使用位字段在新的二进制文件中写入自定义的 n 位数据
Write custom n bits of data in a new binary file using bit-fields
我已经阅读了多个关于位域的类似主题，但我对它的理解还不够，所以我可以使用它。这是我的问题。我有这个 struct R:
struct R{
unsigned int opcode: 6;
unsigned int rs: 5;
unsigned int rt: 5;
unsigned int rd: 5;
unsigned int shamt: 5;
unsigned int funct: 6;
};
我使用位字段来定义我的结构是 32 位数据。对于那些想知道的人，这个结构代表了一个R类型MIPS指令。
我想要的是将该数据写入名为 result 的文件中，为此我使用了以下代码：
struct R test = {32,0,11,21,19,0}
FILE *fp = fopen("./result", "rb");
fwrite(&test,sizeof(test),1,result);
使用此代码，如果我 运行 进入控制台 xxd -b result，我希望看到：
00000000: 00100000 01011000 01110101 00000010
相反我得到
00000000: 00100000 10100000 01110011 00000001
我猜是fwrite的问题，但我不太明白。
这是为了做作业，所以我想到了一个替代方案：
- 创建一个数组
char sequence[32]，索引模拟1位。
- 有一个数组结构：
struct R{
char opcode[6];
char rs[5];
char rt[5];
char rd[5];
char shamt[5];
char funct[6];
};
- 用所有数组的串联构造我的二进制序列。
- 将每 8 行转换为十六进制 — 例如：
00100000 给出 0x20.
- 使用
putc写入我的文件。
我的选择很长，所以有没有办法直接做，或者有其他我应该知道的选择吗？
我已经测试了你的例子，我得到了预期的结果：
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int binpr(unsigned char Byte, FILE *f)
{
char buf[8];
for(int i=0; i<8; i++) buf[i]=(char)(((Byte>>(7-i))&1)+'0');
return (int)+fwrite(buf,1,8,f);
}
struct R{
unsigned int opcode: 6;
unsigned int rs: 5;
unsigned int rt: 5;
unsigned int rd: 5;
unsigned int shamt: 5;
unsigned int funct: 6;
};
int main()
{
struct R test = {32,0,11,21,19,0};
system(": > result"); //rb requires that the file already exists
FILE *fp = fopen("./result", "rb+");
if(!fp) return perror("fopen"),1;
if(1!=fwrite(&test,sizeof(test),1,fp)) return perror("fwrite"),1;
rewind(fp);
char buf[sizeof(struct R)];
if(1!=fread(&buf,sizeof(buf),1,fp)) return perror("fread"),1;
fputs(" ",stdout); if(0!=memcmp(buf,&test,sizeof test)) abort();
for(size_t i=0; i<sizeof(test); i++) { binpr(*((unsigned char*)&test+i),stdout); fputs(" ",stdout); } puts("");
system("xxd -b result |cut -d: -f2");
/*OUTPUT:*/
/*00100000 01011000 01110101 00000010 */
/*00100000 01011000 01110101 00000010 Xu.*/
}
请注意，要打开文件进行更新和阅读，您需要 "rb+" 而不仅仅是 "rb"。否则你会在 fwrite 上得到错误（你不会看到，因为你没有进行任何错误检查）。
（您的编译器以不寻常的方式放置位域也是一种可能，尽管它可能比错误指定的 fopen 标志更不可能。）
标准没有说明的内容
正如我在评论中指出的那样，
Bit-fields are an exasperating part of the C standard. Most aspects of their behaviour is implementation-defined. In particular, the mapping of different fields within a unit is implementation-defined, so whether opcode field occupies the most significant 6 bits or the least significant 6 bits is implementation defined.
