ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<msd>:
  endbr64
  push   %r15
  lea    D8(%rip),%r15
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  lea    D4(%rip),%r13
  push   %r12
  lea    D43(%rip),%r12
  push   %rbp
  lea    D12(%rip),%rbp
  push   %rbx
  sub    $0x218,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
L12:
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %esi,%esi
  mov    $0x200,%edx
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x4,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    $0x1206,%eax
  mov    %rbp,%rdx
  mov    %rbx,%rdi
  mov    $0x1f,%esi
  movl   $0x12330984,0x4(%rsp)
  movb   $0x24,0x8(%rsp)
  movb   $0x80,0xc(%rsp)
  mov    %ax,0xe(%rsp)
  movb   $0x24,0x13(%rsp)
  call   <hex@plt>
  mov    $0x1f,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <send_bulk@plt>
  test   %rax,%rax
  js     L7
  mov    $0x200,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  add    $0x1,%r14d
  call   <memset@plt>
  mov    $0x24,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <recv_bulk@plt>
  mov    %r15,%rdx
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <hex@plt>
  mov    $0x200,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x24,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <recv_bulk@plt>
  lea    D9(%rip),%rdx
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <hex@plt>
  mov    $0x249f0,%edi
  call   <usleep@plt>
  cmp    $0x3,%r14d
  jne    L12
L7:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x218,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D8
datatype: string
name: unpack_data
arguments: data_tag: D4
datatype: string
name: unpack_data
arguments: data_tag: D43
datatype: string
name: unpack_data
arguments: data_tag: D12
datatype: string
name: unpack_data
arguments: data_tag: D9
datatype: string

tool: D8: inquiry result

tool: D4: i=%d


tool: D43: USBC

tool: D12: intquery req

tool: D9: inquiry sense?

assistant: ```c
void msd() {
  unsigned InEndPt = 1;
  unsigned OutEndPt = 1;
  unsigned char abBuf[512];
  int i;
  for (i = 0; i < 3; i++) {
    printf("i=%d\n", i);
    memset(abBuf, 0, sizeof(abBuf));
    memcpy(abBuf, "USBC", 4);
    *(uint32_t *)(&abBuf[4]) = 0x12330984;
    abBuf[8] = 0x24;
    abBuf[0xc] = 0x80;
    abBuf[0xe] = 0x06;
    abBuf[0x0f] = 0x12;
    abBuf[0x13] = 0x24;
    hex(abBuf, 31, "intquery req");
    if (send_bulk(OutEndPt, abBuf, 31) < 0)
      return;
    memset(abBuf, 0, sizeof(abBuf));
    int cb = recv_bulk(InEndPt, abBuf, 36);
    hex(abBuf, cb, "inquiry result");
    memset(abBuf, 0, sizeof(abBuf));
    cb = recv_bulk(InEndPt, abBuf, 36);
    hex(abBuf, cb, "inquiry sense?");
    usleep(150000);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<msd>:
  endbr64
  push   %r15
  lea    D47(%rip),%r15
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  lea    D2(%rip),%r13
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  lea    D14(%rip),%rbp
  push   %rbx
  sub    $0x218,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
L2:
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %esi,%esi
  mov    $0x200,%edx
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x4,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    $0x1206,%eax
  mov    %rbp,%rdx
  mov    %rbx,%rdi
  mov    $0x1f,%esi
  movl   $0x12330984,0x4(%rsp)
  movb   $0x24,0x8(%rsp)
  movb   $0x80,0xc(%rsp)
  mov    %ax,0xe(%rsp)
  movb   $0x24,0x13(%rsp)
  call   <hex@plt>
  mov    $0x1f,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <send_bulk@plt>
  test   %rax,%rax
  js     L0
  mov    $0x200,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  add    $0x1,%r14d
  call   <memset@plt>
  mov    $0x24,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <recv_bulk@plt>
  mov    %r15,%rdx
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <hex@plt>
  mov    $0x200,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x24,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <recv_bulk@plt>
  lea    D13(%rip),%rdx
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <hex@plt>
  mov    $0x249f0,%edi
  call   <usleep@plt>
  cmp    $0x3,%r14d
  jne    L2
L0:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x218,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D47
data_type: string
name: retrieve_data
arguments: label: D2
data_type: string
name: retrieve_data
arguments: label: D7
data_type: string
name: retrieve_data
arguments: label: D14
data_type: string
name: retrieve_data
arguments: label: D13
data_type: string

