ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readNewRecord>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x20,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %edx,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x1018(%rsp)
  xor    %eax,%eax
  mov    %rsi,%rdi
  call   <initRecord@plt>
  cmp    $0x1,%ebp
  je     L4
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  mov    %rsp,%rdi
  mov    %r12,%rdx
  call   <getline@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    $0xffffffff,%edx
  jne    L28
L42:
  mov    0x1018(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L32
  add    $0x1020,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  lea    0x10(%rsp),%rsi
  mov    $0xfff,%edx
  mov    %r12,%rdi
  call   <gzgets@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L42
  lea    0x10(%rsp),%rdi
  mov    %rbx,%rsi
  call   <fromLineToRecord@plt>
  mov    %ebp,%eax
  jmp    L42
L28:
  mov    %rbx,%rsi
  mov    (%rsp),%rdi
  call   <fromLineToRecord@plt>
  mov    $0x1,%eax
  jmp    L42
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readNewRecord(void *fileDescriptor, struct Record *record, int isZip) {
  initRecord(record);
  if (isZip == 1) {
    char *readGz;
    char buffer[0x1000];
    readGz = gzgets(fileDescriptor, buffer, 0x1000 - 1);
    if (readGz != 0) {
      fromLineToRecord(buffer, record);
      return 1;
    }
  } else {
    char *line = NULL;
    size_t len = 0;
    ssize_t read;
    read = getline(&line, &len, fileDescriptor);
    if (read != -1) {
      fromLineToRecord(line, record);
      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
<readNewRecord>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1018(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r12
  mov    %edx,%ebx
  mov    %rsi,%rdi
  mov    %rsi,%rbp
  call   <initRecord@plt>
  cmp    $0x1,%ebx
  je     L47
  lea    0x8(%rsp),%rsi
  mov    %rsp,%rdi
  mov    %r12,%rdx
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  call   <getline@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    $0xffffffff,%r8d
  jne    L30
L49:
  mov    0x1018(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x1028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L30:
  mov    (%rsp),%rdi
  mov    %rbp,%rsi
  call   <fromLineToRecord@plt>
  mov    $0x1,%eax
  jmp    L49
  nopl   0x0(%rax,%rax,1)
L47:
  lea    0x10(%rsp),%r13
  mov    $0xfff,%edx
  mov    %r12,%rdi
  mov    %r13,%rsi
  call   <gzgets@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  je     L49
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <fromLineToRecord@plt>
  mov    $0x1,%eax
  jmp    L49
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readNewRecord(void *fileDescriptor, struct Record *record, int isZip) {
  initRecord(record);
  if (isZip == 1) {
    char *readGz;
    char buffer[0x1000];
    readGz = gzgets(fileDescriptor, buffer, 0x1000 - 1);
    if (readGz != 0) {
      fromLineToRecord(buffer, record);
      return 1;
    }
  } else {
    char *line = NULL;
    size_t len = 0;
    ssize_t read;
    read = getline(&line, &len, fileDescriptor);
    if (read != -1) {
      fromLineToRecord(line, record);
      return 1;
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readNewRecord>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1018(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r12
  mov    %edx,%ebx
  mov    %rsi,%rdi
  mov    %rsi,%rbp
  call   <initRecord@plt>
  cmp    $0x1,%ebx
  je     L31
  lea    0x8(%rsp),%rsi
  mov    %rsp,%rdi
  mov    %r12,%rdx
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  call   <getline@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    $0xffffffff,%r8d
  jne    L3
L7:
  mov    0x1018(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x1028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L3:
  mov    (%rsp),%rdi
  mov    %rbp,%rsi
  call   <fromLineToRecord@plt>
  mov    $0x1,%eax
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L31:
  lea    0x10(%rsp),%r13
