ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D12(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  add    (%rsp),%esi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D12
type: string
name: parse_data
arguments: label: D11
type: string

tool: D12: %d

tool: D11: pause

assistant: ```c
int main() {
  int num, numero;
  scanf("%d", &num);
  scanf("%d", &numero);
  printf("%d", num + numero);
  system("pause");
}
```
user: What is the c source code of the assembly code below?

```asm
<print_dec_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdx
  lea    D8(%rip),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    D3(%rip),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
datatype: string

tool: D8: %lu

assistant: ```c
char *print_dec_1(val)
unsigned long val;
{
  static char vbuf[32];
  sprintf(vbuf, "%lu", val);
  return vbuf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<print_dec_1>:
  endbr64
  push   %rbx
  mov    %rdi,%rdx
  lea    D7(%rip),%rsi
  lea    D6(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

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

tool: D7: %lu

assistant: ```c
char *print_dec_1(val)
unsigned long val;
{
  static char vbuf[32];
  sprintf(vbuf, "%lu", val);
  return vbuf;
}
```
user: What is the c source code of the assembly code below:

```asm
<print_dec_1>:
  endbr64
  push   %r12
  lea    D3(%rip),%r12
  mov    %rdi,%rdx
  xor    %eax,%eax
  mov    %r12,%rdi
  lea    D1(%rip),%rsi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D1
data_type: string

tool: D1: %lu

assistant: ```c
char *print_dec_1(val)
unsigned long val;
{
  static char vbuf[32];
  sprintf(vbuf, "%lu", val);
  return vbuf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_dec_1>:
  endbr64
  push   %r12
  lea    D9(%rip),%r12
  mov    %rdi,%rdx
  xor    %eax,%eax
  mov    %r12,%rdi
  lea    D1(%rip),%rsi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type_infered: string