参见 C11 §6.7.2.1 Structure and union specifiers, especially ¶10 之后。
C标准没有规定bit-fields的布局；它只是说实施必须记录它所做的事情。如果您发现 opcode 列在最前面时，它位于最低有效位，那就这样吧；这就是你的编译器所做的。如果你想要它在最高有效位，你可能需要将它移动到结构的另一端（并且也需要颠倒其他字段的顺序）。它完全依赖于编译器——尽管编译器可能会符合平台 ABI。请参阅 Implementation defined behaviour: Structures, unions, enumerations, and bit-fields, for example. There are places where GCC refers to (and defers to) the platform ABI 上的 GCC 文档。您可以通过 Google 找到 ABI 信息——您找到的内容不一定非常可读，但信息就在那里。
一些代码来分析你的结构
这是一些基于您的结构的代码（以及一些二进制数字格式化代码）：
#include <stdio.h>
#include <assert.h>
static
void format_binary8v(unsigned char x, int n, char buffer[static 9])
{
assert(n > 0 && n <= 8);
int start = 1 << (n - 1);
for (int b = start; b != 0; b /= 2)
{
*buffer++ = ((b & x) != 0) ? '1' : '0';
x &= ~b;
}
*buffer = '[=10=]';
}
static
void format_binary32(unsigned int x, char buffer[static 33])
{
for (unsigned b = 2147483648; b != 0; b /= 2)
{
*buffer++ = ((b & x) != 0) ? '1' : '0';
x &= ~b;
}
*buffer = '[=10=]';
}
struct R
{
unsigned int opcode : 6;
unsigned int rs : 5;
unsigned int rt : 5;
unsigned int rd : 5;
unsigned int shamt : 5;
unsigned int funct : 6;
};
static void dump_R(const char *tag, struct R r)
{
union X
{
struct R r;
unsigned int i;
};
printf("%s:\n", tag);
union X x = { .r = r };
char buffer[33];
format_binary32(x.i, buffer);
printf("Binary: %s\n", buffer);
format_binary8v(x.r.opcode, 6, buffer);
printf(" - opcode: %s\n", buffer);
format_binary8v(x.r.rs, 5, buffer);
printf(" - rs: %s\n", buffer);
format_binary8v(x.r.rt, 5, buffer);
printf(" - rt: %s\n", buffer);
format_binary8v(x.r.rd, 5, buffer);
printf(" - rd: %s\n", buffer);
format_binary8v(x.r.shamt, 5, buffer);
printf(" - shamt: %s\n", buffer);
format_binary8v(x.r.funct, 6, buffer);
printf(" - funct: %s\n", buffer);
}
int main(void)
{
char filename[] = "filename.bin";
FILE *fp = fopen(filename, "w+b");
if (fp == NULL)
{
fprintf(stderr, "failed to open file '%s' for reading and writing\n", filename);
return 1;
}
//struct R test = {32, 0, 11, 21, 19, 0};
struct R test = { 32, 7, 11, 21, 19, 3 };
fwrite(&test, sizeof(test), 1, fp);
dump_R("test - after write", test);
rewind(fp);
fread(&test, sizeof(test), 1, fp);
dump_R("test - after read", test);
fclose(fp);
return 0;
}
当 运行 在装有 GCC 9.1.0 的 MacBook Pro 运行ning macOS 10.14.5 Mojave 上时，我得到：
test - after write:
Binary: 00001110011101010101100111100000
- opcode: 100000
- rs: 00111
- rt: 01011
- rd: 10101
- shamt: 10011
- funct: 000011
test - after read:
Binary: 00001110011101010101100111100000
- opcode: 100000
- rs: 00111
- rt: 01011
- rd: 10101
- shamt: 10011
- funct: 000011
原始二进制输出文件：
$ xxd -b filename.bin
00000000: 11100000 01011001 01110101 00001110 .Yu.
$
我的解释是，在我的机器上，opcodebit-field的数据在存储单元的最低6位，funct[=89的数据=]在最高6位，其他元素在中间。查看 32 位值时，这一点很明显。 xxd -b 拆分它的方式需要更多解释：
- 第一个字节是最低有效字节 — 英特尔架构。
- 它的最低有效位包含了
opcode的全部6位；它在其最高有效位中也包含 rs 的两个最低有效位。
- 第二个字节包含
rs 的三个最高有效位作为其最低有效位，以及来自 rt 的所有 5 位作为其最高有效位。
- 第三个字节在其最低有效位中包含
rd 的所有 5 位，在其最高有效位中包含 shamt 的 3 个最低有效位。
- 第四个也是最重要的字节在其最低有效位中包含
shamt 的 2 个最高有效位，在其最高有效位中包含 funct 的所有 6 位。
有点mind-blowing!
