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

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    $0x10,%esi
  mov    %rbx,%rdi
  call   <realloc@plt>
  mov    %rax,%rbx
  mov    (%rax),%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int **ppi = malloc(sizeof *ppi);
  *ppi = malloc(sizeof **ppi);
  ppi = realloc(ppi, 2 * sizeof *ppi);
  free(*ppi);
  free(ppi);
}
```
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
  push   %rbp
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    $0x4,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    $0x10,%esi
  mov    %rbp,%rdi
  mov    %rax,0x0(%rbp)
  call   <realloc@plt>
  mov    (%rax),%rdi
  mov    %rax,%rbp
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int **ppi = malloc(sizeof *ppi);
  *ppi = malloc(sizeof **ppi);
  ppi = realloc(ppi, 2 * sizeof *ppi);
  free(*ppi);
  free(ppi);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    $0x4,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    $0x10,%esi
  mov    %rbp,%rdi
  mov    %rax,0x0(%rbp)
  call   <realloc@plt>
  mov    (%rax),%rdi
  mov    %rax,%rbp
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int **ppi = malloc(sizeof *ppi);
  *ppi = malloc(sizeof **ppi);
  ppi = realloc(ppi, 2 * sizeof *ppi);
  free(*ppi);
  free(ppi);
}
```
user: What is the c source code of the assembly code below?

```asm
<printNumCpusInfo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <output@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type_infered: string

tool: D2: Number of physical CPUs	0	0	


assistant: ```c
void printNumCpusInfo(const char *cmd) {
  (void)cmd;
  output("Number of physical CPUs\t0\t0\t\n");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<printNumCpusInfo>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <output@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
datatype: string

tool: D8: Number of physical CPUs	0	0	


assistant: ```c
void printNumCpusInfo(const char *cmd) {
  (void)cmd;
  output("Number of physical CPUs\t0\t0\t\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<printNumCpusInfo>:
  endbr64
  lea    D6(%rip),%rdi
  jmp    1050 <output@plt>
```

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

tool: D6: Number of physical CPUs	0	0	


assistant: ```c
void printNumCpusInfo(const char *cmd) {
  (void)cmd;
  output("Number of physical CPUs\t0\t0\t\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<printNumCpusInfo>:
  endbr64
  lea    D4(%rip),%rdi
  jmp    1050 <output@plt>
```

tool_calls: name: retrieve_data
arguments: label: D4
data_type: string

tool: D4: Number of physical CPUs	0	0	


assistant: ```c
void printNumCpusInfo(const char *cmd) {
  (void)cmd;
  output("Number of physical CPUs\t0\t0\t\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int flag = MSG_CMSG_CLOEXEC;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

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

assistant: ```c
int main(void) {
  int flag = MSG_CMSG_CLOEXEC;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
int main(void) {
  int flag = MSG_CMSG_CLOEXEC;
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int flag = MSG_CMSG_CLOEXEC;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<gc_configure>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <_cli@plt>
  mov    $0x6,%esi
  mov    $0x3ce,%edi
  call   <out_port@plt>
  mov    $0x1,%esi
  mov    $0x3cf,%edi
  call   <out_port@plt>
  mov    $0x0,%eax
  call   <_sti@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void gc_configure(void) {
  _cli();
  out_port(0x3ce, 0x6);
  out_port(0x3cf, 0x1);
  _sti();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gc_configure>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <_cli@plt>
  mov    $0x6,%esi
  mov    $0x3ce,%edi
  call   <out_port@plt>
  mov    $0x1,%esi
  mov    $0x3cf,%edi
  call   <out_port@plt>
  mov    $0x0,%eax
  call   <_sti@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void gc_configure(void) {