tool: D47: inquiry result

tool: D2: i=%d


tool: D7: USBC

tool: D14: intquery req

tool: D13: inquiry sense?

assistant: ```c
void msd() {
  unsigned InEndPt = 1;
  unsigned OutEndPt = 1;
  unsigned char abBuf[512];
  int i;
  for (i = 0; i < 3; i++) {
    printf("i=%d\n", i);
    memset(abBuf, 0, sizeof(abBuf));
    memcpy(abBuf, "USBC", 4);
    *(uint32_t *)(&abBuf[4]) = 0x12330984;
    abBuf[8] = 0x24;
    abBuf[0xc] = 0x80;
    abBuf[0xe] = 0x06;
    abBuf[0x0f] = 0x12;
    abBuf[0x13] = 0x24;
    hex(abBuf, 31, "intquery req");
    if (send_bulk(OutEndPt, abBuf, 31) < 0)
      return;
    memset(abBuf, 0, sizeof(abBuf));
    int cb = recv_bulk(InEndPt, abBuf, 36);
    hex(abBuf, cb, "inquiry result");
    memset(abBuf, 0, sizeof(abBuf));
    cb = recv_bulk(InEndPt, abBuf, 36);
    hex(abBuf, cb, "inquiry sense?");
    usleep(150000);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nullArray>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L8:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void nullArray(int arrayLetters[], int size) {
  for (int i = 0; i < size; i++) {
    arrayLetters[i] = 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nullArray>:
  endbr64
  test   %esi,%esi
  jle    L12
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rdx
L19:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L19
L12:
  ret
```

assistant: ```c
void nullArray(int arrayLetters[], int size) {
  for (int i = 0; i < size; i++) {
    arrayLetters[i] = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<nullArray>:
  endbr64
  test   %esi,%esi
  jle    L9
  movslq %esi,%rsi
  lea    0x0(,%rsi,4),%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L9:
  ret
```