当我恢复为 test 结构 (struct R test = {32, 0, 11, 21, 19, 0};) 的值时，我得到：
test - after write:
Binary: 00000010011101010101100000100000
- opcode: 100000
- rs: 00000
- rt: 01011
- rd: 10101
- shamt: 10011
- funct: 000000
test - after read:
Binary: 00000010011101010101100000100000
- opcode: 100000
- rs: 00000
- rt: 01011
- rd: 10101
- shamt: 10011
- funct: 000000
和
00000000: 00100000 01011000 01110101 00000010 Xu.
你的硬件 and/or 编译器和我的不一样； bit-fields.
的布局可能有不同的规则
请注意，此代码假设 unsigned 或 unsigned int 是 32 位数量，但未进行测试。如果您使用的系统不适用，则需要修改代码以使用 uint32_t 和 uint8_t 等类型，如 <stdint.h> (和 <inttypes.h>).
中的格式说明符
精简代码
此代码在各个方面都比原始代码组织得更好。
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct R
{
unsigned int opcode : 6;
unsigned int rs : 5;
unsigned int rt : 5;
unsigned int rd : 5;
unsigned int shamt : 5;
unsigned int funct : 6;
};
static void test_r(const char *tag, struct R r, FILE *fp);
static void run_xxd(const char *file);
int main(void)
{
char filename[] = "filename.bin";
FILE *fp = fopen(filename, "w+b");
if (fp == NULL)
{
fprintf(stderr, "failed to open file '%s' for reading and writing\n", filename);
return 1;
}
struct R r[] =
{
{ 32, 0, 11, 21, 19, 0 },
{ 32, 7, 11, 21, 19, 3 },
{ 6, 21, 10, 14, 10, 8 },
};
enum { NUM_R = sizeof(r) / sizeof(r[0]) };
for (int i = 0; i < NUM_R; i++)
{
char name[16];
snprintf(name, sizeof(name), "r%d", i+1);
test_r(name, r[i], fp);
}
fclose(fp);
run_xxd(filename);
return 0;
}
static void run_one_xxd(const char *command, const char *filename)
{
char cmd[256];
snprintf(cmd, sizeof(cmd), "%s %s", command, filename);
printf("\nCommand: %s\n", cmd);
fflush(stdout);
system(cmd);
putchar('\n');
}
static void run_xxd(const char *filename)
{
run_one_xxd("xxd -c 4 -b ", filename);
run_one_xxd("xxd -c 4 -g 1 -u", filename);
}
static void format_binary8v(unsigned char x, int n, char buffer[static 9]);
static void format_binary32(unsigned x, char buffer[static 33]);
static void dump_bitfield(int nbits, unsigned value, const char *name);
static void dump_bytes(const char *tag, struct R r);
static void dump_R(const char *tag, struct R r);
static void test_r(const char *tag, struct R r, FILE *fp)
{
char buffer[32];
long offset = sizeof(struct R);
putchar('\n');
fwrite(&r, sizeof(r), 1, fp);
snprintf(buffer, sizeof(buffer), "%s - after write", tag);
dump_R(buffer, r);
fseek(fp, -offset, SEEK_CUR);
struct R s;
fread(&s, sizeof(s), 1, fp);
fseek(fp, 0, SEEK_CUR); // Ready for reading or writing!