tool: D1: %lu

assistant: ```c
char *print_dec_1(val)
unsigned long val;
{
  static char vbuf[32];
  sprintf(vbuf, "%lu", val);
  return vbuf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<recvall>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L27
L54:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L56
  mov    $0x0,%eax
  jmp    L34
L56:
  cmpl   $0x0,-0x4(%rbp)
  jne    L26
  mov    $0x0,%eax
  jmp    L34
L26:
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x8(%rbp)
L27:
  mov    -0xc(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L54
  mov    -0xc(%rbp),%eax
L34:
  leave
  ret
```

assistant: ```c
int recvall(int sockfd, void *buf, int len) {
  int total = 0;
  int bytesleft = len;
  int n;
  while (total < len) {
    if ((n = read(sockfd, buf + total, bytesleft)) == -1) {
      return 0;
    }
    if (n == 0)
      return 0;
    total += n;
    bytesleft -= n;
  }
  return total;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<recvall>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edx,%edx
  jle    L4
  mov    %edi,%r13d
  mov    %rsi,%r14
  mov    %edx,%r12d
  mov    %edx,%ebp
  mov    $0x0,%ebx
L26:
  movslq %ebx,%rsi
  add    %r14,%rsi
  mov    %ebp,%edx
  mov    %r13d,%edi
  call   <read@plt>
  lea    0x1(%rax),%edx
  cmp    $0x1,%edx
  jbe    L7
  add    %eax,%ebx
  sub    %eax,%ebp
  cmp    %ebx,%r12d
  jg     L26
  jmp    L14
L4:
  mov    $0x0,%ebx
  jmp    L14
L7:
  mov    $0x0,%ebx
L14:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int recvall(int sockfd, void *buf, int len) {
  int total = 0;
  int bytesleft = len;
  int n;
  while (total < len) {
    if ((n = read(sockfd, buf + total, bytesleft)) == -1) {
      return 0;
    }
    if (n == 0)
      return 0;
    total += n;
    bytesleft -= n;
  }
  return total;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<recvall>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edx,%edx
  jle    L4
  mov    %edi,%ebp
  mov    %rsi,%r13
  mov    %edx,%ebx
  mov    %edx,%r14d
  xor    %r12d,%r12d
  jmp    L2
  nop
L31:
  add    %eax,%r12d
  sub    %eax,%r14d
  cmp    %r12d,%ebx
  jle    L18
L2:
  movslq %r12d,%rsi
  mov    %r14d,%edx
  mov    %ebp,%edi
  add    %r13,%rsi
  call   <read@plt>
  lea    0x1(%rax),%edx
  cmp    $0x1,%edx
  ja     L31
L4:
  xor    %r12d,%r12d
L18:
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int recvall(int sockfd, void *buf, int len) {
  int total = 0;
  int bytesleft = len;
  int n;
  while (total < len) {
    if ((n = read(sockfd, buf + total, bytesleft)) == -1) {
      return 0;
    }
    if (n == 0)
      return 0;
    total += n;
    bytesleft -= n;
  }
  return total;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<recvall>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edx,%edx
  jle    L6
  mov    %edi,%ebp
  mov    %rsi,%r13
  mov    %edx,%ebx
  mov    %edx,%r14d
  xor    %r12d,%r12d
  jmp    L16
  nop
L19:
  add    %eax,%r12d
  sub    %eax,%r14d
  cmp    %r12d,%ebx
  jle    L25
L16:
  movslq %r12d,%rsi
  mov    %r14d,%edx
  mov    %ebp,%edi
  add    %r13,%rsi
  call   <read@plt>
  lea    0x1(%rax),%edx
  cmp    $0x1,%edx
  ja     L19
L6:
  xor    %r12d,%r12d
L25:
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int recvall(int sockfd, void *buf, int len) {
  int total = 0;
  int bytesleft = len;
  int n;
  while (total < len) {
    if ((n = read(sockfd, buf + total, bytesleft)) == -1) {
      return 0;
    }
    if (n == 0)
      return 0;
    total += n;
    bytesleft -= n;
  }
  return total;
}
```
user: What is the c source code of the assembly code below?

```asm
<skipone>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L50:
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L50
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L50
  jmp    L97
L52:
  addl   $0x1,-0x4(%rbp)
L97:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L26
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L26
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L26
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L52
  jmp    L26
L55:
  addl   $0x1,-0x4(%rbp)
L26:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L55
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L55
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int skipone(char *line) {
  int i = 0;
  while (line[i] == ' ' || line[i] == '\t')
    i++;
  while (line[i] != ' ' && line[i] != '\t' && line[i] != '\n' && line[i] != '\0')
    i++;
  while (line[i] == ' ' || line[i] == '\t')
    i++;
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<skipone>:
  endbr64
  mov    %rdi,%rsi
  movzbl (%rdi),%eax
  cmp    $0x20,%al
  je     L126
  cmp    $0x9,%al
  jne    L138
L126:
  mov    $0x1,%eax
L10:
  mov    %eax,%ecx
  movzbl (%rsi,%rax,1),%edx
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L10
  cmp    $0x9,%dl
  je     L10
  movslq %ecx,%rax
  movzbl (%rsi,%rax,1),%eax
  lea    -0x9(%rax),%edx
  cmp    $0x1,%dl
  jbe    L5
  test   $0xdf,%al
  je     L5
L87:
  add    $0x1,%ecx
  movslq %ecx,%rdx
L102:
  mov    %edx,%ecx
  movzbl (%rsi,%rdx,1),%eax
  add    $0x1,%rdx
  test   $0xdf,%al
  je     L5
  sub    $0x9,%eax
  cmp    $0x1,%al
  ja     L102
L5:
  movslq %ecx,%rax
  movzbl (%rsi,%rax,1),%eax
  cmp    $0x20,%al
  je     L139
  cmp    $0x9,%al
  jne    L104
L139:
  add    $0x1,%ecx
  movslq %ecx,%rax
L96:
  mov    %eax,%ecx
  movzbl (%rsi,%rax,1),%edx
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L96
  cmp    $0x9,%dl
  je     L96
L104:
  mov    %ecx,%eax
  ret
L138:
  movzbl (%rdi),%eax
  lea    -0x9(%rax),%edx
  cmp    $0x1,%dl
  jbe    L71
  mov    $0x0,%ecx
  test   $0xdf,%al
  jne    L87
L71:
  mov    $0x0,%ecx
  jmp    L104
```

