ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<list_files_in_dir>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <opendir@plt>
  mov    %rax,%r12
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x1,%r14d
  lea    D9(%rip),%r13
  test   %r12,%r12
  jne    L12
  jmp    L47
L25:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
L12:
  mov    %r12,%rdi
  call   <readdir@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L44
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    $0x380,%eax
  jle    L25
  add    $0x1,%r14d
  mov    %r14d,%esi
  shl    $0xa,%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
  jmp    L25
L44:
  mov    %r12,%rdi
  call   <closedir@plt>
L47:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
type_infered: string

tool: D9: 


assistant: ```c
char *list_files_in_dir(char *dir_name) {
  DIR *d;
  struct dirent *dir;
  d = opendir(dir_name);
  char *buffer = malloc(sizeof(char) * 1024);
  int buffer_mul = 1;
  if (d) {
    while ((dir = readdir(d)) != NULL) {
      if ((strlen(buffer) + 128) > 1024) {
        buffer_mul++;
        buffer = (char *)realloc(buffer, sizeof(char) * buffer_mul * 1024);
      }
      strcat(buffer, dir->d_name);
      strcat(buffer, "\n");
    }
    closedir(d);
  }
  return buffer;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<list_files_in_dir>:
  endbr64
  push   %r14
  mov    $0x1,%r14d
  push   %r13
  lea    D8(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <opendir@plt>
  mov    $0x400,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r12,%r12
  jne    L22
  jmp    L11
  nopl   (%rax)
L30:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
L22:
  mov    %r12,%rdi
  call   <readdir@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L6
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    $0x380,%eax
  jle    L30
  add    $0x1,%r14d
  mov    %rbp,%rdi
  mov    %r14d,%esi
  shl    $0xa,%esi
  call   <realloc@plt>
  mov    %rax,%rbp
  jmp    L30
  nopl   0x0(%rax)
L6:
  mov    %r12,%rdi
  call   <closedir@plt>
L11:
  pop    %rbx
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

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

tool: D8: 


assistant: ```c
char *list_files_in_dir(char *dir_name) {
  DIR *d;
  struct dirent *dir;
  d = opendir(dir_name);
  char *buffer = malloc(sizeof(char) * 1024);
  int buffer_mul = 1;
  if (d) {
    while ((dir = readdir(d)) != NULL) {
      if ((strlen(buffer) + 128) > 1024) {
        buffer_mul++;
        buffer = (char *)realloc(buffer, sizeof(char) * buffer_mul * 1024);
      }
      strcat(buffer, dir->d_name);
      strcat(buffer, "\n");
    }
    closedir(d);
  }
  return buffer;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<list_files_in_dir>:
  endbr64
  push   %r14
  mov    $0x1,%r14d
  push   %r13
  lea    D2(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <opendir@plt>
  mov    $0x400,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r12,%r12
  jne    L42
  jmp    L24
  nopl   (%rax)
L34:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
L42:
  mov    %r12,%rdi
  call   <readdir@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L17
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    $0x380,%eax
  jle    L34
  add    $0x1,%r14d
  mov    %rbp,%rdi
  mov    %r14d,%esi
  shl    $0xa,%esi
  call   <realloc@plt>
  mov    %rax,%rbp
  jmp    L34
  nopl   0x0(%rax)
L17:
  mov    %r12,%rdi
  call   <closedir@plt>
L24:
  pop    %rbx
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: tag: D2
data_type: string

tool: D2: 


assistant: ```c
char *list_files_in_dir(char *dir_name) {
  DIR *d;
  struct dirent *dir;
  d = opendir(dir_name);
  char *buffer = malloc(sizeof(char) * 1024);
  int buffer_mul = 1;
  if (d) {
    while ((dir = readdir(d)) != NULL) {
      if ((strlen(buffer) + 128) > 1024) {
        buffer_mul++;
        buffer = (char *)realloc(buffer, sizeof(char) * buffer_mul * 1024);
      }
      strcat(buffer, dir->d_name);
      strcat(buffer, "\n");
    }
    closedir(d);
  }
  return buffer;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <tap_skip_all@plt>
  mov    $0x0,%eax
  call   <tap_finish@plt>
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
datatype: string

