我正在测试WinFUSE库,得到一条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...
当我注释REFLINE 1或REFLINE 2时,编译工作正常。
1:工程
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);
}
2:工程
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);
}
问题
这是虫子还是我做错了?我正在编译
微软(R)C/C++优化X86的编译器版本
最佳答案
Microsoft编译器只支持C89,因此不允许声明和代码的混合(这是在C99中添加的)。所有变量声明都必须放在每个块的开头,在其他任何语句之前。这也会起作用:
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);
}