assistant: ```c
int skipone(char *line) {
  int i = 0;
  while (line[i] == ' ' || line[i] == '\t')
    i++;
  while (line[i] != ' ' && line[i] != '\t' && line[i] != '\n' && line[i] != '\0')
    i++;
  while (line[i] == ' ' || line[i] == '\t')
    i++;
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<skipone>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  je     L7
  cmp    $0x9,%dl
  jne    L120
L7:
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L131:
  movzbl (%rdi,%rax,1),%edx
  mov    %eax,%r8d
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L131
  cmp    $0x9,%dl
  je     L131
L39:
  lea    -0x9(%rdx),%eax
  cmp    $0x1,%al
  jbe    L98
  and    $0xdf,%edx
  lea    0x1(%r8),%eax
  setne  %dl
  cltq
  test   %dl,%dl
  jne    L78
  jmp    L98
  nopl   0x0(%rax)
L34:
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  jbe    L44
L78:
  movzbl (%rdi,%rax,1),%edx
  mov    %eax,%r8d
  add    $0x1,%rax
  test   $0xdf,%dl
  jne    L34
L44:
  cmp    $0x20,%dl
  je     L1
  cmp    $0x9,%dl
  jne    L98
L1:
  add    $0x1,%r8d
  movslq %r8d,%rax
  nopl   0x0(%rax)
L106:
  movzbl (%rdi,%rax,1),%edx
  mov    %eax,%r8d
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L106
  cmp    $0x9,%dl
  je     L106
L98:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L120:
  xor    %r8d,%r8d
  jmp    L39
```

assistant: ```c
int skipone(char *line) {
  int i = 0;
  while (line[i] == ' ' || line[i] == '\t')
    i++;
  while (line[i] != ' ' && line[i] != '\t' && line[i] != '\n' && line[i] != '\0')
    i++;
  while (line[i] == ' ' || line[i] == '\t')
    i++;
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<skipone>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  je     L20
  cmp    $0x9,%dl
  jne    L71
L20:
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  movzbl (%rdi,%rax,1),%edx
  mov    %eax,%r8d
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L17
  cmp    $0x9,%dl
  je     L17
L120:
  lea    -0x9(%rdx),%eax
  cmp    $0x1,%al
  jbe    L131
  and    $0xdf,%edx
  je     L131
  add    $0x1,%r8d
  movslq %r8d,%rax
  jmp    L128
  nopl   (%rax)
L78:
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  jbe    L25
L128:
  movzbl (%rdi,%rax,1),%edx
  mov    %eax,%r8d
  add    $0x1,%rax
  test   $0xdf,%dl
  jne    L78
L25:
  cmp    $0x20,%dl
  je     L52
  cmp    $0x9,%dl
  jne    L131
L52:
  add    $0x1,%r8d
  movslq %r8d,%rax
  nopl   0x0(%rax)
L7:
  movzbl (%rdi,%rax,1),%edx
  mov    %eax,%r8d
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L7
  cmp    $0x9,%dl
  je     L7
L131:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L71:
  xor    %r8d,%r8d
  jmp    L120
```

assistant: ```c
int skipone(char *line) {
  int i = 0;
  while (line[i] == ' ' || line[i] == '\t')
    i++;
  while (line[i] != ' ' && line[i] != '\t' && line[i] != '\n' && line[i] != '\0')
    i++;
  while (line[i] == ' ' || line[i] == '\t')
    i++;
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<putstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L14
L11:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchr@plt>
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L11
  nop
  nop
  leave
  ret
```

assistant: ```c
void putstr(char *s) {
  while (*s)
    putchr(*s++);
}
```
user: What is the c source code of the assembly code below?

```asm
<putstr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L18
L3:
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <putchr@plt>
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L3
L18:
  pop    %rbx
  ret
```

assistant: ```c
void putstr(char *s) {
  while (*s)
    putchr(*s++);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<putstr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L7
L13:
  call   <putchr@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L13
L7:
  pop    %rbx
  ret
```

assistant: ```c
void putstr(char *s) {
  while (*s)
    putchr(*s++);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<putstr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L11
L4:
  call   <putchr@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L4
L11:
  pop    %rbx
  ret
```