tool: D0: need to find a way to force readdir to return entries in unsorted order

assistant: ```c
int main(void) {
  tap_skip_all("need to find a way to force readdir to return entries in unsorted order");
  return tap_finish();
}
```
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
  lea    D3(%rip),%rdi
  call   <tap_skip_all@plt>
  mov    $0x0,%eax
  call   <tap_finish@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
type_guessed: string

tool: D3: need to find a way to force readdir to return entries in unsorted order

assistant: ```c
int main(void) {
  tap_skip_all("need to find a way to force readdir to return entries in unsorted order");
  return tap_finish();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <tap_skip_all@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <tap_finish@plt>
```

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

tool: D8: need to find a way to force readdir to return entries in unsorted order

assistant: ```c
int main(void) {
  tap_skip_all("need to find a way to force readdir to return entries in unsorted order");
  return tap_finish();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <tap_skip_all@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <tap_finish@plt>
```

tool_calls: name: read_data
arguments: tag: D0
datatype: string

tool: D0: need to find a way to force readdir to return entries in unsorted order

assistant: ```c
int main(void) {
  tap_skip_all("need to find a way to force readdir to return entries in unsorted order");
  return tap_finish();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<RneFromLLA>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm1
  movss  D55(%rip),%xmm0
  mulss  %xmm0,%xmm1
  movd   %xmm1,%eax
  movd   %eax,%xmm0
  call   <sinf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  movss  %xmm0,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  movss  D55(%rip),%xmm0
  mulss  %xmm0,%xmm1
  movd   %xmm1,%eax
  movd   %eax,%xmm0
  call   <sinf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm1
  movss  D55(%rip),%xmm0
  mulss  %xmm0,%xmm1
  movd   %xmm1,%eax
  movd   %eax,%xmm0
  call   <cosf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  movss  %xmm0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  movss  D55(%rip),%xmm0
  mulss  %xmm0,%xmm1
  movd   %xmm1,%eax
  movd   %eax,%xmm0
  call   <cosf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x10(%rbp),%xmm0
  movss  D10(%rip),%xmm1
  xorps  %xmm1,%xmm0
  mulss  -0x4(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movss  %xmm0,(%rax)
  movss  -0x10(%rbp),%xmm0
  movss  D10(%rip),%xmm1
  xorps  %xmm1,%xmm0
  mulss  -0xc(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movss  %xmm0,0x4(%rax)
  mov    -0x20(%rbp),%rax
  movss  -0x8(%rbp),%xmm0
  movss  %xmm0,0x8(%rax)
  mov    -0x20(%rbp),%rax
  add    $0xc,%rax
  movss  -0xc(%rbp),%xmm0
  movss  D10(%rip),%xmm1
  xorps  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0xc,%rax
  movss  -0x4(%rbp),%xmm0
  movss  %xmm0,0x4(%rax)
  mov    -0x20(%rbp),%rax
  add    $0xc,%rax
  pxor   %xmm0,%xmm0
  movss  %xmm0,0x8(%rax)
  movss  -0x8(%rbp),%xmm0
  movss  D10(%rip),%xmm1
  xorps  %xmm1,%xmm0
  mov    -0x20(%rbp),%rax
  add    $0x18,%rax
  mulss  -0x4(%rbp),%xmm0
  movss  %xmm0,(%rax)
  movss  -0x8(%rbp),%xmm0
  movss  D10(%rip),%xmm1
  xorps  %xmm1,%xmm0
  mov    -0x20(%rbp),%rax
  add    $0x18,%rax
  mulss  -0xc(%rbp),%xmm0
  movss  %xmm0,0x4(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x18,%rax
  movss  -0x10(%rbp),%xmm0
  movss  D10(%rip),%xmm1
  xorps  %xmm1,%xmm0
  movss  %xmm0,0x8(%rax)
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D55
type: byte[4]