  mov    $0xfff,%edx
  mov    %r12,%rdi
  mov    %r13,%rsi
  call   <gzgets@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  je     L7
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <fromLineToRecord@plt>
  mov    $0x1,%eax
  jmp    L7
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readNewRecord(void *fileDescriptor, struct Record *record, int isZip) {
  initRecord(record);
  if (isZip == 1) {
    char *readGz;
    char buffer[0x1000];
    readGz = gzgets(fileDescriptor, buffer, 0x1000 - 1);
    if (readGz != 0) {
      fromLineToRecord(buffer, record);
      return 1;
    }
  } else {
    char *line = NULL;
    size_t len = 0;
    ssize_t read;
    read = getline(&line, &len, fileDescriptor);
    if (read != -1) {
      fromLineToRecord(line, record);
      return 1;
    }
  }
  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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x24(%rbp)
  lea    -0x24(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x5,-0x1c(%rbp)
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movl   $0x5,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int main(void) {
  int *p;
  {
    int x = 5;
    p = &x;
  }
  int y = 0;
  { int y = 5; }
  y;
  p = malloc(sizeof(int));
  *p = 5;
  free(p);
  int *q;
  {
    int x;
    q = &x;
  }
  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    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movl   $0x5,(%rax)
  call   <free@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int *p;
  {
    int x = 5;
    p = &x;
  }
  int y = 0;
  { int y = 5; }
  y;
  p = malloc(sizeof(int));
  *p = 5;
  free(p);
  int *q;
  {
    int x;
    q = &x;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <malloc@plt>
  movl   $0x5,(%rax)
  mov    %rax,%rdi
  call   <free@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int *p;
  {
    int x = 5;
    p = &x;
  }
  int y = 0;
  { int y = 5; }
  y;
  p = malloc(sizeof(int));
  *p = 5;
  free(p);
  int *q;
  {
    int x;
    q = &x;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <malloc@plt>
  movl   $0x5,(%rax)
  mov    %rax,%rdi
  call   <free@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int *p;
  {
    int x = 5;
    p = &x;
  }
  int y = 0;
  { int y = 5; }
  y;
  p = malloc(sizeof(int));
  *p = 5;
  free(p);
  int *q;
  {
    int x;
    q = &x;
  }
  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
  mov    $0x0,%eax
  call   <test_get_line@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  test_get_line();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test_get_line@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test_get_line();
  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
  xor    %eax,%eax
  call   <test_get_line@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

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

assistant: ```c
int main() {
  test_get_line();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_head_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x4,%eax
  jg     L52
  mov    $0x1,%eax
  jmp    L19
L52:
  nop
L27:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  je     L53
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L27
L53:
  addq   $0x1,-0x8(%rbp)
  nop
L33:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  je     L23
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L33
L23:
  mov    $0x0,%eax
L19:
  leave
  ret
```

assistant: ```c
int parse_head_line(const char *src, char *method, char *filepath) {
  if (strlen(src) < 5)
    return 1;
  while ((*method++ = *src++) && *src != ' ')
    ;
  src++;
  while ((*filepath++ = *src++) && *src != ' ')
    ;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<parse_head_line>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  cmp    $0x4,%edx
  jle    L39
L2:
  mov    %rbx,%rax
  add    $0x1,%rbx
  add    $0x1,%r12
  movzbl -0x1(%rbx),%edx
  mov    %dl,-0x1(%r12)
  test   %dl,%dl
  je     L17
  cmpb   $0x20,(%rbx)
  jne    L2
L17:
  add    $0x2,%rax
L41:
  add    $0x1,%rax
  add    $0x1,%rbp
  movzbl -0x1(%rax),%edx
  mov    %dl,-0x1(%rbp)
  test   %dl,%dl
  je     L46
  cmpb   $0x20,(%rax)
  jne    L41
  mov    $0x0,%eax
  jmp    L39
L46:
  mov    $0x0,%eax
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parse_head_line(const char *src, char *method, char *filepath) {
  if (strlen(src) < 5)
    return 1;
  while ((*method++ = *src++) && *src != ' ')
    ;
  src++;
  while ((*filepath++ = *src++) && *src != ' ')
    ;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<parse_head_line>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  cmp    $0x4,%r8d
  jle    L51
  movzbl (%rbx),%eax
  jmp    L58
  nopl   0x0(%rax)
L65:
  movzbl (%rbx),%eax
  cmp    $0x20,%al
  je     L28
L58:
  mov    %rbx,%rsi
  add    $0x1,%rbp
  add    $0x1,%rbx
  mov    %al,-0x1(%rbp)
  test   %al,%al
  jne    L65
L28:
  movzbl 0x2(%rsi),%eax
  xor    %ecx,%ecx
  jmp    L48
  nopl   0x0(%rax)
L60:
  add    $0x1,%rcx
  movzbl 0x2(%rsi,%rcx,1),%eax
  cmp    $0x20,%al
  je     L45
L48:
  mov    %al,(%r12,%rcx,1)
  test   %al,%al
  jne    L60
L45:
  xor    %eax,%eax
L51:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parse_head_line(const char *src, char *method, char *filepath) {
  if (strlen(src) < 5)
    return 1;
  while ((*method++ = *src++) && *src != ' ')
    ;
  src++;
  while ((*filepath++ = *src++) && *src != ' ')
    ;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<parse_head_line>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  cmp    $0x4,%r8d
  jg     L57
  jmp    L17
  cs nopw 0x0(%rax,%rax,1)
L4:
  cmpb   $0x20,(%rbx)
  je     L58
L57:
  mov    %rbx,%rax
  movzbl (%rbx),%edx
  add    $0x1,%rbx
  add    $0x1,%r12
  mov    %dl,-0x1(%r12)
  test   %dl,%dl
  jne    L4
L58:
  add    $0x2,%rax
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L44:
  cmpb   $0x20,(%rax)
  je     L33
L27:
  movzbl (%rax),%edx
  add    $0x1,%rax
  add    $0x1,%rbp
  mov    %dl,-0x1(%rbp)
  test   %dl,%dl
  jne    L44
L33:
  xor    %eax,%eax
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parse_head_line(const char *src, char *method, char *filepath) {
  if (strlen(src) < 5)
    return 1;
  while ((*method++ = *src++) && *src != ' ')
    ;
  src++;
  while ((*filepath++ = *src++) && *src != ' ')
    ;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sprintfstype>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  lea    D15(%rip),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    D9(%rip),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
type_guessed: string

tool: D15: %#x

assistant: ```c
const char *sprintfstype(magic)
int magic;
{
  static char buf[32];
  sprintf(buf, "%#x", magic);
  return 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
<sprintfstype>:
  endbr64
  push   %rbx
  mov    %edi,%edx
  lea    D6(%rip),%rsi
  lea    D4(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type: string

tool: D6: %#x

assistant: ```c
const char *sprintfstype(magic)
int magic;
{
  static char buf[32];
  sprintf(buf, "%#x", magic);
  return buf;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sprintfstype>:
  endbr64
  push   %r12
  lea    D5(%rip),%r12
  mov    %edi,%edx
  xor    %eax,%eax
  mov    %r12,%rdi
  lea    D13(%rip),%rsi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D13
data_type: string

tool: D13: %#x

assistant: ```c
const char *sprintfstype(magic)
int magic;
{
  static char buf[32];
  sprintf(buf, "%#x", magic);
  return buf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sprintfstype>:
  endbr64
  push   %r12
  lea    D7(%rip),%r12
  mov    %edi,%edx
  xor    %eax,%eax
  mov    %r12,%rdi
  lea    D11(%rip),%rsi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D11
data_type: string

tool: D11: %#x

assistant: ```c
const char *sprintfstype(magic)
int magic;
{
  static char buf[32];
  sprintf(buf, "%#x", magic);
  return buf;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  lea    D25(%rip),%rdi
  lea    D74(%rip),%r12
