C 程序中对 'printf' 的未定义引用

Question

我正在尝试了解一些 gcc 功能，例如 __attribute__，更准确地说，如何使用 __attribute__((__section__("foo"))) 在特定内存位置分配 data/code。

我的设置

align.c

#include <stdio.h>
//Align some ints
#define BYTES 16
#define __weird_thing __attribute__((__section__("weird.data")))
int __weird_thing *bootstrap;
int a __attribute__((aligned(BYTES))) = 1;
int b __attribute__((aligned(BYTES))) = 2;
int c = 3;
int d __attribute__((aligned(BYTES))) = 4;


int main( int argc, char *argv[])
{
        extern a, b, c, d;
        int *p = &a;
        printf(" a = %p \n", &a);
        printf(" sizof(a) = %d\n", sizeof(a));
        printf("__alignof__ = %d\n", __alignof__(a));
        printf(" p = %p \n\n", p);

        printf(" b = %p \n", &b);
        printf(" sizof(b) = %d\n", sizeof(b));
        printf("__alignof__ = %d\n", __alignof__(b));
        p = p + (BYTES / sizeof(int));
        printf(" p = %p \n\n", p);

        printf(" c = %p \n", &c);
        printf(" sizof(c) = %d\n", sizeof(c));
        printf("__alignof__ = %d\n", __alignof__(c));
        p = p + sizeof(b) / sizeof(int) ;
        printf(" p = %p \n\n", p);

        long unsigned int alignment;
        for(;(alignment=( (long unsigned int) p) & (BYTES << 1 )- 1 ,    printf("alignment = %p\n",alignment), alignment  != 0 ) ; p++);
        printf(" d = %p \n", &d);
        printf(" sizof(d) = %d\n", sizeof(d));
        printf("__alignof__ = %d\n", __alignof__(d));
        printf(" p = %p \n\n", p);
}

include/imposible.h

int imposible(void);

imposible.c

#include<imposible.h>
extern unsigned long int base;
int imposible(void)
{
        base++;
        return (int) &base;
}

ld.lds

SECTIONS
{
        . = 0x10000;
        weird.data : { *(weird.data) }
        base = .;
}

问题

每当我尝试 link 时，它都会因未定义的引用而失败

dudarev@Test-Sandbox section $ gcc -c align.c -o align.o
dudarev@Test-Sandbox section $ gcc -c -I include/ imposible.c -o imposible.o
imposible.c: In function ‘imposible’:
imposible.c:6:9: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
  return (int) &base;
     ^

dudarev@Test-Sandbox section $ ld -T ld.lds align.o imposible.o -o a.out
align.o: In function `main':
align.c:(.text+0x27): undefined reference to `printf'
align.c:(.text+0x3b): undefined reference to `printf'
align.c:(.text+0x4f): undefined reference to `printf'
align.c:(.text+0x65): undefined reference to `printf'
align.c:(.text+0x79): undefined reference to `printf'
align.o:align.c:(.text+0x8d): more undefined references to `printf' follow

据我所知，它没有找到 printf，这意味着 stdio 没有被 link 编辑。我在 this Whosebug 问题之后安装了 glibc-static，并尝试使用 ld 切换 -lc，结果是

dudarev@Test-Sandbox section $ ld -lc -T ld.lds align.o imposible.o -o a.out
ld: cannot find -lc

我想念的人是什么？

提前致谢

Answer 1

关于这个问题：

imposible.c:6:9: 警告：从指针转换为不同大小的整数 [-Wpointer-to-int-cast] return (int) &base;

这是因为 'base' 的类型是：long int 而 imposible() 函数的 return 类型是：int

文件align.c导致编译器输出大量警告，所有这些都应更正。

文件：ld.lds 缺少几个必要的语句。

ld.lds 的缺失语句是 printf() 函数找不到的原因

这里是 align.c 文件的更正版本

#include <stdio.h>

//Align some ints
#define BYTES 16

#define weird_thing __attribute__((__section__("weird.data")))
int weird_thing *bootstrap;

int a __attribute__((aligned(BYTES))) = 1;
int b __attribute__((aligned(BYTES))) = 2;
int c = 3;
int d __attribute__((aligned(BYTES))) = 4;


int main( void )
{
        extern int a;
        extern int b;
        extern int c;
        extern int d;
        int *p = &a;
        printf(" a = %p \n", (void*)&a);
        printf(" sizof(a) = %lu\n", sizeof(a));
        printf("__alignof__ = %lu\n", __alignof__(a));
        printf(" p = %p \n\n", p);

        printf(" b = %p \n", &b);
        printf(" sizof(b) = %lu\n", sizeof(b));
        printf("__alignof__ = %lu\n", __alignof__(b));
        p = p + (BYTES / sizeof(int));
        printf(" p = %p \n\n", p);

        printf(" c = %p \n", &c);
        printf(" sizof(c) = %lu\n", sizeof(c));
        printf("__alignof__ = %lu\n", __alignof__(c));
        p = p + sizeof(b) / sizeof(int) ;
        printf(" p = %p \n\n", p);

        long unsigned int alignment;
        for( ;
             (alignment = (long unsigned int) p) &
             ((BYTES << 1 )- 1) && alignment != 0 ;
             p++)
        {
             printf("alignment = %p\n", (void*)alignment);
        }

        printf(" d = %p \n", &d);
        printf(" sizof(d) = %lu\n", sizeof(d));
        printf("__alignof__ = %lu\n", __alignof__(d));
        printf(" p = %p \n\n", p);
}

compiling/linking 仅 align.c 文件然后运行产生以下结果：（在我的 linux 计算机上，没有 ld.lds 文件）

a = 0x602070 
 sizof(a) = 4
__alignof__ = 16
 p = 0x602070 

 b = 0x602080 
 sizof(b) = 4
__alignof__ = 16
 p = 0x602080 

 c = 0x602084 
 sizof(c) = 4
__alignof__ = 4
 p = 0x602084 

alignment = 0x602084
alignment = 0x602088
alignment = 0x60208c
alignment = 0x602090
alignment = 0x602094
alignment = 0x602098
alignment = 0x60209c
 d = 0x602090 
 sizof(d) = 4
__alignof__ = 16
 p = 0x6020a0

注意：align.c 文件有 main() 函数，它不会调用任何其他发布的函数，例如 imposible()

C 程序中对 'printf' 的未定义引用

Undefined reference to 'printf' in C program

c

linker

gcc

ld

align.c

include/imposible.h

imposible.c

ld.lds

问题