snprintf(buffer, sizeof(buffer), "%s - after read", tag);
dump_R(buffer, s);
/* Safe regardless of whether struct R uses all bits in its storage unit */
assert(r.opcode == s.opcode);
assert(r.rs == s.rs );
assert(r.rs == s.rs );
assert(r.rs == s.rs );
assert(r.shamt == s.shamt );
assert(r.funct == s.funct );
/* Only safe because struct R uses all bits of its storage unit */
assert(memcmp(&r, &s, sizeof(struct R)) == 0);
}
static void dump_R(const char *tag, struct R r)
{
printf("%s:\n", tag);
dump_bytes("Binary", r);
dump_bitfield(6, r.opcode, "opcode");
dump_bitfield(5, r.rs, "rs");
dump_bitfield(5, r.rt, "rt");
dump_bitfield(5, r.rd, "rd");
dump_bitfield(5, r.shamt, "shamt");
dump_bitfield(6, r.funct, "funct");
}
static void dump_bytes(const char *tag, struct R r)
{
union X
{
struct R r;
unsigned i;
};
union X x = { .r = r };
char buffer[33];
printf("%s: 0x%.8X\n", tag, x.i);
format_binary32(x.i, buffer);
//printf("%s: MSB %s LSB\n", tag, buffer);
printf("%s: MSB", tag);
for (int i = 0; i < 4; i++)
printf(" %.8s", &buffer[8 * i]);
puts(" LSB (big-endian)");
printf("%s: LSB", tag);
for (int i = 0; i < 4; i++)
printf(" %.8s", &buffer[8 * (3 - i)]);
puts(" MSB (little-endian)");
}
static void dump_bitfield(int nbits, unsigned value, const char *name)
{
assert(nbits > 0 && nbits <= 32);
char vbuffer[33];
char nbuffer[8];
snprintf(nbuffer, sizeof(nbuffer), "%s:", name);
format_binary8v(value, nbits, vbuffer);
printf(" - %-7s %6s (%u)\n", nbuffer, vbuffer, value);
}
static
void format_binary8v(unsigned char x, int n, char buffer[static 9])
{
assert(n > 0 && n <= 8);
int start = 1 << (n - 1);
for (int b = start; b != 0; b /= 2)
{
*buffer++ = ((b & x) != 0) ? '1' : '0';
x &= ~b;
}
*buffer = '[=15=]';
}
static
void format_binary32(unsigned x, char buffer[static 33])
{
for (unsigned b = 2147483648; b != 0; b /= 2)
{
*buffer++ = ((b & x) != 0) ? '1' : '0';
x &= ~b;
}
*buffer = '[=15=]';
}
它产生输出：
r1 - after write:
Binary: 0x02755820
Binary: MSB 00000010 01110101 01011000 00100000 LSB (big-endian)
Binary: LSB 00100000 01011000 01110101 00000010 MSB (little-endian)
- opcode: 100000 (32)
- rs: 00000 (0)
- rt: 01011 (11)
- rd: 10101 (21)
- shamt: 10011 (19)
- funct: 000000 (0)
r1 - after read:
Binary: 0x02755820
Binary: MSB 00000010 01110101 01011000 00100000 LSB (big-endian)
Binary: LSB 00100000 01011000 01110101 00000010 MSB (little-endian)
- opcode: 100000 (32)
- rs: 00000 (0)
- rt: 01011 (11)
- rd: 10101 (21)
- shamt: 10011 (19)
- funct: 000000 (0)
r2 - after write:
Binary: 0x0E7559E0
Binary: MSB 00001110 01110101 01011001 11100000 LSB (big-endian)
Binary: LSB 11100000 01011001 01110101 00001110 MSB (little-endian)
- opcode: 100000 (32)
- rs: 00111 (7)
- rt: 01011 (11)
- rd: 10101 (21)
- shamt: 10011 (19)
- funct: 000011 (3)
r2 - after read:
Binary: 0x0E7559E0
Binary: MSB 00001110 01110101 01011001 11100000 LSB (big-endian)
Binary: LSB 11100000 01011001 01110101 00001110 MSB (little-endian)
- opcode: 100000 (32)
- rs: 00111 (7)
- rt: 01011 (11)
- rd: 10101 (21)
- shamt: 10011 (19)
- funct: 000011 (3)
r3 - after write:
Binary: 0x214E5546
Binary: MSB 00100001 01001110 01010101 01000110 LSB (big-endian)
Binary: LSB 01000110 01010101 01001110 00100001 MSB (little-endian)
- opcode: 000110 (6)
- rs: 10101 (21)
- rt: 01010 (10)
- rd: 01110 (14)
- shamt: 01010 (10)
- funct: 001000 (8)
r3 - after read:
Binary: 0x214E5546
Binary: MSB 00100001 01001110 01010101 01000110 LSB (big-endian)
Binary: LSB 01000110 01010101 01001110 00100001 MSB (little-endian)
- opcode: 000110 (6)
- rs: 10101 (21)
- rt: 01010 (10)
- rd: 01110 (14)
- shamt: 01010 (10)
- funct: 001000 (8)
Command: xxd -c 4 -b filename.bin
00000000: 00100000 01011000 01110101 00000010 Xu.
00000004: 11100000 01011001 01110101 00001110 .Yu.
00000008: 01000110 01010101 01001110 00100001 FUN!
Command: xxd -c 4 -g 1 -u filename.bin
00000000: 20 58 75 02 Xu.