name: unpack_data
arguments: data_label: D10
type: byte[4]

tool: {"D55": [0x35,0xfa,0x8e,0x3c]}

tool: {"D10": [0x00,0x00,0x00,0x80]}

assistant: ```c
void RneFromLLA(float LLA[3], float Rne[3][3]) {
  float sinLat, sinLon, cosLat, cosLon;
  sinLat = (float)sinf((3.14159265358979323846f / 180.0f) * LLA[0]);
  sinLon = (float)sinf((3.14159265358979323846f / 180.0f) * LLA[1]);
  cosLat = (float)cosf((3.14159265358979323846f / 180.0f) * LLA[0]);
  cosLon = (float)cosf((3.14159265358979323846f / 180.0f) * LLA[1]);
  Rne[0][0] = -sinLat * cosLon;
  Rne[0][1] = -sinLat * sinLon;
  Rne[0][2] = cosLat;
  Rne[1][0] = -sinLon;
  Rne[1][1] = cosLon;
  Rne[1][2] = 0;
  Rne[2][0] = -cosLat * cosLon;
  Rne[2][1] = -cosLat * sinLon;
  Rne[2][2] = -sinLat;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RneFromLLA>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbx
  movss  D48(%rip),%xmm0
  mulss  (%rdi),%xmm0
  call   <sinf@plt>
  mov    %rax,%r13
  movss  D48(%rip),%xmm0
  mulss  0x4(%r12),%xmm0
  call   <sinf@plt>
  pxor   %xmm4,%xmm4
  cvtsi2ss %rax,%xmm4
  movss  %xmm4,0xc(%rsp)
  movss  D48(%rip),%xmm0
  mulss  (%r12),%xmm0
  call   <cosf@plt>
  pxor   %xmm5,%xmm5
  cvtsi2ss %rax,%xmm5
  movd   %xmm5,%ebp
  movss  D48(%rip),%xmm0
  mulss  0x4(%r12),%xmm0
  call   <cosf@plt>
  pxor   %xmm2,%xmm2
  cvtsi2ss %rax,%xmm2
  pxor   %xmm1,%xmm1
  cvtsi2ss %r13,%xmm1
  movss  D47(%rip),%xmm0
  xorps  %xmm0,%xmm1
  movaps %xmm1,%xmm3
  mulss  %xmm2,%xmm3
  movss  %xmm3,(%rbx)
  movss  0xc(%rsp),%xmm4
  movaps %xmm4,%xmm3
  mulss  %xmm1,%xmm3
  movss  %xmm3,0x4(%rbx)
  mov    %ebp,0x8(%rbx)
  movaps %xmm4,%xmm3
  xorps  %xmm0,%xmm3
  movss  %xmm3,0xc(%rbx)
  movss  %xmm2,0x10(%rbx)
  movl   $0x0,0x14(%rbx)
  movd   %ebp,%xmm6
  xorps  %xmm6,%xmm0
  mulss  %xmm0,%xmm2
  movss  %xmm2,0x18(%rbx)
  mulss  %xmm4,%xmm0
  movss  %xmm0,0x1c(%rbx)
  movss  %xmm1,0x20(%rbx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D48
datatype: byte[4]
name: extract_symbolic_data
arguments: label: D47
datatype: byte[4]

tool: {"D48": [0x35,0xfa,0x8e,0x3c]}

tool: {"D47": [0x00,0x00,0x00,0x80]}

assistant: ```c
void RneFromLLA(float LLA[3], float Rne[3][3]) {
  float sinLat, sinLon, cosLat, cosLon;
  sinLat = (float)sinf((3.14159265358979323846f / 180.0f) * LLA[0]);
  sinLon = (float)sinf((3.14159265358979323846f / 180.0f) * LLA[1]);
  cosLat = (float)cosf((3.14159265358979323846f / 180.0f) * LLA[0]);
  cosLon = (float)cosf((3.14159265358979323846f / 180.0f) * LLA[1]);
  Rne[0][0] = -sinLat * cosLon;
  Rne[0][1] = -sinLat * sinLon;
  Rne[0][2] = cosLat;
  Rne[1][0] = -sinLon;
  Rne[1][1] = cosLon;
  Rne[1][2] = 0;
  Rne[2][0] = -cosLat * cosLon;
  Rne[2][1] = -cosLat * sinLon;
  Rne[2][2] = -sinLat;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RneFromLLA>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  movss  D38(%rip),%xmm0
  mulss  (%rdi),%xmm0
  call   <sinf@plt>
  movss  D38(%rip),%xmm0
  mulss  0x4(%rbp),%xmm0
  mov    %rax,%r12
  call   <sinf@plt>
  movss  D38(%rip),%xmm0
  pxor   %xmm4,%xmm4
  mulss  0x0(%rbp),%xmm0
  cvtsi2ss %rax,%xmm4
  movss  %xmm4,0xc(%rsp)
  call   <cosf@plt>
  movss  D38(%rip),%xmm0
  pxor   %xmm1,%xmm1
  mulss  0x4(%rbp),%xmm0
  cvtsi2ss %rax,%xmm1
  movss  %xmm1,0x8(%rsp)
  call   <cosf@plt>
  pxor   %xmm2,%xmm2
  pxor   %xmm3,%xmm3
  movss  D31(%rip),%xmm0
  cvtsi2ss %r12,%xmm2
  movss  0xc(%rsp),%xmm4
  movss  0x8(%rsp),%xmm1
  movl   $0x0,0x14(%rbx)
  cvtsi2ss %rax,%xmm3
  movss  %xmm1,0x8(%rbx)
  xorps  %xmm0,%xmm2
  movaps %xmm2,%xmm5
  movss  %xmm3,0x10(%rbx)
  mulss  %xmm3,%xmm5
  movss  %xmm2,0x20(%rbx)
  movss  %xmm5,(%rbx)
  movaps %xmm2,%xmm5
  mulss  %xmm4,%xmm5
  movss  %xmm5,0x4(%rbx)