assistant: ```c
void nullArray(int arrayLetters[], int size) {
  for (int i = 0; i < size; i++) {
    arrayLetters[i] = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<nullArray>:
  endbr64
  test   %esi,%esi
  jle    L5
  movslq %esi,%rsi
  lea    0x0(,%rsi,4),%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L5:
  ret
```

assistant: ```c
void nullArray(int arrayLetters[], int size) {
  for (int i = 0; i < size; i++) {
    arrayLetters[i] = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sbuf_printf_uuid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x26,%esi
  mov    %rax,%rdi
  call   <snprintf_uuid@plt>
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  lea    D6(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sbuf_printf@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type: string

tool: D6: %s

assistant: ```c
int __attribute__(()) sbuf_printf_uuid(struct sbuf *sb, struct uuid *uuid) {
  char buf[38];
  snprintf_uuid(buf, sizeof(buf), uuid);
  return (sbuf_printf(sb, "%s", buf));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sbuf_printf_uuid>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0x26,%esi
  mov    %rbp,%rdi
  call   <snprintf_uuid@plt>
  mov    %rbp,%rdx
  lea    D3(%rip),%rsi
  mov    %rbx,%rdi
  call   <sbuf_printf@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D3
type: string

tool: D3: %s

assistant: ```c
int __attribute__(()) sbuf_printf_uuid(struct sbuf *sb, struct uuid *uuid) {
  char buf[38];
  snprintf_uuid(buf, sizeof(buf), uuid);
  return (sbuf_printf(sb, "%s", buf));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sbuf_printf_uuid>:
  endbr64
  push   %r12
  mov    %rsi,%rdx
  mov    $0x26,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <snprintf_uuid@plt>
  mov    %r12,%rdx
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <sbuf_printf@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D9
type: string

tool: D9: %s

assistant: ```c
int __attribute__(()) sbuf_printf_uuid(struct sbuf *sb, struct uuid *uuid) {
  char buf[38];
  snprintf_uuid(buf, sizeof(buf), uuid);
  return (sbuf_printf(sb, "%s", buf));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sbuf_printf_uuid>:
  endbr64
  push   %r12
  mov    %rsi,%rdx
  mov    $0x26,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <snprintf_uuid@plt>
  mov    %r12,%rdx
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <sbuf_printf@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D9
datatype: string

tool: D9: %s

assistant: ```c
int __attribute__(()) sbuf_printf_uuid(struct sbuf *sb, struct uuid *uuid) {
  char buf[38];
  snprintf_uuid(buf, sizeof(buf), uuid);
  return (sbuf_printf(sb, "%s", buf));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<init_rand_port>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jle    L39
  cmpq   $0x7ffffffe,-0x18(%rbp)
  jle    L33
L39:
  mov    $0x0,%eax
  call   <get_init_rand_port@plt>
  cltq
  mov    %rax,-0x18(%rbp)
L33:
  mov    D28(%rip),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  movl   $0x0,-0x4(%rbp)
  jmp    L10
L22:
  mov    D28(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <genr_rand_port@plt>
  movslq %eax,%rdx
  mov    D28(%rip),%rax
  mov    %rdx,(%rax)
  addl   $0x1,-0x4(%rbp)
L10:
  cmpl   $0xf,-0x4(%rbp)
  jle    L22
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
long init_rand_port(long seed) {
  extern long rand_num;
  int i;
  if (seed < 1 || seed > (2147483647L - 1)) {
    seed = get_init_rand_port();
  }
  rand_num = seed;
  for (i = 0; i < 16; i++) {
    rand_num = genr_rand_port(rand_num);
  }
  return seed;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<init_rand_port>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  lea    -0x1(%rdi),%rax
  cmp    $0x7ffffffd,%rax
  ja     L10
L9:
  mov    D5(%rip),%rax
  mov    %r12,(%rax)
  mov    $0x10,%ebx
  mov    %rax,%rbp
L24:
  mov    0x0(%rbp),%rdi
  mov    $0x0,%eax
  call   <genr_rand_port@plt>
  cltq
  mov    %rax,0x0(%rbp)
  sub    $0x1,%ebx
  jne    L24
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  mov    $0x0,%eax
  call   <get_init_rand_port@plt>
  movslq %eax,%r12
  jmp    L9
```

assistant: ```c
long init_rand_port(long seed) {
  extern long rand_num;
  int i;
  if (seed < 1 || seed > (2147483647L - 1)) {
    seed = get_init_rand_port();
  }
  rand_num = seed;
  for (i = 0; i < 16; i++) {
    rand_num = genr_rand_port(rand_num);
  }
  return seed;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<init_rand_port>:
  endbr64
  lea    -0x1(%rdi),%rax
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  cmp    $0x7ffffffd,%rax
  ja     L15
L17:
  mov    D1(%rip),%rbp
  mov    %r12,%rdi
  mov    $0x10,%ebx
  mov    %r12,0x0(%rbp)
  nopw   0x0(%rax,%rax,1)
L22:
  xor    %eax,%eax
  call   <genr_rand_port@plt>
  movslq %eax,%rdi
  mov    %rdi,0x0(%rbp)
  sub    $0x1,%ebx
  jne    L22
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  call   <get_init_rand_port@plt>
  movslq %eax,%r12
  jmp    L17
```

assistant: ```c
long init_rand_port(long seed) {
  extern long rand_num;
  int i;
  if (seed < 1 || seed > (2147483647L - 1)) {
    seed = get_init_rand_port();
  }
  rand_num = seed;
  for (i = 0; i < 16; i++) {
    rand_num = genr_rand_port(rand_num);
  }
  return seed;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<init_rand_port>:
  endbr64
  lea    -0x1(%rdi),%rax
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  cmp    $0x7ffffffd,%rax
  ja     L28
L25:
  mov    D2(%rip),%rbp
  mov    %r12,%rdi
  mov    $0x10,%ebx
  mov    %r12,0x0(%rbp)
  nopw   0x0(%rax,%rax,1)
L13:
  xor    %eax,%eax
  call   <genr_rand_port@plt>
  movslq %eax,%rdi
  mov    %rdi,0x0(%rbp)
  sub    $0x1,%ebx
  jne    L13
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L28:
  xor    %eax,%eax
  call   <get_init_rand_port@plt>
  movslq %eax,%r12
  jmp    L25
```