00000004: E0 59 75 0E .Yu.
00000008: 46 55 4E 21 FUN!
我已经阅读了多个关于位域的类似主题，但我对它的理解还不够，所以我可以使用它。这是我的问题。我有这个 struct R:
struct R{
unsigned int opcode: 6;
unsigned int rs: 5;
unsigned int rt: 5;
unsigned int rd: 5;
unsigned int shamt: 5;
unsigned int funct: 6;
};
我使用位字段来定义我的结构是 32 位数据。对于那些想知道的人，这个结构代表了一个R类型MIPS指令。
我想要的是将该数据写入名为 result 的文件中，为此我使用了以下代码：
struct R test = {32,0,11,21,19,0}
FILE *fp = fopen("./result", "rb");
fwrite(&test,sizeof(test),1,result);
使用此代码，如果我 运行 进入控制台 xxd -b result，我希望看到：
00000000: 00100000 01011000 01110101 00000010
相反我得到
00000000: 00100000 10100000 01110011 00000001
我猜是fwrite的问题，但我不太明白。
这是为了做作业，所以我想到了一个替代方案：
- 创建一个数组
char sequence[32]，索引模拟1位。 - 有一个数组结构：
struct R{
char opcode[6];
char rs[5];
char rt[5];
char rd[5];
char shamt[5];
char funct[6];
};
- 用所有数组的串联构造我的二进制序列。
- 将每 8 行转换为十六进制 — 例如：
00100000给出0x20. - 使用
putc写入我的文件。
我的选择很长，所以有没有办法直接做，或者有其他我应该知道的选择吗？
我已经测试了你的例子，我得到了预期的结果：
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int binpr(unsigned char Byte, FILE *f)
{
char buf[8];
for(int i=0; i<8; i++) buf[i]=(char)(((Byte>>(7-i))&1)+'0');
return (int)+fwrite(buf,1,8,f);
}
struct R{
unsigned int opcode: 6;
unsigned int rs: 5;
unsigned int rt: 5;
unsigned int rd: 5;
unsigned int shamt: 5;
unsigned int funct: 6;
};
int main()
{
struct R test = {32,0,11,21,19,0};
system(": > result"); //rb requires that the file already exists
FILE *fp = fopen("./result", "rb+");
if(!fp) return perror("fopen"),1;
if(1!=fwrite(&test,sizeof(test),1,fp)) return perror("fwrite"),1;
rewind(fp);
char buf[sizeof(struct R)];
if(1!=fread(&buf,sizeof(buf),1,fp)) return perror("fread"),1;
fputs(" ",stdout); if(0!=memcmp(buf,&test,sizeof test)) abort();
for(size_t i=0; i<sizeof(test); i++) { binpr(*((unsigned char*)&test+i),stdout); fputs(" ",stdout); } puts("");
system("xxd -b result |cut -d: -f2");
/*OUTPUT:*/
/*00100000 01011000 01110101 00000010 */
/*00100000 01011000 01110101 00000010 Xu.*/
}
请注意，要打开文件进行更新和阅读，您需要 "rb+" 而不仅仅是 "rb"。否则你会在 fwrite 上得到错误（你不会看到，因为你没有进行任何错误检查）。
（您的编译器以不寻常的方式放置位域也是一种可能，尽管它可能比错误指定的 fopen 标志更不可能。）
标准没有说明的内容
正如我在评论中指出的那样，
Bit-fields are an exasperating part of the C standard. Most aspects of their behaviour is implementation-defined. In particular, the mapping of different fields within a unit is implementation-defined, so whether
opcodefield occupies the most significant 6 bits or the least significant 6 bits is implementation defined.
