ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<respond>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  lea    D14(%rip),%rsi
  push   %r12
  mov    %edx,%r12d
  mov    %edi,%edx
  push   %rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <write@plt>
  mov    %r13,%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    D4(%rip),%rsi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <write@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x90,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
datatype: string
name: extract_symbolic_data
arguments: tag: D4
datatype: string

tool: D14: HTTP/1.1 200 OK
Server: nanorestfile
Content-Length: %d


tool: D4: Connection: close
Content-Type: text/%s



assistant: ```c
void respond(int len, char *mime, int socketfd) {
  char buffer[128];
  sprintf(buffer, "HTTP/1.1 200 OK\r\nServer: nanorestfile\r\nContent-Length: %d\r\n", len);
  write(socketfd, buffer, strlen(buffer));
  sprintf(buffer, "Connection: close\r\nContent-Type: text/%s\r\n\r\n", mime);
  write(socketfd, buffer, strlen(buffer));
}
```
user: Translate the below assembly code into C programming language:

```asm
<respond>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  lea    D8(%rip),%rsi
  push   %r12
  mov    %edx,%r12d
  mov    %edi,%edx
  push   %rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <write@plt>
  mov    %r13,%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    D4(%rip),%rsi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <write@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x90,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D8
type_guessed: string
name: read_data
arguments: data_label: D4
type_guessed: string

tool: D8: HTTP/1.1 200 OK
Server: nanorestfile
Content-Length: %d


tool: D4: Connection: close
Content-Type: text/%s



assistant: ```c
void respond(int len, char *mime, int socketfd) {
  char buffer[128];
  sprintf(buffer, "HTTP/1.1 200 OK\r\nServer: nanorestfile\r\nContent-Length: %d\r\n", len);
  write(socketfd, buffer, strlen(buffer));
  sprintf(buffer, "Connection: close\r\nContent-Type: text/%s\r\n\r\n", mime);
  write(socketfd, buffer, strlen(buffer));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<string_match_end>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jbe    L0
  mov    -0x10(%rbp),%rax
  sub    -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L0
  mov    $0x1,%eax
  jmp    L9
L0:
  mov    $0x0,%eax
L9:
  leave
  ret
```

assistant: ```c
bool string_match_end(const char *str, const char *pat) {
  const size_t str_len = strlen(str);
  const size_t pat_len = strlen(pat);
  return (str_len > pat_len && strncmp(&str[str_len - pat_len], pat, pat_len) == 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<string_match_end>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %rax,%rbp
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  cmp    %rdx,%rbp
  ja     L5
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  sub    %rdx,%rbp
  lea    (%rbx,%rbp,1),%rdi
  mov    %r12,%rsi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L2
```

assistant: ```c
bool string_match_end(const char *str, const char *pat) {
  const size_t str_len = strlen(str);
  const size_t pat_len = strlen(pat);
  return (str_len > pat_len && strncmp(&str[str_len - pat_len], pat, pat_len) == 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<string_match_end>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    %rax,%rdx
  xor    %eax,%eax
  cmp    %rdx,%rbp
  ja     L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L1:
  sub    %rdx,%rbp
  mov    %r12,%rsi
  lea    (%rbx,%rbp,1),%rdi
  call   <strncmp@plt>
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool string_match_end(const char *str, const char *pat) {
  const size_t str_len = strlen(str);
  const size_t pat_len = strlen(pat);
  return (str_len > pat_len && strncmp(&str[str_len - pat_len], pat, pat_len) == 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
<string_match_end>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    %rax,%rdx
  xor    %eax,%eax
  cmp    %rdx,%rbp
  ja     L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L4:
  sub    %rdx,%rbp
  mov    %r12,%rsi
  lea    (%rbx,%rbp,1),%rdi
  call   <strncmp@plt>
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool string_match_end(const char *str, const char *pat) {
  const size_t str_len = strlen(str);
  const size_t pat_len = strlen(pat);
  return (str_len > pat_len && strncmp(&str[str_len - pat_len], pat, pat_len) == 0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x40,%edi
  call   <rec@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
datatype: string
name: read_data
arguments: data_tag: D6
datatype: string

tool: D2: lsframe2: start

tool: D6: lsframe2: done, result is %d


assistant: ```c
int main(void) {
  int sum;
  printf("lsframe2: start\n");
  sum = rec(64);
  printf("lsframe2: done, result is %d\n", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x40,%edi
  call   <rec@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D14
type_infered: string
name: parse_data
arguments: data_label: D12
type_infered: string

