ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<bintooct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x1,-0x18(%rbp)
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L25
L27:
  mov    -0x28(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  imul   -0x18(%rbp),%rax
  add    %rax,-0x20(%rbp)
  shlq   -0x18(%rbp)
  mov    -0x28(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x28(%rbp)
L25:
  mov    -0x28(%rbp),%rax
  test   %rax,%rax
  jne    L27
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D23
datatype: string
name: read_data
arguments: label: D14
datatype: string
name: read_data
arguments: label: D29
datatype: string

tool: D23: Enter binary value: 

tool: D14: %ld

tool: D29: Octal: %lo

assistant: ```c
int bintooct() {
  long int binum, octnum = 0, j = 1, rem;
  printf("Enter binary value: ");
  scanf("%ld", &binum);
  while (binum != 0) {
    rem = binum % 10;
    octnum += rem * j;
    j *= 2;
    binum /= 10;
  }
  printf("Octal: %lo", octnum);
  return octnum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bintooct>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D19(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L9
  mov    $0x1,%esi
  mov    $0x0,%ebx
  movabs $0x6666666666666667,%rdi
L1:
  mov    %rcx,%rax
  imul   %rdi
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  imul   %rsi,%rcx
  add    %rcx,%rbx
  add    %rsi,%rsi
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L1
  movq   $0x0,(%rsp)
L15:
  mov    %rbx,%rsi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x10,%rsp
  pop    %rbx
  ret
L9:
  mov    %rcx,%rbx
  jmp    L15
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D14
data_type: string
name: disassemble_data
arguments: data_label: D19
data_type: string
name: disassemble_data
arguments: data_label: D4
data_type: string

tool: D14: Enter binary value: 

tool: D19: %ld

tool: D4: Octal: %lo

assistant: ```c
int bintooct() {
  long int binum, octnum = 0, j = 1, rem;
  printf("Enter binary value: ");
  scanf("%ld", &binum);
  while (binum != 0) {
    rem = binum % 10;
    octnum += rem * j;
    j *= 2;
    binum /= 10;
  }
  printf("Octal: %lo", octnum);
  return octnum;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bintooct>:
  endbr64
  push   %r12
  lea    D10(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L14
  mov    $0x1,%esi
  xor    %r8d,%r8d
  movabs $0x6666666666666667,%rdi
  nopl   0x0(%rax)
L39:
  mov    %rcx,%rax
  imul   %rdi
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  imul   %rsi,%rcx
  add    %rsi,%rsi
  add    %rcx,%r8
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L39
  movq   $0x0,(%rsp)
  mov    %r8d,%r12d
L16:
  xor    %eax,%eax
  mov    %r8,%rsi
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  xchg   %ax,%ax
L14:
  xor    %r8d,%r8d
  xor    %r12d,%r12d
  jmp    L16
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D10
datatype: string
name: disassemble_data
arguments: tag: D6
datatype: string
name: disassemble_data
arguments: tag: D27
datatype: string

tool: D10: Enter binary value: 

tool: D6: %ld

tool: D27: Octal: %lo

assistant: ```c
int bintooct() {
  long int binum, octnum = 0, j = 1, rem;
  printf("Enter binary value: ");
  scanf("%ld", &binum);
  while (binum != 0) {
    rem = binum % 10;
    octnum += rem * j;
    j *= 2;
    binum /= 10;
  }
  printf("Octal: %lo", octnum);
  return octnum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bintooct>:
  endbr64
  push   %r12
  lea    D1(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L32
  mov    $0x1,%esi
  xor    %r8d,%r8d
  movabs $0x6666666666666667,%rdi
  nopl   0x0(%rax)
L20:
  mov    %rcx,%rax
  imul   %rdi
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  imul   %rsi,%rcx
  add    %rsi,%rsi
  add    %rcx,%r8
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L20
  movq   $0x0,(%rsp)
  mov    %r8d,%r12d
L22:
  xor    %eax,%eax
  mov    %r8,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  xchg   %ax,%ax
L32:
  xor    %r8d,%r8d
  xor    %r12d,%r12d
  jmp    L22
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D1
type: string
name: parse_data
arguments: data_tag: D18
type: string
name: parse_data
arguments: data_tag: D9
type: string

tool: D1: Enter binary value: 

tool: D18: %ld

tool: D9: Octal: %lo

assistant: ```c
int bintooct() {
  long int binum, octnum = 0, j = 1, rem;
  printf("Enter binary value: ");
  scanf("%ld", &binum);
  while (binum != 0) {
    rem = binum % 10;
    octnum += rem * j;
    j *= 2;
    binum /= 10;
  }
  printf("Octal: %lo", octnum);
  return octnum;
}
```
user: What is the c source code of the assembly code below?