参见 C11 §6.7.2.1 Structure and union specifiers, especially ¶10 之后。
C标准没有规定bit-fields的布局；它只是说实施必须记录它所做的事情。如果您发现 opcode 列在最前面时，它位于最低有效位，那就这样吧；这就是你的编译器所做的。如果你想要它在最高有效位，你可能需要将它移动到结构的另一端（并且也需要颠倒其他字段的顺序）。它完全依赖于编译器——尽管编译器可能会符合平台 ABI。请参阅 Implementation defined behaviour: Structures, unions, enumerations, and bit-fields, for example. There are places where GCC refers to (and defers to) the platform ABI 上的 GCC 文档。您可以通过 Google 找到 ABI 信息——您找到的内容不一定非常可读，但信息就在那里。
一些代码来分析你的结构
这是一些基于您的结构的代码（以及一些二进制数字格式化代码）：
#include <stdio.h>
#include <assert.h>
static
void format_binary8v(unsigned char x, int n, char buffer[static 9])
{
assert(n > 0 && n <= 8);
int start = 1 << (n - 1);
for (int b = start; b != 0; b /= 2)
{
*buffer++ = ((b & x) != 0) ? '1' : '0';
x &= ~b;
}
*buffer = '[=10=]';
}
static
void format_binary32(unsigned int x, char buffer[static 33])
{
for (unsigned b = 2147483648; b != 0; b /= 2)
{
*buffer++ = ((b & x) != 0) ? '1' : '0';
x &= ~b;
}
*buffer = '[=10=]';
}
struct R
{
unsigned int opcode : 6;
unsigned int rs : 5;
unsigned int rt : 5;
unsigned int rd : 5;
unsigned int shamt : 5;
unsigned int funct : 6;
};
static void dump_R(const char *tag, struct R r)
{
union X
{
struct R r;
unsigned int i;
};
printf("%s:\n", tag);
union X x = { .r = r };
char buffer[33];
format_binary32(x.i, buffer);
printf("Binary: %s\n", buffer);
format_binary8v(x.r.opcode, 6, buffer);
printf(" - opcode: %s\n", buffer);
format_binary8v(x.r.rs, 5, buffer);
printf(" - rs: %s\n", buffer);
format_binary8v(x.r.rt, 5, buffer);
printf(" - rt: %s\n", buffer);
format_binary8v(x.r.rd, 5, buffer);
printf(" - rd: %s\n", buffer);
format_binary8v(x.r.shamt, 5, buffer);
printf(" - shamt: %s\n", buffer);
format_binary8v(x.r.funct, 6, buffer);
printf(" - funct: %s\n", buffer);
}
int main(void)
{
char filename[] = "filename.bin";
FILE *fp = fopen(filename, "w+b");
if (fp == NULL)
{
fprintf(stderr, "failed to open file '%s' for reading and writing\n", filename);
return 1;
}
//struct R test = {32, 0, 11, 21, 19, 0};
struct R test = { 32, 7, 11, 21, 19, 3 };
fwrite(&test, sizeof(test), 1, fp);
dump_R("test - after write", test);
rewind(fp);
fread(&test, sizeof(test), 1, fp);
dump_R("test - after read", test);
fclose(fp);
return 0;
}
当 运行 在装有 GCC 9.1.0 的 MacBook Pro 运行ning macOS 10.14.5 Mojave 上时，我得到：
test - after write:
Binary: 00001110011101010101100111100000
- opcode: 100000
- rs: 00111
- rt: 01011
- rd: 10101
- shamt: 10011
- funct: 000011
test - after read:
Binary: 00001110011101010101100111100000
- opcode: 100000
- rs: 00111
- rt: 01011
- rd: 10101
- shamt: 10011
- funct: 000011
原始二进制输出文件：
$ xxd -b filename.bin
00000000: 11100000 01011001 01110101 00001110 .Yu.
$
我的解释是，在我的机器上，opcodebit-field的数据在存储单元的最低6位，funct[=89的数据=]在最高6位，其他元素在中间。查看 32 位值时，这一点很明显。 xxd -b 拆分它的方式需要更多解释：
- 第一个字节是最低有效字节 — 英特尔架构。
- 它的最低有效位包含了
opcode的全部6位；它在其最高有效位中也包含rs的两个最低有效位。 - 第二个字节包含
rs的三个最高有效位作为其最低有效位，以及来自rt的所有 5 位作为其最高有效位。 - 第三个字节在其最低有效位中包含
rd的所有 5 位，在其最高有效位中包含shamt的 3 个最低有效位。 - 第四个也是最重要的字节在其最低有效位中包含
shamt的 2 个最高有效位，在其最高有效位中包含funct的所有 6 位。
有点mind-blowing!