assistant: ```c
void putstr(char *s) {
  while (*s)
    putchr(*s++);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<GetcFile>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <FindFptr@plt>
  mov    %rax,-0x8(%rbp)
  mov    D39(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  jne    L16
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <gengetchar@plt>
  mov    %eax,-0xc(%rbp)
  jmp    L14
L16:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,-0xc(%rbp)
L14:
  mov    D39(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  jne    L20
  mov    D36(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L20
  mov    D39(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <clearerr@plt>
L20:
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int GetcFile(void *theEnv, const char *logicalName) {
  FILE *fptr;
  int theChar;
  fptr = FindFptr(theEnv, logicalName);
  if (fptr == stdin) {
    theChar = gengetchar(theEnv);
  } else {
    theChar = getc(fptr);
  }
  if ((fptr == stdin) && (theChar == EOF))
    clearerr(stdin);
  return (theChar);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GetcFile>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <FindFptr@plt>
  mov    %rax,%rbx
  mov    D3(%rip),%rax
  cmp    %rbx,(%rax)
  je     L37
  mov    %rbx,%rdi
  call   <getc@plt>
  mov    %eax,%ebp
L6:
  mov    D3(%rip),%rax
  cmp    %rbx,(%rax)
  je     L43
L46:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L37:
  mov    %rbp,%rdi
  call   <gengetchar@plt>
  mov    %eax,%ebp
  jmp    L6
L43:
  mov    D0(%rip),%rax
  cmp    %ebp,(%rax)
  jne    L46
  mov    %rbx,%rdi
  call   <clearerr@plt>
  jmp    L46
```

assistant: ```c
int GetcFile(void *theEnv, const char *logicalName) {
  FILE *fptr;
  int theChar;
  fptr = FindFptr(theEnv, logicalName);
  if (fptr == stdin) {
    theChar = gengetchar(theEnv);
  } else {
    theChar = getc(fptr);
  }
  if ((fptr == stdin) && (theChar == EOF))
    clearerr(stdin);
  return (theChar);
}
```
user: What is the c source code of the assembly code below:

```asm
<GetcFile>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <FindFptr@plt>
  mov    D18(%rip),%rbx
  mov    %rax,%rbp
  cmp    %rax,(%rbx)
  je     L3
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,%r12d
  cmp    %rbp,(%rbx)
  je     L26
L19:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %r12,%rdi
  call   <gengetchar@plt>
  mov    %eax,%r12d
  cmp    %rbp,(%rbx)
  jne    L19
L26:
  mov    D14(%rip),%rax
  cmp    %r12d,(%rax)
  jne    L19
  mov    %rbp,%rdi
  call   <clearerr@plt>
  jmp    L19
```