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    D47(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  pxor   %xmm4,%xmm4
  movsd  D33(%rip),%xmm3
  movss  D56(%rip),%xmm5
  comisd %xmm4,%xmm0
  ja     L59
  jmp    L10
  xchg   %ax,%ax
L47:
  add    $0x1,%ebx
  comisd %xmm4,%xmm0
  jbe    L10
L59:
  comisd %xmm3,%xmm0
  jbe    L10
  pxor   %xmm2,%xmm2
  movaps %xmm5,%xmm1
  cvtsi2ss %ebx,%xmm2
  divss  %xmm2,%xmm1
  cvtss2sd %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jb     L47
  subsd  %xmm1,%xmm0
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  movsd  %xmm0,(%rsp)
  call   <printf@plt>
  mov    D33(%rip),%rax
  movsd  (%rsp),%xmm0
  pxor   %xmm4,%xmm4
  movss  D56(%rip),%xmm5
  movq   %rax,%xmm3
  comisd %xmm3,%xmm0
  jbe    L47
  comisd %xmm4,%xmm0
  jbe    L47
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    D33(%rip),%rax
  movsd  (%rsp),%xmm0
  pxor   %xmm4,%xmm4
  movss  D56(%rip),%xmm5
  movq   %rax,%xmm3
  jmp    L47
  nopw   0x0(%rax,%rax,1)
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L82:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D25
datatype: string
name: unpack_data
arguments: data_tag: D74
datatype: string
name: unpack_data
arguments: data_tag: D1
datatype: string
name: unpack_data
arguments: data_tag: D47
datatype: string
name: unpack_data
arguments: data_tag: D33
datatype: byte[8]
name: unpack_data
arguments: data_tag: D56
datatype: byte[4]

tool: D25: Problem 11: Egyptian fraction sum with least operands of real number 

tool: D74: , 

tool: D1: 1/%d

tool: D47: %lf

tool: {"D33": [0x00,0x00,0x00,0xe0,0xe2,0x36,0x1a,0x3f]}

tool: {"D56": [0x00,0x00,0x80,0x3f]}

assistant: ```c
void main() {
  double real;
  printf("Problem 11: Egyptian fraction sum with least operands of real number \n");
  scanf("%lf", &real);
  int i = 2;
  double remover;
  while (real > 0 && real > 0.0001f) {
    remover = 1 / (float)i;
    if (remover <= real) {
      real -= remover;
      printf("1/%d", i);
      if (real > 0.0001f && real > 0)
        printf(", ");
    }
    i++;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  lea    D82(%rip),%rdi
  lea    D66(%rip),%r12
  push   %rbp
  lea    D38(%rip),%rbp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    D44(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  pxor   %xmm4,%xmm4
  movsd  D42(%rip),%xmm3
  movss  D15(%rip),%xmm5
  comisd %xmm4,%xmm0
  ja     L22
  jmp    L82
  xchg   %ax,%ax
L41:
  add    $0x1,%ebx
  comisd %xmm4,%xmm0
  jbe    L82
L22:
  comisd %xmm3,%xmm0
  jbe    L82
  pxor   %xmm2,%xmm2
  movaps %xmm5,%xmm1
  cvtsi2ss %ebx,%xmm2
  divss  %xmm2,%xmm1
  cvtss2sd %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jb     L41
  subsd  %xmm1,%xmm0
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  movsd  %xmm0,(%rsp)
  call   <printf@plt>
  mov    D42(%rip),%rax
  movsd  (%rsp),%xmm0
  pxor   %xmm4,%xmm4
  movss  D15(%rip),%xmm5
  movq   %rax,%xmm3
  comisd %xmm3,%xmm0
  jbe    L41
  comisd %xmm4,%xmm0
  jbe    L41
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    D42(%rip),%rax
  movsd  (%rsp),%xmm0
  pxor   %xmm4,%xmm4
  movss  D15(%rip),%xmm5
  movq   %rax,%xmm3
  jmp    L41
  nopw   0x0(%rax,%rax,1)
L82:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D82
type_infered: string
name: unpack_data
arguments: data_tag: D66
type_infered: string
name: unpack_data
arguments: data_tag: D38
type_infered: string
name: unpack_data
arguments: data_tag: D44
type_infered: string
name: unpack_data
arguments: data_tag: D42
type_infered: byte[8]
name: unpack_data
arguments: data_tag: D15
type_infered: byte[4]

tool: D82: Problem 11: Egyptian fraction sum with least operands of real number 

tool: D66: , 

tool: D38: 1/%d

tool: D44: %lf

tool: {"D42": [0x00,0x00,0x00,0xe0,0xe2,0x36,0x1a,0x3f]}

tool: {"D15": [0x00,0x00,0x80,0x3f]}

assistant: ```c
void main() {
  double real;
  printf("Problem 11: Egyptian fraction sum with least operands of real number \n");
  scanf("%lf", &real);
  int i = 2;