当我恢复为 test 结构 (struct R test = {32, 0, 11, 21, 19, 0};) 的值时，我得到：
test - after write:
Binary: 00000010011101010101100000100000
- opcode: 100000
- rs: 00000
- rt: 01011
- rd: 10101
- shamt: 10011
- funct: 000000
test - after read:
Binary: 00000010011101010101100000100000
- opcode: 100000
- rs: 00000
- rt: 01011
- rd: 10101
- shamt: 10011
- funct: 000000
和
00000000: 00100000 01011000 01110101 00000010 Xu.
你的硬件 and/or 编译器和我的不一样； bit-fields.
的布局可能有不同的规则请注意，此代码假设 unsigned 或 unsigned int 是 32 位数量，但未进行测试。如果您使用的系统不适用，则需要修改代码以使用 uint32_t 和 uint8_t 等类型，如 <stdint.h> (和 <inttypes.h>).
精简代码
此代码在各个方面都比原始代码组织得更好。
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct R
{
unsigned int opcode : 6;
unsigned int rs : 5;
unsigned int rt : 5;
unsigned int rd : 5;
unsigned int shamt : 5;
unsigned int funct : 6;
};
static void test_r(const char *tag, struct R r, FILE *fp);
static void run_xxd(const char *file);
int main(void)
{
char filename[] = "filename.bin";
FILE *fp = fopen(filename, "w+b");
if (fp == NULL)
{
fprintf(stderr, "failed to open file '%s' for reading and writing\n", filename);
return 1;
}
struct R r[] =
{
{ 32, 0, 11, 21, 19, 0 },
{ 32, 7, 11, 21, 19, 3 },
{ 6, 21, 10, 14, 10, 8 },
};
enum { NUM_R = sizeof(r) / sizeof(r[0]) };
for (int i = 0; i < NUM_R; i++)
{
char name[16];
snprintf(name, sizeof(name), "r%d", i+1);
test_r(name, r[i], fp);
}
fclose(fp);
run_xxd(filename);
return 0;
}
static void run_one_xxd(const char *command, const char *filename)
{
char cmd[256];
snprintf(cmd, sizeof(cmd), "%s %s", command, filename);
printf("\nCommand: %s\n", cmd);
fflush(stdout);
system(cmd);
putchar('\n');
}
static void run_xxd(const char *filename)
{
run_one_xxd("xxd -c 4 -b ", filename);
run_one_xxd("xxd -c 4 -g 1 -u", filename);
}
static void format_binary8v(unsigned char x, int n, char buffer[static 9]);
static void format_binary32(unsigned x, char buffer[static 33]);
static void dump_bitfield(int nbits, unsigned value, const char *name);
static void dump_bytes(const char *tag, struct R r);
static void dump_R(const char *tag, struct R r);
static void test_r(const char *tag, struct R r, FILE *fp)
{
char buffer[32];
long offset = sizeof(struct R);
putchar('\n');
fwrite(&r, sizeof(r), 1, fp);
snprintf(buffer, sizeof(buffer), "%s - after write", tag);
dump_R(buffer, r);
fseek(fp, -offset, SEEK_CUR);
struct R s;
fread(&s, sizeof(s), 1, fp);
fseek(fp, 0, SEEK_CUR); // Ready for reading or writing!