tool: D14: lsframe2: start

tool: D12: lsframe2: done, result is %d


assistant: ```c
int main(void) {
  int sum;
  printf("lsframe2: start\n");
  sum = rec(64);
  printf("lsframe2: done, result is %d\n", sum);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <puts@plt>
  mov    $0x40,%edi
  call   <rec@plt>
  lea    D18(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_guessed: string
name: parse_data
arguments: data_tag: D18
type_guessed: string

tool: D3: lsframe2: start

tool: D18: lsframe2: done, result is %d


assistant: ```c
int main(void) {
  int sum;
  printf("lsframe2: start\n");
  sum = rec(64);
  printf("lsframe2: done, result is %d\n", sum);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <puts@plt>
  mov    $0x40,%edi
  call   <rec@plt>
  lea    D11(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_infered: string
name: parse_data
arguments: data_label: D11
type_infered: string

tool: D2: lsframe2: start

tool: D11: lsframe2: done, result is %d


assistant: ```c
int main(void) {
  int sum;
  printf("lsframe2: start\n");
  sum = rec(64);
  printf("lsframe2: done, result is %d\n", sum);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<input_map_a2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x62,-0x4(%rbp)
  jne    L2
  mov    $0x1,%eax
  jmp    L9
L2:
  mov    $0x0,%eax
L9:
  pop    %rbp
  ret
```

assistant: ```c
int input_map_a2(char c) {
  if (c == 'b')
    return 1;
  else
    return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<input_map_a2>:
  endbr64
  cmp    $0x62,%dil
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int input_map_a2(char c) {
  if (c == 'b')
    return 1;
  else
    return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<input_map_a2>:
  endbr64
  xor    %eax,%eax
  cmp    $0x62,%dil
  sete   %al
  ret
```