  _cli();
  out_port(0x3ce, 0x6);
  out_port(0x3cf, 0x1);
  _sti();
}
```
user: What is the c source code of the assembly code below:

```asm
<gc_configure>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <_cli@plt>
  mov    $0x6,%esi
  mov    $0x3ce,%edi
  call   <out_port@plt>
  mov    $0x1,%esi
  mov    $0x3cf,%edi
  call   <out_port@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <_sti@plt>
```

assistant: ```c
void gc_configure(void) {
  _cli();
  out_port(0x3ce, 0x6);
  out_port(0x3cf, 0x1);
  _sti();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gc_configure>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <_cli@plt>
  mov    $0x6,%esi
  mov    $0x3ce,%edi
  call   <out_port@plt>
  mov    $0x1,%esi
  mov    $0x3cf,%edi
  call   <out_port@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <_sti@plt>
```

assistant: ```c
void gc_configure(void) {
  _cli();
  out_port(0x3ce, 0x6);
  out_port(0x3cf, 0x1);
  _sti();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<req_login>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  lea    D18(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  mov    %eax,-0x34(%rbp)
  cmpl   $0x1,-0x34(%rbp)
  jne    L11
  lea    -0x30(%rbp),%rax
  mov    $0x20,%edx
  lea    D2(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L11
  mov    $0x1,%eax
  jmp    L20
L11:
  mov    $0x0,%eax
L20:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D18
data_type: string
name: retrieve_data
arguments: tag: D2
data_type: string

tool: D18: %*[^=]%*[^&]%*[^=]%*[^&]&%s

tool: D2: commit=Login

assistant: ```c
int req_login(char *data) {
  int c;
  char cl[32];
  c = sscanf(data, "%*[^=]%*[^&]%*[^=]%*[^&]&%s", cl);
  return c == 1 && !strncmp(cl, "commit=Login", 32);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<req_login>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  lea    D16(%rip),%rsi
  call   <sscanf@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    $0x1,%edx
  je     L14
L1:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x38,%rsp
  ret
L14:
  mov    %rsp,%rdi
  mov    $0x20,%edx
  lea    D19(%rip),%rsi
  call   <strncmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L1
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D16
datatype: string
name: disassemble_data
arguments: tag: D19
datatype: string

tool: D16: %*[^=]%*[^&]%*[^=]%*[^&]&%s

tool: D19: commit=Login

assistant: ```c
int req_login(char *data) {
  int c;
  char cl[32];
  c = sscanf(data, "%*[^=]%*[^&]%*[^=]%*[^&]&%s", cl);
  return c == 1 && !strncmp(cl, "commit=Login", 32);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<req_login>:
  endbr64
  push   %rbp
  lea    D15(%rip),%rsi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdx
  call   <sscanf@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    $0x1,%r8d
  je     L24
L1:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x30,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax)
L24:
  mov    $0x20,%edx
  lea    D3(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L1
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D15
type_guessed: string
name: unpack_data
arguments: data_label: D3
type_guessed: string

tool: D15: %*[^=]%*[^&]%*[^=]%*[^&]&%s

tool: D3: commit=Login

assistant: ```c
int req_login(char *data) {
  int c;
  char cl[32];
  c = sscanf(data, "%*[^=]%*[^&]%*[^=]%*[^&]&%s", cl);
  return c == 1 && !strncmp(cl, "commit=Login", 32);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<req_login>:
  endbr64
  push   %rbp
  lea    D16(%rip),%rsi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdx
  call   <sscanf@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    $0x1,%r8d
  je     L19
L1:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x30,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax)
L19:
  mov    $0x20,%edx
  lea    D10(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L1
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D16