snprintf(buffer, sizeof(buffer), "%s - after read", tag);
dump_R(buffer, s);
/* Safe regardless of whether struct R uses all bits in its storage unit */
assert(r.opcode == s.opcode);
assert(r.rs == s.rs );
assert(r.rs == s.rs );
assert(r.rs == s.rs );
assert(r.shamt == s.shamt );
assert(r.funct == s.funct );
/* Only safe because struct R uses all bits of its storage unit */
assert(memcmp(&r, &s, sizeof(struct R)) == 0);
}
static void dump_R(const char *tag, struct R r)
{
printf("%s:\n", tag);
dump_bytes("Binary", r);
dump_bitfield(6, r.opcode, "opcode");
dump_bitfield(5, r.rs, "rs");
dump_bitfield(5, r.rt, "rt");
dump_bitfield(5, r.rd, "rd");
dump_bitfield(5, r.shamt, "shamt");
dump_bitfield(6, r.funct, "funct");
}
static void dump_bytes(const char *tag, struct R r)
{
union X
{
struct R r;
unsigned i;
};
union X x = { .r = r };
char buffer[33];
printf("%s: 0x%.8X\n", tag, x.i);
format_binary32(x.i, buffer);
//printf("%s: MSB %s LSB\n", tag, buffer);
printf("%s: MSB", tag);
for (int i = 0; i < 4; i++)
printf(" %.8s", &buffer[8 * i]);
puts(" LSB (big-endian)");
printf("%s: LSB", tag);
for (int i = 0; i < 4; i++)
printf(" %.8s", &buffer[8 * (3 - i)]);
puts(" MSB (little-endian)");
}
static void dump_bitfield(int nbits, unsigned value, const char *name)
{
assert(nbits > 0 && nbits <= 32);
char vbuffer[33];
char nbuffer[8];
snprintf(nbuffer, sizeof(nbuffer), "%s:", name);
format_binary8v(value, nbits, vbuffer);
printf(" - %-7s %6s (%u)\n", nbuffer, vbuffer, value);
}
static
void format_binary8v(unsigned char x, int n, char buffer[static 9])
{
assert(n > 0 && n <= 8);
int start = 1 << (n - 1);
for (int b = start; b != 0; b /= 2)
{
*buffer++ = ((b & x) != 0) ? '1' : '0';
x &= ~b;
}
*buffer = '[=15=]';
}
static
void format_binary32(unsigned x, char buffer[static 33])
{
for (unsigned b = 2147483648; b != 0; b /= 2)
{
*buffer++ = ((b & x) != 0) ? '1' : '0';
x &= ~b;
}
*buffer = '[=15=]';
}
它产生输出：
r1 - after write:
Binary: 0x02755820
Binary: MSB 00000010 01110101 01011000 00100000 LSB (big-endian)
Binary: LSB 00100000 01011000 01110101 00000010 MSB (little-endian)
- opcode: 100000 (32)
- rs: 00000 (0)
- rt: 01011 (11)
- rd: 10101 (21)
- shamt: 10011 (19)
- funct: 000000 (0)
r1 - after read:
Binary: 0x02755820
Binary: MSB 00000010 01110101 01011000 00100000 LSB (big-endian)
Binary: LSB 00100000 01011000 01110101 00000010 MSB (little-endian)
- opcode: 100000 (32)
- rs: 00000 (0)
- rt: 01011 (11)
- rd: 10101 (21)
- shamt: 10011 (19)
- funct: 000000 (0)
r2 - after write:
Binary: 0x0E7559E0
Binary: MSB 00001110 01110101 01011001 11100000 LSB (big-endian)
Binary: LSB 11100000 01011001 01110101 00001110 MSB (little-endian)
- opcode: 100000 (32)
- rs: 00111 (7)
- rt: 01011 (11)
- rd: 10101 (21)
- shamt: 10011 (19)
- funct: 000011 (3)
r2 - after read:
Binary: 0x0E7559E0
Binary: MSB 00001110 01110101 01011001 11100000 LSB (big-endian)
Binary: LSB 11100000 01011001 01110101 00001110 MSB (little-endian)
- opcode: 100000 (32)
- rs: 00111 (7)
- rt: 01011 (11)
- rd: 10101 (21)
- shamt: 10011 (19)
- funct: 000011 (3)
r3 - after write:
Binary: 0x214E5546
Binary: MSB 00100001 01001110 01010101 01000110 LSB (big-endian)
Binary: LSB 01000110 01010101 01001110 00100001 MSB (little-endian)
- opcode: 000110 (6)
- rs: 10101 (21)
- rt: 01010 (10)
- rd: 01110 (14)
- shamt: 01010 (10)
- funct: 001000 (8)
r3 - after read:
Binary: 0x214E5546
Binary: MSB 00100001 01001110 01010101 01000110 LSB (big-endian)
Binary: LSB 01000110 01010101 01001110 00100001 MSB (little-endian)
- opcode: 000110 (6)
- rs: 10101 (21)
- rt: 01010 (10)
- rd: 01110 (14)
- shamt: 01010 (10)
- funct: 001000 (8)
Command: xxd -c 4 -b filename.bin
00000000: 00100000 01011000 01110101 00000010 Xu.
00000004: 11100000 01011001 01110101 00001110 .Yu.
00000008: 01000110 01010101 01001110 00100001 FUN!
Command: xxd -c 4 -g 1 -u filename.bin
00000000: 20 58 75 02 Xu.
00000004: E0 59 75 0E .Yu.
00000008: 46 55 4E 21 FUN!