assistant: ```c
int input_map_a2(char c) {
  if (c == 'b')
    return 1;
  else
    return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<input_map_a2>:
  endbr64
  xor    %eax,%eax
  cmp    $0x62,%dil
  sete   %al
  ret
```

assistant: ```c
int input_map_a2(char c) {
  if (c == 'b')
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x4(%rbp)
  jmp    L13
L4:
  shll   -0x4(%rbp)
L13:
  cmpl   $0xf,-0x4(%rbp)
  jle    L4
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D7
data_type: string

tool: D7: %d

assistant: ```c
int main(void) {
  int i;
  i = 1;
  while (i < 16)
    i *= 2;
  printf("%d", i);
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x10,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D1
data_type: string

tool: D1: %d

assistant: ```c
int main(void) {
  int i;
  i = 1;
  while (i < 16)
    i *= 2;
  printf("%d", i);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x10,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
datatype: string

tool: D7: %d

assistant: ```c
int main(void) {
  int i;
  i = 1;
  while (i < 16)
    i *= 2;
  printf("%d", i);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x10,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
type_infered: string

tool: D8: %d

assistant: ```c
int main(void) {
  int i;
  i = 1;
  while (i < 16)
    i *= 2;
  printf("%d", i);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<h2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L48
L80:
  movl   $0x0,-0x4(%rbp)
  jmp    L10
L35:
  cmpl   $0x0,-0x10(%rbp)
  jne    L8
  cmpl   $0x0,-0x4(%rbp)
  je     L24
L8:
  cmpl   $0xf,-0x8(%rbp)
  jne    L55
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  test   %eax,%eax
  je     L7
L55:
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  mov    -0x8(%rbp),%edx
  add    $0x1,%edx
  cmp    %edx,%eax
  je     L7
  addl   $0x1,-0xc(%rbp)
L7:
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  test   %eax,%eax
  je     L24
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  mov    %eax,-0x8(%rbp)
L24:
  addl   $0x1,-0x4(%rbp)
L10:
  cmpl   $0x3,-0x4(%rbp)
  jle    L35
  addl   $0x1,-0x10(%rbp)
L48:
  cmpl   $0x3,-0x10(%rbp)
  jle    L80
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int h2(int m[4][4]) {
  int i;
  int counter = 0;
  int previous = m[0][0];
  for (i = 0; i < 4; i++) {
    int j;
    for (j = 0; j < 4; j++) {
      if (!((i == 0) && (j == 0))) {
        if (!((previous == 15) && (m[j][i] == 0))) {
          if (m[j][i] != previous + 1) {
            counter++;
          }
        }
        if (m[j][i] != 0) {
          previous = m[j][i];
        }
      }
    }
  }
  return counter;
}
```
user: What is the c source code of the assembly code below:

```asm
<h2>:
  endbr64
  mov    (%rdi),%r8d
  mov    $0x0,%r9d
  mov    $0x0,%esi
  jmp    L92
L51:
  mov    (%rcx),%edx
  test   %edx,%edx
  je     L26
  cmp    $0x10,%edx
  je     L40
L0:
  add    $0x1,%r9d
L10:
  mov    (%rcx),%edx
  test   %edx,%edx
  cmovne %edx,%r8d
L26:
  cmp    $0x3,%eax
  jg     L107
L61:
  add    $0x1,%eax
  add    $0x10,%rcx
L3:
  lea    -0x1(%rax),%edx
  or     %esi,%edx
  je     L61
  cmp    $0xf,%r8d
  je     L51
  lea    0x1(%r8),%edx
  cmp    %edx,(%rcx)
  jne    L0
  jmp    L10
L107:
  add    $0x1,%esi
  add    $0x4,%rdi
  cmp    $0x4,%esi
  je     L32
L92:
  mov    %rdi,%rcx
  mov    $0x1,%eax
  jmp    L3
L40:
  mov    %edx,%r8d
  jmp    L26
L32:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int h2(int m[4][4]) {
  int i;
  int counter = 0;
  int previous = m[0][0];
  for (i = 0; i < 4; i++) {
    int j;
    for (j = 0; j < 4; j++) {
      if (!((i == 0) && (j == 0))) {
        if (!((previous == 15) && (m[j][i] == 0))) {
          if (m[j][i] != previous + 1) {
            counter++;
          }
        }
        if (m[j][i] != 0) {
          previous = m[j][i];
        }
      }
    }
  }
  return counter;
}
```
user: What is the c source code of the assembly code below?

```asm
<h2>:
  endbr64
  mov    (%rdi),%r8d
  xor    %r9d,%r9d
  xor    %esi,%esi
L26:
  mov    %rdi,%rdx
  xor    %eax,%eax
L23:
  mov    %esi,%ecx
  or     %eax,%ecx
  add    $0x1,%eax
  test   %ecx,%ecx
  je     L62
  mov    (%rdx),%ecx
  cmp    $0xf,%r8d
  je     L38
  lea    0x1(%r8),%r10d
  cmp    %ecx,%r10d
  je     L16
L30:
  add    $0x1,%r9d
L16:
  test   %ecx,%ecx
  cmovne %ecx,%r8d
L8:
  cmp    $0x4,%eax
  jne    L62
  add    $0x1,%esi
  add    $0x4,%rdi
  cmp    $0x4,%esi
  jne    L26
  mov    %r9d,%eax
  ret
  nopl   0x0(%rax)
L62:
  add    $0x10,%rdx
  jmp    L23
  cs nopw 0x0(%rax,%rax,1)
L38:
  test   %ecx,%ecx
  je     L8
  cmp    $0x10,%ecx
  jne    L30
  mov    $0x10,%r8d
  jmp    L8
```