datatype: string
name: retrieve_data
arguments: data_tag: D10
datatype: string

tool: D16: %*[^=]%*[^&]%*[^=]%*[^&]&%s

tool: D10: commit=Login

assistant: ```c
int req_login(char *data) {
  int c;
  char cl[32];
  c = sscanf(data, "%*[^=]%*[^&]%*[^=]%*[^&]&%s", cl);
  return c == 1 && !strncmp(cl, "commit=Login", 32);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bot_response>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    $0x20,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x20,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    -0x38(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  jne    L65
  mov    $0xffffffff,%eax
  jmp    L6
L65:
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x1,%eax
  cltq
  movb   $0x0,-0x30(%rbp,%rax,1)
  lea    -0x30(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L45
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L60
L45:
  mov    $0xfffffffe,%eax
  jmp    L6
L60:
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    -0x38(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jg     L68
  mov    $0x1,%eax
  jmp    L6
L68:
  mov    $0x0,%eax
L6:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D6
type: string

tool: D6: exit

assistant: ```c
int bot_response(int *res) {
  char buffer[32];
  bzero(buffer, 32);
  (*res) = read(0, buffer, 32);
  if ((*res) == -1)
    return (-1);
  buffer[(*res) - 1] = '\0';
  if (!strcmp(buffer, "exit") || (*res) == 0)
    return (-2);
  (*res) = atoi(buffer);
  if ((*res) <= 0)
    return (1);
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bot_response>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x20,%esi
  mov    %rbx,%rdi
  call   <bzero@plt>
  mov    $0x20,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %eax,%ebx
  mov    %eax,0x0(%rbp)
  cmp    $0xffffffff,%eax
  je     L9
  lea    -0x1(%rax),%eax
  cltq
  movb   $0x0,(%rsp,%rax,1)
  mov    %rsp,%rdi
  lea    D0(%rip),%rsi
  call   <strcmp@plt>
  test   %ebx,%ebx
  je     L29
  test   %eax,%eax
  je     L29
  mov    %rsp,%rdi
  call   <atoi@plt>
  mov    %eax,0x0(%rbp)
  test   %eax,%eax
  setle  %bl
  movzbl %bl,%ebx
L9:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  mov    %ebx,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L29:
  mov    $0xfffffffe,%ebx
  jmp    L9
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D0
data_type: string

tool: D0: exit

assistant: ```c
int bot_response(int *res) {
  char buffer[32];
  bzero(buffer, 32);
  (*res) = read(0, buffer, 32);
  if ((*res) == -1)
    return (-1);
  buffer[(*res) - 1] = '\0';
  if (!strcmp(buffer, "exit") || (*res) == 0)
    return (-2);
  (*res) = atoi(buffer);
  if ((*res) <= 0)
    return (1);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bot_response>:
  endbr64
  push   %rbp
  mov    $0x20,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <bzero@plt>
  mov    $0x20,%edx
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <read@plt>
  mov    %eax,(%rbx)
  cmp    $0xffffffff,%eax
  je     L23
  lea    -0x1(%rax),%edx
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  cmpl   $0x74697865,(%rsp)
  je     L42
L24:
  mov    $0x1,%edx
L51:
  test   %eax,%eax
  je     L14
  test   %edx,%edx
  je     L14
  mov    %rbp,%rdi
  call   <atoi@plt>
  test   %eax,%eax
  mov    %eax,(%rbx)
  setle  %al
  movzbl %al,%eax
L23:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L42:
  xor    %edx,%edx
  cmpb   $0x0,0x4(%rsp)
  je     L51
  jmp    L24
  nopl   0x0(%rax,%rax,1)
L14:
  mov    $0xfffffffe,%eax
  jmp    L23
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bot_response(int *res) {
  char buffer[32];
  bzero(buffer, 32);
  (*res) = read(0, buffer, 32);
  if ((*res) == -1)
    return (-1);
  buffer[(*res) - 1] = '\0';
  if (!strcmp(buffer, "exit") || (*res) == 0)
    return (-2);
  (*res) = atoi(buffer);
  if ((*res) <= 0)
    return (1);
  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
<bot_response>:
  endbr64
  push   %rbp
  mov    $0x20,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <bzero@plt>
  mov    $0x20,%edx
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <read@plt>
  mov    %eax,(%rbx)
  cmp    $0xffffffff,%eax
  je     L27
  lea    -0x1(%rax),%edx
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  cmpl   $0x74697865,(%rsp)
  je     L33
L58:
  mov    $0x1,%edx
L68:
  test   %eax,%eax
  je     L56
  test   %edx,%edx
  je     L56
  mov    %rbp,%rdi
  call   <atoi@plt>
  test   %eax,%eax
  mov    %eax,(%rbx)
  setle  %al
  movzbl %al,%eax
L27:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L60
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L33:
  xor    %edx,%edx
  cmpb   $0x0,0x4(%rsp)
  je     L68
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L56:
  mov    $0xfffffffe,%eax
  jmp    L27
L60:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bot_response(int *res) {
  char buffer[32];
  bzero(buffer, 32);
  (*res) = read(0, buffer, 32);
  if ((*res) == -1)
    return (-1);
  buffer[(*res) - 1] = '\0';
  if (!strcmp(buffer, "exit") || (*res) == 0)
    return (-2);
  (*res) = atoi(buffer);
  if ((*res) <= 0)
    return (1);
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<computeChecksum>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbp
  and    $0x3fffffc0,%ebp
  mov    %rdi,%r14
  shr    $0x1f,%r14
  and    $0x1,%r14d
  mov    %rdi,%r13
  shr    $0x1e,%r13
  and    $0x1,%r13d
  test   %esi,%esi
  jne    L7
L18:
  mov    %rbp,%r15
  xor    $0x3fffffc0,%r15
  test   %r13,%r13
  cmove  %rbp,%r15
  mov    %rbp,%rdi
  and    $0x3b1f3480,%edi
  call   <countBits@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  and    $0x1d8f9a40,%edi
  call   <countBits@plt>
  mov    %rax,%r12
  mov    %rbp,%rdi
  and    $0x2ec7cd00,%edi
  call   <countBits@plt>
  mov    %rax,0x8(%rsp)
  mov    %rbp,%rdi
  and    $0x1763e680,%edi
  call   <countBits@plt>
  mov    %rax,0x10(%rsp)
  mov    %rbp,%rdi
  and    $0x2bb1f340,%edi
  call   <countBits@plt>
  mov    %rax,0x18(%rsp)
  mov    %rbp,%rdi
  and    $0xb7a89c0,%edi
  call   <countBits@plt>
  add    %r14,%rbx
  shl    $0x5,%rbx
  and    $0x20,%ebx
  add    %r14,%rax
  and    $0x1,%eax
  or     %r15,%rax
  or     %rax,%rbx
  add    %r13,%r12
  shl    $0x4,%r12
  and    $0x10,%r12d
  or     %r12,%rbx
  mov    0x8(%rsp),%rcx
  add    %r14,%rcx
  mov    %rcx,%rax
  shl    $0x3,%rax
  and    $0x8,%eax
  or     %rax,%rbx
  mov    0x10(%rsp),%rax
  add    %r13,%rax
  shl    $0x2,%rax
  and    $0x4,%eax
  or     %rax,%rbx
  mov    0x18(%rsp),%rax
  add    %r13,%rax
  add    %rax,%rax
  and    $0x2,%eax
  or     %rbx,%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  and    $0x2bb1f340,%edi
  call   <countBits@plt>
  add    %r13,%rax
  mov    %rbp,%rdx
  xor    $0x40,%rdx
  test   $0x1,%al
  cmovne %rdx,%rbp
  mov    %rbp,%rdi
  and    $0xb7a89c0,%edi
  call   <countBits@plt>
  add    %r14,%rax
  mov    %rbp,%rdx
  xor    $0x80,%dl
  test   $0x1,%al
  cmovne %rdx,%rbp
  jmp    L18
```

assistant: ```c
unsigned long computeChecksum(unsigned long source, int nib) {
  unsigned long bmask[6] = {0x3B1F3480UL, 0x1D8F9A40UL, 0x2EC7CD00UL, 0x1763E680UL, 0x2BB1F340UL, 0x0B7A89C0UL};
  unsigned long D;
  unsigned long d = source & 0x3FFFFFC0UL;
  unsigned long D29 = (source >> 31) & 0x1UL;
  unsigned long D30 = (source >> 30) & 0x1UL;
  if (nib) {
    if ((D30 + countBits(bmask[4] & d)) % 2)
      d ^= (0x1UL << 6);
    if ((D29 + countBits(bmask[5] & d)) % 2)
      d ^= (0x1UL << 7);
  }
  D = d;
  if (D30)
    D ^= 0x3FFFFFC0UL;
  D |= ((D29 + countBits(bmask[0] & d)) % 2) << 5;
  D |= ((D30 + countBits(bmask[1] & d)) % 2) << 4;
  D |= ((D29 + countBits(bmask[2] & d)) % 2) << 3;
  D |= ((D30 + countBits(bmask[3] & d)) % 2) << 2;
  D |= ((D30 + countBits(bmask[4] & d)) % 2) << 1;
  D |= ((D29 + countBits(bmask[5] & d)) % 2);
  D &= 0x3FFFFFFFUL;
  return (D);
}
```
user: What is the c source code of the assembly code below?

