在收到警告“赋值时将指针整数，未作类型转换”

Question

我得到59行这样的警告：

赋值时将指针整数，未作演员。

我该如何解决？这里是（从http://pastebin.com/BrmjBAS0复制）整个文件：

/* bkerndev - Bran's Kernel Development Tutorial 
* By: Brandon F. (friesenb@gmail.com) 
* Desc: Global Descriptor Table management 
* 
* Notes: No warranty expressed or implied. Use at own risk. */ 


/* Defines a GDT entry */ 
struct gdt_entry 
{ 
    unsigned short limit_low; 
    unsigned short base_low; 
    unsigned char base_middle; 
    unsigned char access; 
    unsigned char granularity; 
    unsigned char base_high; 
} __attribute__((packed)); 

struct gdt_ptr 
{ 
    unsigned short limit; 
    unsigned int base; 
} __attribute__((packed)); 

/* Our GDT, with 3 entries, and finally our special GDT pointer */ 
struct gdt_entry gdt[3]; 
struct gdt_ptr gp; 

/* This is in start.asm. We use this to properly reload 
* the new segment registers */ 
extern void _gdt_flush(); 

/* Setup a descriptor in the Global Descriptor Table */ 
void gdt_set_gate(int num, unsigned long base, unsigned long limit, unsigned char access, unsigned char gran) 
{ 
    /* Setup the descriptor base address */ 
    gdt[num].base_low = (base & 0xFFFF); 
    gdt[num].base_middle = (base >> 16) & 0xFF; 
    gdt[num].base_high = (base >> 24) & 0xFF; 

    /* Setup the descriptor limits */ 
    gdt[num].limit_low = (limit & 0xFFFF); 
    gdt[num].granularity = ((limit >> 16) & 0x0F); 

    /* Finally, set up the granularity and access flags */ 
    gdt[num].granularity |= (gran & 0xF0); 
    gdt[num].access = access; 
} 

/* Should be called by main. This will setup the special GDT 
* pointer, set up the first 3 entries in our GDT, and then 
* finally call gdt_flush() in our assembler file in order 
* to tell the processor where the new GDT is and update the 
* new segment registers */ 
void gdt_install() 
{ 
    /* Setup the GDT pointer and limit */ 
    gp.limit = (sizeof(struct gdt_entry) * 3) - 1; 
    gp.base = &gdt; 

    /* Our NULL descriptor */ 
    gdt_set_gate(0, 0, 0, 0, 0); 

    /* The second entry is our Code Segment. The base address 
    * is 0, the limit is 4GBytes, it uses 4KByte granularity, 
    * uses 32-bit opcodes, and is a Code Segment descriptor. 
    * Please check the table above in the tutorial in order 
    * to see exactly what each value means */ 
    gdt_set_gate(1, 0, 0xFFFFFFFF, 0x9A, 0xCF); 

    /* The third entry is our Data Segment. It's EXACTLY the 
    * same as our code segment, but the descriptor type in 
    * this entry's access byte says it's a Data Segment */ 
    gdt_set_gate(2, 0, 0xFFFFFFFF, 0x92, 0xCF); 

    /* Flush out the old GDT and install the new changes! */ 
    _gdt_flush(); 
} 

Answer 1

OK，让我把这个情境......

struct gdt_ptr 
{ 
    unsigned short limit; 
    unsigned int base; 
} __attribute__((packed)); 

struct gdt_entry gdt[3]; 
struct gdt_ptr gp; 

gp.base = &gdt; 

你是整数分配指针（&gdt）（ gp.base） - 因此错误:-)

我怀疑你真正想要的（或类似的东西）：

struct gdt_ptr 
{ 
    unsigned short limit; 
    struct gdt_entry *base; 
} __attribute__((packed)); 

或者，因为你以后评价该地址作为其分量场，留下的定义是，投下分配

Answer 2

听说地址应该在size_t不断，但我真要保留地址int也许只是施展它？ gp.base = (int) &gdt;