  movaps %xmm4,%xmm5
  xorps  %xmm0,%xmm5
  xorps  %xmm1,%xmm0
  mulss  %xmm0,%xmm3
  movss  %xmm5,0xc(%rbx)
  mulss  %xmm4,%xmm0
  movss  %xmm3,0x18(%rbx)
  movss  %xmm0,0x1c(%rbx)
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D38
type_infered: byte[4]
name: read_data
arguments: label: D31
type_infered: byte[4]

tool: {"D38": [0x35,0xfa,0x8e,0x3c]}

tool: {"D31": [0x00,0x00,0x00,0x80]}

assistant: ```c
void RneFromLLA(float LLA[3], float Rne[3][3]) {
  float sinLat, sinLon, cosLat, cosLon;
  sinLat = (float)sinf((3.14159265358979323846f / 180.0f) * LLA[0]);
  sinLon = (float)sinf((3.14159265358979323846f / 180.0f) * LLA[1]);
  cosLat = (float)cosf((3.14159265358979323846f / 180.0f) * LLA[0]);
  cosLon = (float)cosf((3.14159265358979323846f / 180.0f) * LLA[1]);
  Rne[0][0] = -sinLat * cosLon;
  Rne[0][1] = -sinLat * sinLon;
  Rne[0][2] = cosLat;
  Rne[1][0] = -sinLon;
  Rne[1][1] = cosLon;
  Rne[1][2] = 0;
  Rne[2][0] = -cosLat * cosLon;
  Rne[2][1] = -cosLat * sinLon;
  Rne[2][2] = -sinLat;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RneFromLLA>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  movss  D45(%rip),%xmm0
  mulss  (%rdi),%xmm0
  call   <sinf@plt>
  movss  D45(%rip),%xmm0
  mulss  0x4(%rbp),%xmm0
  mov    %rax,%r12
  call   <sinf@plt>
  movss  D45(%rip),%xmm0
  pxor   %xmm4,%xmm4
  mulss  0x0(%rbp),%xmm0
  cvtsi2ss %rax,%xmm4
  movss  %xmm4,0xc(%rsp)
  call   <cosf@plt>
  movss  D45(%rip),%xmm0
  pxor   %xmm1,%xmm1
  mulss  0x4(%rbp),%xmm0
  cvtsi2ss %rax,%xmm1
  movss  %xmm1,0x8(%rsp)
  call   <cosf@plt>
  pxor   %xmm2,%xmm2
  pxor   %xmm3,%xmm3
  movss  D34(%rip),%xmm5
  cvtsi2ss %r12,%xmm2
  movss  0xc(%rsp),%xmm4
  movss  0x8(%rsp),%xmm1
  movl   $0x0,0x14(%rbx)
  cvtsi2ss %rax,%xmm3
  movaps %xmm4,%xmm8
  movaps %xmm1,%xmm6
  xorps  %xmm5,%xmm1
  xorps  %xmm5,%xmm8
  xorps  %xmm5,%xmm2
  unpcklps %xmm8,%xmm6
  movaps %xmm2,%xmm0
  movaps %xmm2,%xmm7
  movss  %xmm3,0x10(%rbx)
  mulss  %xmm3,%xmm0
  movss  %xmm2,0x20(%rbx)
  mulss  %xmm4,%xmm7
  mulss  %xmm1,%xmm3
  unpcklps %xmm7,%xmm0
  movlhps %xmm6,%xmm0
  movss  %xmm3,0x18(%rbx)
  movups %xmm0,(%rbx)
  movaps %xmm1,%xmm0
  mulss  %xmm4,%xmm0
  movss  %xmm0,0x1c(%rbx)
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D45
type_infered: byte[4]
name: parse_data
arguments: tag: D34
type_infered: byte[4]

tool: {"D45": [0x35,0xfa,0x8e,0x3c]}

tool: {"D34": [0x00,0x00,0x00,0x80]}

assistant: ```c
void RneFromLLA(float LLA[3], float Rne[3][3]) {
  float sinLat, sinLon, cosLat, cosLon;
  sinLat = (float)sinf((3.14159265358979323846f / 180.0f) * LLA[0]);
  sinLon = (float)sinf((3.14159265358979323846f / 180.0f) * LLA[1]);
  cosLat = (float)cosf((3.14159265358979323846f / 180.0f) * LLA[0]);
  cosLon = (float)cosf((3.14159265358979323846f / 180.0f) * LLA[1]);
  Rne[0][0] = -sinLat * cosLon;
  Rne[0][1] = -sinLat * sinLon;
  Rne[0][2] = cosLat;
  Rne[1][0] = -sinLon;
  Rne[1][1] = cosLon;
  Rne[1][2] = 0;
  Rne[2][0] = -cosLat * cosLon;
  Rne[2][1] = -cosLat * sinLon;
  Rne[2][2] = -sinLat;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<do_div>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x1c(%rbp),%ecx
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x1c(%rbp),%esi
  mov    $0x0,%edx
  div    %rsi
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long long do_div(unsigned long long *n, unsigned base) {
  unsigned long long res;
  res = (*n) % (unsigned long long)base;
  *n = (*n) / (unsigned long long)base;
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<do_div>:
  endbr64
  mov    (%rdi),%rax
  mov    %esi,%esi
  mov    $0x0,%edx
  div    %rsi
  mov    %rax,(%rdi)
  mov    %rdx,%rax
  ret
```