```asm
<computeChecksum>:
  endbr64
  push   %r15
  mov    %rdi,%rcx
  mov    %rdi,%rdx
  push   %r14
  shr    $0x1f,%rcx
  shr    $0x1e,%rdx
  push   %r13
  and    $0x1,%ecx
  and    $0x1,%edx
  push   %r12
  mov    %rcx,%r13
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  and    $0x3fffffc0,%ebp
  sub    $0x28,%rsp
  test   %esi,%esi
  jne    L26
  and    $0xb7a89c0,%edi
  mov    %rdi,%r12
L17:
  mov    %rbp,%r15
  mov    %rbp,%rdi
  mov    %rdx,0x18(%rsp)
  xor    $0x3fffffc0,%r15
  test   %rdx,%rdx
  cmove  %rbp,%r15
  and    $0x3b1f3480,%edi
  call   <countBits@plt>
  mov    %rbp,%rdi
  and    $0x1d8f9a40,%edi
  mov    %rax,%rbx
  call   <countBits@plt>
  mov    %rbp,%rdi
  add    %r13,%rbx
  and    $0x2ec7cd00,%edi
  mov    %rax,%r14
  shl    $0x5,%rbx
  call   <countBits@plt>
  mov    %rbp,%rdi
  and    $0x20,%ebx
  and    $0x1763e680,%edi
  mov    %rax,0x8(%rsp)
  call   <countBits@plt>
  mov    %rbp,%rdi
  and    $0x2bb1f340,%edi
  mov    %rax,0x10(%rsp)
  call   <countBits@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <countBits@plt>
  mov    0x18(%rsp),%rdx
  mov    0x10(%rsp),%r12
  add    %r13,%rax
  add    0x8(%rsp),%r13
  and    $0x1,%eax
  add    %rdx,%r14
  shl    $0x3,%r13
  add    %rdx,%r12
  shl    $0x4,%r14
  or     %r15,%rax
  and    $0x8,%r13d
  shl    $0x2,%r12
  or     %rax,%rbx
  and    $0x10,%r14d
  lea    0x0(%rbp,%rdx,1),%rax
  and    $0x4,%r12d
  or     %r14,%rbx
  add    %rax,%rax
  or     %r13,%rbx
  and    $0x2,%eax
  or     %r12,%rbx
  or     %rbx,%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  and    $0x2bb1f340,%edi
  mov    %rdx,0x8(%rsp)
  call   <countBits@plt>
  mov    0x8(%rsp),%rdx
  mov    %rbp,%rsi
  xor    $0x40,%rsi
  add    %rdx,%rax
  test   $0x1,%al
  cmovne %rsi,%rbp
  mov    %rbp,%r12
  and    $0xb7a89c0,%r12d
  mov    %r12,%rdi
  call   <countBits@plt>
  mov    0x8(%rsp),%rdx
  add    %r13,%rax
  test   $0x1,%al
  je     L17
  xor    $0x80,%bpl
  mov    %rbp,%r12
  and    $0xb7a89c0,%r12d
  jmp    L17
```

assistant: ```c
unsigned long computeChecksum(unsigned long source, int nib) {
  unsigned long bmask[6] = {0x3B1F3480UL, 0x1D8F9A40UL, 0x2EC7CD00UL, 0x1763E680UL, 0x2BB1F340UL, 0x0B7A89C0UL};
  unsigned long D;
  unsigned long d = source & 0x3FFFFFC0UL;
  unsigned long D29 = (source >> 31) & 0x1UL;
  unsigned long D30 = (source >> 30) & 0x1UL;
  if (nib) {
    if ((D30 + countBits(bmask[4] & d)) % 2)
      d ^= (0x1UL << 6);
    if ((D29 + countBits(bmask[5] & d)) % 2)
      d ^= (0x1UL << 7);
  }
  D = d;
  if (D30)
    D ^= 0x3FFFFFC0UL;
  D |= ((D29 + countBits(bmask[0] & d)) % 2) << 5;
  D |= ((D30 + countBits(bmask[1] & d)) % 2) << 4;
  D |= ((D29 + countBits(bmask[2] & d)) % 2) << 3;
  D |= ((D30 + countBits(bmask[3] & d)) % 2) << 2;
  D |= ((D30 + countBits(bmask[4] & d)) % 2) << 1;
  D |= ((D29 + countBits(bmask[5] & d)) % 2);
  D &= 0x3FFFFFFFUL;
  return (D);
}
```