assistant: ```c
int h2(int m[4][4]) {
  int i;
  int counter = 0;
  int previous = m[0][0];
  for (i = 0; i < 4; i++) {
    int j;
    for (j = 0; j < 4; j++) {
      if (!((i == 0) && (j == 0))) {
        if (!((previous == 15) && (m[j][i] == 0))) {
          if (m[j][i] != previous + 1) {
            counter++;
          }
        }
        if (m[j][i] != 0) {
          previous = m[j][i];
        }
      }
    }
  }
  return counter;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<h2>:
  endbr64
  mov    (%rdi),%eax
  lea    0x4(%rdi),%rdx
  xor    %r8d,%r8d
  add    $0x10,%rdi
  mov    $0xf,%r9d
  mov    0xc(%rdx),%ecx
  cmp    $0xf,%eax
  je     L50
L273:
  lea    0x1(%rax),%esi
  cmp    %ecx,%esi
  je     L308
  add    $0x1,%r8d
L308:
  test   %ecx,%ecx
  je     L305
  mov    0x1c(%rdx),%eax
  cmp    $0xf,%ecx
  je     L280
L12:
  lea    0x1(%rcx),%esi
  cmp    %esi,%eax
  je     L143
  add    $0x1,%r8d
L143:
  test   %eax,%eax
  je     L232
  mov    0x2c(%rdx),%esi
  cmp    $0xf,%eax
  je     L328
L318:
  lea    0x1(%rax),%ecx
  cmp    %ecx,%esi
  je     L59
L320:
  add    $0x1,%r8d
L59:
  test   %esi,%esi
  cmovne %esi,%eax
  cmp    %rdx,%rdi
  je     L75
  mov    (%rdx),%ecx
  cmp    $0xf,%eax
  je     L123
L51:
  lea    0x1(%rax),%esi
  cmp    %ecx,%esi
  je     L207
L15:
  add    $0x1,%r8d
L207:
  test   %ecx,%ecx
  cmovne %ecx,%eax
  add    $0x4,%rdx
  mov    0xc(%rdx),%ecx
  cmp    $0xf,%eax
  jne    L273
L50:
  test   %ecx,%ecx
  je     L9
L304:
  cmp    $0x10,%ecx
  je     L85
  add    $0x1,%r8d
  test   %ecx,%ecx
  mov    0x1c(%rdx),%eax
  cmove  %r9d,%ecx
  cmp    $0xf,%ecx
  jne    L12
L280:
  test   %eax,%eax
  je     L3
L247:
  cmp    $0x10,%eax
  je     L222
  add    $0x1,%r8d
  test   %eax,%eax
  mov    0x2c(%rdx),%esi
  cmove  %r9d,%eax
  cmp    $0xf,%eax
  jne    L318
  nop
L328:
  test   %esi,%esi
  jne    L25
L87:
  cmp    %rdx,%rdi
  je     L75
  mov    (%rdx),%ecx
L123:
  test   %ecx,%ecx
  je     L32
  mov    $0xf,%eax
  cmp    $0x10,%ecx
  jne    L15
  add    $0x4,%rdx
  mov    $0x10,%eax
  mov    0xc(%rdx),%ecx
  jmp    L273
  nopw   0x0(%rax,%rax,1)
L75:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L305:
  mov    %eax,%ecx
L85:
  mov    0x1c(%rdx),%eax
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L232:
  mov    0x2c(%rdx),%esi
  mov    %ecx,%eax
  jmp    L318
  nopw   0x0(%rax,%rax,1)
L3:
  mov    0x2c(%rdx),%esi
  test   %esi,%esi
  je     L87
L25:
  mov    $0xf,%eax
  cmp    $0x10,%esi
  jne    L320
  cmp    %rdx,%rdi
  je     L75
  mov    (%rdx),%ecx
  mov    $0x10,%eax
  jmp    L51
  xchg   %ax,%ax
L32:
  mov    0x10(%rdx),%ecx
  add    $0x4,%rdx
  test   %ecx,%ecx
  jne    L304
L9:
  mov    0x1c(%rdx),%eax
  test   %eax,%eax
  jne    L247
  jmp    L3
L222:
  mov    0x2c(%rdx),%esi
  jmp    L318
```

assistant: ```c
int h2(int m[4][4]) {
  int i;
  int counter = 0;
  int previous = m[0][0];
  for (i = 0; i < 4; i++) {
    int j;
    for (j = 0; j < 4; j++) {
      if (!((i == 0) && (j == 0))) {
        if (!((previous == 15) && (m[j][i] == 0))) {
          if (m[j][i] != previous + 1) {
            counter++;
          }
        }
        if (m[j][i] != 0) {
          previous = m[j][i];
        }
      }
    }
  }
  return counter;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<last_one12>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L8
  xor    %edx,%edx
  tzcnt  -0x8(%rbp),%edx
  mov    $0x50,%eax
  sub    %edx,%eax
  jmp    L14
L8:
  xor    %edx,%edx
  tzcnt  -0x4(%rbp),%edx
  mov    $0x35,%eax
  sub    %edx,%eax
L14:
  pop    %rbp
  ret
```