assistant: ```c
long init_rand_port(long seed) {
  extern long rand_num;
  int i;
  if (seed < 1 || seed > (2147483647L - 1)) {
    seed = get_init_rand_port();
  }
  rand_num = seed;
  for (i = 0; i < 16; i++) {
    rand_num = genr_rand_port(rand_num);
  }
  return seed;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<simple_strchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  jmp    L30
L18:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L8
  mov    $0x0,%eax
  jmp    L32
L8:
  addq   $0x1,-0x8(%rbp)
L30:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    -0xc(%rbp),%edx
  cmp    %dl,%al
  jne    L18
  mov    -0x8(%rbp),%rax
L32:
  pop    %rbp
  ret
```

assistant: ```c
char *simple_strchr(const char *s, int c) {
  for (; *s != (char)c; ++s)
    if (*s == '\0')
      return NULL;
  return (char *)s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<simple_strchr>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %sil,%al
  je     L3
L1:
  test   %al,%al
  je     L29
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  cmp    %dl,%al
  jne    L1
  mov    %rdi,%rax
  ret
L3:
  mov    %rdi,%rax
  ret
L29:
  mov    $0x0,%eax
  ret
```

assistant: ```c
char *simple_strchr(const char *s, int c) {
  for (; *s != (char)c; ++s)
    if (*s == '\0')
      return NULL;
  return (char *)s;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<simple_strchr>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %sil,%al
  jne    L27
  jmp    L37
L8:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  cmp    %dl,%al
  je     L37
L27:
  test   %al,%al
  jne    L8
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L37:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *simple_strchr(const char *s, int c) {
  for (; *s != (char)c; ++s)
    if (*s == '\0')
      return NULL;
  return (char *)s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<simple_strchr>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %sil,%al
  jne    L30
  jmp    L26
L32:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  cmp    %dl,%al
  je     L26
L30:
  test   %al,%al
  jne    L32
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *simple_strchr(const char *s, int c) {
  for (; *s != (char)c; ++s)
    if (*s == '\0')
      return NULL;
  return (char *)s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L3
L13:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L3:
  cmpl   $0x9,-0x8(%rbp)
  jle    L13
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int j;
  for (i = 1; i < 10; ++i)
    j = i;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int j;
  for (i = 1; i < 10; ++i)
    j = i;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int j;
  for (i = 1; i < 10; ++i)
    j = i;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read_in>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L4:
  lea    -0x9(%rbp),%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x3,%edi
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L4
  lea    -0x9(%rbp),%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L4
```

assistant: ```c
void *read_in(void *ignore) {
  char c;
  for (;;) {
    if (read(3, &c, sizeof(char)) == 1)
      write(1, &c, sizeof(char));
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<read_in>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rbx
L2:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x3,%edi
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L2
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L2
```