assistant: ```c
unsigned long long do_div(unsigned long long *n, unsigned base) {
  unsigned long long res;
  res = (*n) % (unsigned long long)base;
  *n = (*n) / (unsigned long long)base;
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<do_div>:
  endbr64
  mov    (%rdi),%rax
  mov    %esi,%esi
  xor    %edx,%edx
  div    %rsi
  mov    %rax,(%rdi)
  mov    %rdx,%rax
  ret
```

assistant: ```c
unsigned long long do_div(unsigned long long *n, unsigned base) {
  unsigned long long res;
  res = (*n) % (unsigned long long)base;
  *n = (*n) / (unsigned long long)base;
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<do_div>:
  endbr64
  mov    (%rdi),%rax
  mov    %esi,%esi
  xor    %edx,%edx
  div    %rsi
  mov    %rax,(%rdi)
  mov    %rdx,%rax
  ret
```

assistant: ```c
unsigned long long do_div(unsigned long long *n, unsigned base) {
  unsigned long long res;
  res = (*n) % (unsigned long long)base;
  *n = (*n) / (unsigned long long)base;
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<smoothstep_inv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
  movsd  D13(%rip),%xmm0
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <asin@plt>
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%edi
  call   <sin@plt>
  movsd  D10(%rip),%xmm1
  subsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D13
data_type: byte[8]
name: extract_symbolic_data
arguments: label: D10
data_type: f64