assistant: ```c
int last_one12(unsigned int u1, unsigned int u2) {
  if (u2) {
    return 80 - __builtin_ctz(u2);
  }
  return 53 - __builtin_ctz(u1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<last_one12>:
  endbr64
  test   %esi,%esi
  jne    L4
  tzcnt  %edi,%edi
  mov    $0x35,%eax
  sub    %edi,%eax
  ret
L4:
  tzcnt  %esi,%esi
  mov    $0x50,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int last_one12(unsigned int u1, unsigned int u2) {
  if (u2) {
    return 80 - __builtin_ctz(u2);
  }
  return 53 - __builtin_ctz(u1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<last_one12>:
  endbr64
  test   %esi,%esi
  jne    L5
  tzcnt  %edi,%edi
  mov    $0x35,%eax
  sub    %edi,%eax
  ret
  nopl   0x0(%rax)
L5:
  tzcnt  %esi,%esi
  mov    $0x50,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int last_one12(unsigned int u1, unsigned int u2) {
  if (u2) {
    return 80 - __builtin_ctz(u2);
  }
  return 53 - __builtin_ctz(u1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<last_one12>:
  endbr64
  test   %esi,%esi
  jne    L4
  tzcnt  %edi,%edi
  mov    $0x35,%eax
  sub    %edi,%eax
  ret
  nopl   0x0(%rax)
L4:
  tzcnt  %esi,%esi
  mov    $0x50,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int last_one12(unsigned int u1, unsigned int u2) {
  if (u2) {
    return 80 - __builtin_ctz(u2);
  }
  return 53 - __builtin_ctz(u1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<f1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %ax,-0x4(%rbp)
L17:
  cmpw   $0x0,-0x4(%rbp)
  jg     L0
  mov    $0x64,%eax
  jmp    L19
L0:
  movzwl -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %ax,-0x4(%rbp)
  cmpw   $0x0,-0x4(%rbp)
  jne    L17
  mov    $0xc8,%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
short f1(short x) {
  do {
    if (x <= 0)
      return 100;
  } while (--x);
  return 200;
}
```
user: What is the c source code of the assembly code below?

```asm
<f1>:
  endbr64
L10:
  test   %di,%di
  jle    L3
  sub    $0x1,%di
  jne    L10
  mov    $0xc8,%eax
  ret
L3:
  mov    $0x64,%eax
  ret
```

assistant: ```c
short f1(short x) {
  do {
    if (x <= 0)
      return 100;
  } while (--x);
  return 200;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<f1>:
  endbr64
  nopl   0x0(%rax)
L18:
  test   %di,%di
  jle    L13
  sub    $0x1,%di
  jne    L18
  mov    $0xc8,%eax
  ret
  nopl   0x0(%rax)
L13:
  mov    $0x64,%eax
  ret
```

assistant: ```c
short f1(short x) {
  do {
    if (x <= 0)
      return 100;
  } while (--x);
  return 200;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f1>:
  endbr64
  nopl   0x0(%rax)
L14:
  test   %di,%di
  jle    L4
  sub    $0x1,%di
  jne    L14
  mov    $0xc8,%eax
  ret
  nopl   0x0(%rax)
L4:
  mov    $0x64,%eax
  ret
```

assistant: ```c
short f1(short x) {
  do {
    if (x <= 0)
      return 100;
  } while (--x);
  return 200;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ADCSequenceDataGet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x18(%rbp),%eax
  add    $0x2,%eax
  shl    $0x5,%eax
  add    %eax,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L9:
  mov    -0x14(%rbp),%eax
  add    $0x8,%eax
  cltq
  mov    %rax,%rcx
  mov    -0x20(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    (%rcx),%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L7:
  mov    -0x14(%rbp),%eax
  add    $0xc,%eax
  cltq
  mov    (%rax),%eax
  and    $0x100,%eax
  test   %eax,%eax
  jne    L0
  cmpl   $0x7,-0x4(%rbp)
  jle    L9
L0:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int32_t ADCSequenceDataGet(uint32_t ui32Base, uint32_t ui32SequenceNum, uint32_t *pui32Buffer) {
  uint32_t ui32Count;
  ;
  ;
  ui32Base += (0x00000040) + ((0x00000060 - 0x00000040) * ui32SequenceNum);
  ui32Count = 0;
  while (!((*((volatile uint32_t *)(ui32Base + (0x0000004C - 0x00000040)))) & 0x00000100) && (ui32Count < 8)) {
    *pui32Buffer++ = (*((volatile uint32_t *)(ui32Base + (0x00000048 - 0x00000040))));
    ui32Count++;
  }
  return (ui32Count);
}
```