assistant: ```c
void *read_in(void *ignore) {
  char c;
  for (;;) {
    if (read(3, &c, sizeof(char)) == 1)
      write(1, &c, sizeof(char));
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<read_in>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rbx
  xchg   %ax,%ax
L11:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x3,%edi
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L11
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L11
```

assistant: ```c
void *read_in(void *ignore) {
  char c;
  for (;;) {
    if (read(3, &c, sizeof(char)) == 1)
      write(1, &c, sizeof(char));
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<read_in>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rbx
  xchg   %ax,%ax
L7:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x3,%edi
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L7
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L7
```

assistant: ```c
void *read_in(void *ignore) {
  char c;
  for (;;) {
    if (read(3, &c, sizeof(char)) == 1)
      write(1, &c, sizeof(char));
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<rearmar_bitmap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L9
L5:
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  movq   $0x0,-0x10(%rbp)
  jmp    L25
L0:
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    (%rax,%rax,1),%ecx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    -0x28(%rbp),%rdx
  add    %rdx,%rax
  movzbl (%rax),%eax
  add    %eax,%ecx
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
L25:
  cmpq   $0x7,-0x10(%rbp)
  jbe    L0
  addq   $0x1,-0x18(%rbp)
L9:
  mov    -0x18(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jb     L5
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *rearmar_bitmap(char *bitmap, size_t tam) {
  char *bloque = malloc(tam);
  size_t z = 0;
  for (size_t i = 0; i < tam; i++) {
    bloque[i] = 0;
    for (size_t j = 0; j < 8; j++) {
      bloque[i] = bloque[i] * 2 + bitmap[z++];
    }
  }
  return bloque;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rearmar_bitmap>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rsi,%rdi
  call   <malloc@plt>
  test   %rbp,%rbp
  je     L3
  mov    %rax,%r8
  lea    0x8(%rbx),%rdi
  lea    (%rdi,%rbp,8),%r11
  mov    $0x0,%r10d
  jmp    L2
L1:
  mov    %cl,(%r9)
  add    $0x1,%r8
  add    $0x8,%rdi
  cmp    %r11,%rdi
  je     L3
L2:
  mov    %r8,%r9
  lea    -0x8(%rdi),%rdx
  mov    %r10d,%ecx
L29:
  movzbl (%rdx),%esi
  lea    (%rsi,%rcx,2),%ecx
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L29
  jmp    L1
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *rearmar_bitmap(char *bitmap, size_t tam) {
  char *bloque = malloc(tam);
  size_t z = 0;
  for (size_t i = 0; i < tam; i++) {
    bloque[i] = 0;
    for (size_t j = 0; j < 8; j++) {
      bloque[i] = bloque[i] * 2 + bitmap[z++];
    }
  }
  return bloque;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rearmar_bitmap>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x1,%esi
  push   %rbx
  mov    %rdi,%rbx
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <calloc@plt>
  test   %rbp,%rbp
  je     L2
  lea    0x8(%rbx),%rdi
  mov    %rax,%r8
  lea    (%rdi,%rbp,8),%r9
  nopl   (%rax)
L29:
  lea    -0x8(%rdi),%rdx
  xor    %ecx,%ecx
  cs nopw 0x0(%rax,%rax,1)
L1:
  movzbl (%rdx),%esi
  add    $0x1,%rdx
  lea    (%rsi,%rcx,2),%ecx
  cmp    %rdx,%rdi
  jne    L1
  add    $0x8,%rdi
  mov    %cl,(%r8)
  add    $0x1,%r8
  cmp    %r9,%rdi
  jne    L29
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *rearmar_bitmap(char *bitmap, size_t tam) {
  char *bloque = malloc(tam);
  size_t z = 0;
  for (size_t i = 0; i < tam; i++) {
    bloque[i] = 0;
    for (size_t j = 0; j < 8; j++) {
      bloque[i] = bloque[i] * 2 + bitmap[z++];
    }
  }
  return bloque;
}
```