  double remover;
  while (real > 0 && real > 0.0001f) {
    remover = 1 / (float)i;
    if (remover <= real) {
      real -= remover;
      printf("1/%d", i);
      if (real > 0.0001f && real > 0)
        printf(", ");
    }
    i++;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<net_test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L6
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L6:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <nc_slave@plt>
  mov    $0x0,%eax
  call   <cyg_test_exit@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_guessed: string

tool: D0: Start Network Characterization - SLAVE


assistant: ```c
void net_test(int param) {
  if (param == 0) {
    printf("Start Network Characterization - SLAVE\n");
  }
  nc_slave(param);
  cyg_test_exit();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<net_test>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  je     L15
L10:
  mov    %ebx,%edi
  call   <nc_slave@plt>
  mov    $0x0,%eax
  call   <cyg_test_exit@plt>
  pop    %rbx
  ret
L15:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L10
```

tool_calls: name: read_data
arguments: data_tag: D4
data_type: string

tool: D4: Start Network Characterization - SLAVE


assistant: ```c
void net_test(int param) {
  if (param == 0) {
    printf("Start Network Characterization - SLAVE\n");
  }
  nc_slave(param);
  cyg_test_exit();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<net_test>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  test   %edi,%edi
  je     L3
  mov    %ebp,%edi
  call   <nc_slave@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1090 <cyg_test_exit@plt>
  nopw   0x0(%rax,%rax,1)
L3:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  call   <nc_slave@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1090 <cyg_test_exit@plt>
```

tool_calls: name: unpack_data
arguments: tag: D1
datatype: string

tool: D1: Start Network Characterization - SLAVE


assistant: ```c
void net_test(int param) {
  if (param == 0) {
    printf("Start Network Characterization - SLAVE\n");
  }
  nc_slave(param);
  cyg_test_exit();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<net_test>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  test   %edi,%edi
  je     L2
  mov    %ebp,%edi
  call   <nc_slave@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1090 <cyg_test_exit@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  call   <nc_slave@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1090 <cyg_test_exit@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
data_type: string

tool: D7: Start Network Characterization - SLAVE


assistant: ```c
void net_test(int param) {
  if (param == 0) {
    printf("Start Network Characterization - SLAVE\n");
  }
  nc_slave(param);
  cyg_test_exit();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bl2_long4_cvt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  sar    $0x8,%rax
  movzbl %al,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x10,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  sar    $0x10,%rax
  movzbl %al,%eax
  shl    $0x8,%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%rax
  sar    $0x18,%rax
  movzbl %al,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int bl2_long4_cvt(long val) {
  int val4;
  unsigned int res;
  res = ((val & 255) * 256) + ((val >> 8) & 255);
  res = (res << 16) + (((val >> 16) & 255) * 256) + ((val >> 24) & 255);
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bl2_long4_cvt>:
  endbr64
  mov    %rdi,%rdx
  movzbl %dil,%ecx
  shl    $0x8,%ecx
  movzbl %dh,%eax
  add    %eax,%ecx
  shl    $0x10,%ecx
  mov    %rdi,%rax
  sar    $0x10,%rax
  movzbl %al,%eax
  shl    $0x8,%eax
  shr    $0x18,%edx
  add    %edx,%eax
  add    %ecx,%eax
  ret
```

assistant: ```c
unsigned int bl2_long4_cvt(long val) {
  int val4;
  unsigned int res;
  res = ((val & 255) * 256) + ((val >> 8) & 255);
  res = (res << 16) + (((val >> 16) & 255) * 256) + ((val >> 24) & 255);
  return res;
}
```