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);
}

08-07 18:10
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