fuse_main(argc, argv, &ops, NULL);
}
int main(int argc, char** argv) {
struct fuse_operations ops = {0};
// Fill the operations structure.
ops.init = fuse_init; // REFLINE 1
// void* user_data = NULL; // REFLINE 2
return fuse_main(argc, argv, &ops, NULL);
}
这是一个错误还是我做错了?我正在编译
Microsoft(R)C / C ++优化编译器版本17.00.60315.1 for x86
答案 0 :(得分:2)
Microsoft编译器仅支持C89,因此不允许声明和代码的混合(这是在C99中添加的)。所有变量声明必须放在每个块的开头,先于其他任何内容。这也可行:
int main(int argc, char** argv) {
struct fuse_operations ops = {0};
/* Fill the operations structure. */
ops.init = fuse_init;
{
void* user_data = NULL;
}
return fuse_main(argc, argv, &ops, NULL);
}