tool: {"D13": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D10: 0.5

assistant: ```c
double smoothstep_inv(double x) {
  (void)(!isnan(x));
  return (0.5 - sin(asin(1 - 2 * x) / 3));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<smoothstep_inv>:
  endbr64
  sub    $0x8,%rsp
  addsd  %xmm0,%xmm0
  movsd  D12(%rip),%xmm1
  subsd  %xmm0,%xmm1
  cvttsd2si %xmm1,%edi
  call   <asin@plt>
  movslq %eax,%rdi
  imul   $0x55555556,%rdi,%rdi
  shr    $0x20,%rdi
  sar    $0x1f,%eax
  sub    %eax,%edi
  call   <sin@plt>
  movapd %xmm0,%xmm1
  movsd  D1(%rip),%xmm0
  subsd  %xmm1,%xmm0
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D12
type_guessed: byte[8]
name: read_data
arguments: label: D1
type_guessed: f64

tool: {"D12": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D1: 0.5

assistant: ```c
double smoothstep_inv(double x) {
  (void)(!isnan(x));
  return (0.5 - sin(asin(1 - 2 * x) / 3));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<smoothstep_inv>:
  endbr64
  addsd  %xmm0,%xmm0
  sub    $0x8,%rsp
  movsd  D10(%rip),%xmm1
  subsd  %xmm0,%xmm1
  cvttsd2si %xmm1,%edi
  call   <asin@plt>
  movslq %eax,%rdi
  sar    $0x1f,%eax
  imul   $0x55555556,%rdi,%rdi
  shr    $0x20,%rdi
  sub    %eax,%edi
  call   <sin@plt>
  movapd %xmm0,%xmm1
  movsd  D3(%rip),%xmm0
  add    $0x8,%rsp
  subsd  %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: tag: D10
data_type: byte[8]
name: unpack_data
arguments: tag: D3
data_type: f64

tool: {"D10": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D3: 0.5

assistant: ```c
double smoothstep_inv(double x) {
  (void)(!isnan(x));
  return (0.5 - sin(asin(1 - 2 * x) / 3));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<smoothstep_inv>:
  endbr64
  addsd  %xmm0,%xmm0
  sub    $0x8,%rsp
  movsd  D10(%rip),%xmm1
  subsd  %xmm0,%xmm1
  cvttsd2si %xmm1,%edi
  call   <asin@plt>
  movslq %eax,%rdi
  sar    $0x1f,%eax
  imul   $0x55555556,%rdi,%rdi
  shr    $0x20,%rdi
  sub    %eax,%edi
  call   <sin@plt>
  movapd %xmm0,%xmm1
  movsd  D6(%rip),%xmm0
  add    $0x8,%rsp
  subsd  %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D10
type_guessed: byte[8]
name: parse_data
arguments: data_tag: D6
type_guessed: f64

tool: {"D10": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D6: 0.5

assistant: ```c
double smoothstep_inv(double x) {
  (void)(!isnan(x));
  return (0.5 - sin(asin(1 - 2 * x) / 3));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <g@plt>
  mov    D12(%rip),%eax
  sub    $0x1,%eax
  mov    %eax,D14(%rip)
  nop
  leave
  ret
```

assistant: ```c
void f(int n) {
  static int x = 1;
  g(n);
  x--;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f>:
  endbr64
  sub    $0x8,%rsp
  call   <g@plt>
  subl   $0x1,D6(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void f(int n) {
  static int x = 1;
  g(n);
  x--;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<f>:
  endbr64
  sub    $0x8,%rsp
  call   <g@plt>
  subl   $0x1,D4(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void f(int n) {
  static int x = 1;
  g(n);
  x--;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<f>:
  endbr64
  sub    $0x8,%rsp
  call   <g@plt>
  subl   $0x1,D8(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void f(int n) {
  static int x = 1;
  g(n);
  x--;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L21
L1:
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L21:
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <check_palindrome@plt>
  test   %eax,%eax
  je     L1
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
type: string
name: unpack_data
arguments: label: D17
type: string

tool: D7: %d

tool: D17: %X


assistant: ```c
int main() {
  int InputNumber;
  scanf("%d", &InputNumber);
  while (!check_palindrome(InputNumber)) {
    InputNumber++;
  }
  printf("%X\n", InputNumber);
  return 0;
}
```