assistant: ```c
int GetcFile(void *theEnv, const char *logicalName) {
  FILE *fptr;
  int theChar;
  fptr = FindFptr(theEnv, logicalName);
  if (fptr == stdin) {
    theChar = gengetchar(theEnv);
  } else {
    theChar = getc(fptr);
  }
  if ((fptr == stdin) && (theChar == EOF))
    clearerr(stdin);
  return (theChar);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<GetcFile>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <FindFptr@plt>
  mov    D1(%rip),%rbx
  mov    %rax,%rbp
  cmp    %rax,(%rbx)
  je     L44
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,%r12d
  cmp    %rbp,(%rbx)
  je     L13
L6:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L44:
  mov    %r12,%rdi
  call   <gengetchar@plt>
  mov    %eax,%r12d
  cmp    %rbp,(%rbx)
  jne    L6
L13:
  mov    D17(%rip),%rax
  cmp    %r12d,(%rax)
  jne    L6
  mov    %rbp,%rdi
  call   <clearerr@plt>
  jmp    L6
```

assistant: ```c
int GetcFile(void *theEnv, const char *logicalName) {
  FILE *fptr;
  int theChar;
  fptr = FindFptr(theEnv, logicalName);
  if (fptr == stdin) {
    theChar = gengetchar(theEnv);
  } else {
    theChar = getc(fptr);
  }
  if ((fptr == stdin) && (theChar == EOF))
    clearerr(stdin);
  return (theChar);
}
```
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
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  char *ptr;
  ptr = (char *)malloc(sizeof(char) * 10);
  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    $0xa,%edi
  call   <malloc@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  char *ptr;
  ptr = (char *)malloc(sizeof(char) * 10);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <malloc@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  char *ptr;
  ptr = (char *)malloc(sizeof(char) * 10);
  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    $0xa,%edi
  call   <malloc@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  char *ptr;
  ptr = (char *)malloc(sizeof(char) * 10);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<paw_bisection>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  movq   %xmm0,%r14
  movq   %xmm1,%rbx
  movq   %xmm2,%r12
  movapd %xmm1,%xmm0
  call   *%rdi
  movq   %xmm0,%r13
  movq   %r14,%xmm0
  call   *%rbp
  movsd  %xmm0,(%rsp)
  movq   %r13,%xmm5
  mulsd  %xmm0,%xmm5
  comisd D5(%rip),%xmm5
  jae    L13
L16:
  pxor   %xmm0,%xmm0
  comisd (%rsp),%xmm0
  ja     L54
  movq   %r14,%xmm5
  movq   %rbx,%xmm4
  subsd  %xmm4,%xmm5
  movsd  %xmm5,0x8(%rsp)
  mov    %rbx,(%rsp)
L24:
  mov    $0x190,%ebx
L12:
  movsd  D22(%rip),%xmm3
  mulsd  0x8(%rsp),%xmm3
  movsd  %xmm3,0x8(%rsp)
  movapd %xmm3,%xmm4
  addsd  (%rsp),%xmm4
  movapd %xmm4,%xmm0
  movq   %xmm4,%r14
  call   *%rbp
  movapd %xmm0,%xmm6
  pxor   %xmm7,%xmm7
  cmpnlesd %xmm7,%xmm6
  movapd %xmm6,%xmm1
  movsd  (%rsp),%xmm2
  andpd  %xmm6,%xmm2
  movq   %r14,%xmm6
  andnpd %xmm6,%xmm1
  movapd %xmm1,%xmm7
  orpd   %xmm2,%xmm7
  movsd  %xmm7,(%rsp)
  movsd  0x8(%rsp),%xmm1
  andpd  D36(%rip),%xmm1
  movq   %r12,%xmm4
  comisd %xmm1,%xmm4
  ja     L53
  ucomisd D5(%rip),%xmm0
  jp     L32
  je     L53
L32:
  sub    $0x1,%ebx
  jne    L12
  lea    D56(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L53:
  movsd  (%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  lea    D31(%rip),%rdi
  call   <printf@plt>
  mov    $0x63,%edi
  call   <exit@plt>
  jmp    L16
L54:
  movq   %rbx,%xmm5
  movq   %r14,%xmm6
  subsd  %xmm6,%xmm5
  movsd  %xmm5,0x8(%rsp)
  mov    %r14,(%rsp)
  jmp    L24
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type_infered: f64
name: unpack_data
arguments: data_tag: D22
type_infered: f64
name: unpack_data
arguments: data_tag: D36
type_infered: qword[2]
name: unpack_data
arguments: data_tag: D56
type_infered: string
name: unpack_data
arguments: data_tag: D31
type_infered: string

tool: D5: 0

tool: D22: 0.5

tool: {"D36": [0x7fffffffffffffff,0x0000000000000000]}

tool: D56: unable to find root within alotted number of iterations

tool: D31:  paw_bisection.c : root must be bracketed

assistant: ```c
double paw_bisection(double (*my_func)(double), double x1, double x2, double eps) {
  int j;
  int root_found;
  double dx, f, fmid, xmid, root;
  root_found = 0;
  fmid = my_func(x2);
  f = my_func(x1);
  if (f * fmid >= 0.0) {
    printf(" paw_bisection.c : root must be bracketed");
    exit(99);
  }
  if (f < 0.0) {
    root = x1;
    dx = x2 - x1;
  } else {
    root = x2;
    dx = x1 - x2;
  }
  for (j = 0; j < 400; ++j) {
    dx = 0.5 * dx;
    xmid = root + dx;
    fmid = my_func(xmid);
    if (fmid <= 0.0)
      root = xmid;
    if (fabs(dx) < eps || fmid == 0.0) {
      root_found = 1;
      break;
    }
  }
  if (!(root_found)) {
    printf("unable to find root within alotted number of iterations");
    exit(1);
  }
  return root;
}
```