```asm
<interchop>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  movss  %xmm0,-0x24(%rbp)
  movss  %xmm1,-0x28(%rbp)
  mov    %rdi,-0x30(%rbp)
  mov    %rsi,-0x38(%rbp)
  movl   $0x10000,-0x10(%rbp)
  movss  -0x28(%rbp),%xmm0
  subss  -0x24(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ssl -0x10(%rbp),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movss  -0x24(%rbp),%xmm0
  movss  %xmm0,-0x1c(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x38(%rbp),%rdx
  movd   %eax,%xmm0
  call   *%rdx
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L14
L16:
  mov    -0x1c(%rbp),%eax
  mov    -0x30(%rbp),%rdx
  movd   %eax,%xmm0
  call   *%rdx
  movd   %xmm0,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    -0x38(%rbp),%rdx
  movd   %eax,%xmm0
  call   *%rdx
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  movss  -0x1c(%rbp),%xmm0
  addss  -0xc(%rbp),%xmm0
  movss  %xmm0,-0x1c(%rbp)
  movss  -0x18(%rbp),%xmm0
  addss  -0x4(%rbp),%xmm0
  movss  %xmm0,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    -0x38(%rbp),%rdx
  movd   %eax,%xmm0
  call   *%rdx
  movd   %xmm0,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x18(%rbp),%eax
  mov    -0x38(%rbp),%rdx
  movd   %eax,%xmm0
  call   *%rdx
  movd   %xmm0,%eax
  mov    %eax,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
L14:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L16
  movss  -0x18(%rbp),%xmm0
  leave
  ret
```

assistant: ```c
float interchop(float min, float max, float (*fcn)(float x), float (*chop)(float x)) {
  float f, g, h;
  float x, s, y;
  int i, n;
  n = 65536;
  s = (max - min) / n;
  x = min;
  s = chop(s);
  y = 0;
  for (i = 0; i < n; i++) {
    f = fcn(x);
    g = chop(f);
    x = x + s;
    y = y + g;
    x = chop(x);
    y = chop(y);
  }
  return (y);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<interchop>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movss  %xmm0,0x4(%rsp)
  mov    %rdi,%r12
  mov    %rsi,%rbx
  subss  %xmm0,%xmm1
  mulss  D9(%rip),%xmm1
  movaps %xmm1,%xmm0
  call   *%rsi
  movss  %xmm0,0xc(%rsp)
  mov    $0x10000,%ebp
  pxor   %xmm4,%xmm4
  movss  %xmm4,0x8(%rsp)
L5:
  movss  0x4(%rsp),%xmm0
  call   *%r12
  call   *%rbx
  movss  0x4(%rsp),%xmm1
  addss  0xc(%rsp),%xmm1
  addss  0x8(%rsp),%xmm0
  movss  %xmm0,0x8(%rsp)
  movaps %xmm1,%xmm0
  call   *%rbx
  movss  %xmm0,0x4(%rsp)
  movss  0x8(%rsp),%xmm0
  call   *%rbx
  movss  %xmm0,0x8(%rsp)
  sub    $0x1,%ebp
  jne    L5
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: label: D9
data_type: byte[4]

tool: {"D9": [0x00,0x00,0x80,0x37]}

assistant: ```c
float interchop(float min, float max, float (*fcn)(float x), float (*chop)(float x)) {
  float f, g, h;
  float x, s, y;
  int i, n;
  n = 65536;
  s = (max - min) / n;
  x = min;
  s = chop(s);
  y = 0;
  for (i = 0; i < n; i++) {
    f = fcn(x);
    g = chop(f);
    x = x + s;
    y = y + g;
    x = chop(x);
    y = chop(y);
  }
  return (y);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<interchop>:
  endbr64
  push   %r12
  subss  %xmm0,%xmm1
  mov    %rdi,%r12
  push   %rbp
  mov    $0x10000,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mulss  D6(%rip),%xmm1
  movss  %xmm0,0x4(%rsp)
  movaps %xmm1,%xmm0
  call   *%rsi
  pxor   %xmm4,%xmm4
  movss  %xmm0,0xc(%rsp)
  movss  %xmm4,0x8(%rsp)
  xchg   %ax,%ax
L5:
  movss  0x4(%rsp),%xmm0
  call   *%r12
  call   *%rbx
  movss  0x8(%rsp),%xmm1
  movss  0x4(%rsp),%xmm2
  addss  0xc(%rsp),%xmm2
  addss  %xmm0,%xmm1
  movaps %xmm2,%xmm0
  movss  %xmm1,0x8(%rsp)
  call   *%rbx
  movss  0x8(%rsp),%xmm1
  movss  %xmm0,0x4(%rsp)
  movaps %xmm1,%xmm0
  call   *%rbx
  movss  %xmm0,0x8(%rsp)
  sub    $0x1,%ebp
  jne    L5
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type_infered: byte[4]

tool: {"D6": [0x00,0x00,0x80,0x37]}

assistant: ```c
float interchop(float min, float max, float (*fcn)(float x), float (*chop)(float x)) {
  float f, g, h;
  float x, s, y;
  int i, n;
  n = 65536;
  s = (max - min) / n;
  x = min;
  s = chop(s);
  y = 0;
  for (i = 0; i < n; i++) {
    f = fcn(x);
    g = chop(f);
    x = x + s;
    y = y + g;
    x = chop(x);
    y = chop(y);
  }
  return (y);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<interchop>:
  endbr64
  push   %r12
  subss  %xmm0,%xmm1
  mov    %rdi,%r12
  push   %rbp
  mov    $0x10000,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mulss  D5(%rip),%xmm1
  movss  %xmm0,0x4(%rsp)
  movaps %xmm1,%xmm0
  call   *%rsi
  pxor   %xmm4,%xmm4
  movss  %xmm0,0xc(%rsp)
  movss  %xmm4,0x8(%rsp)
  xchg   %ax,%ax
L5:
  movss  0x4(%rsp),%xmm0
  call   *%r12
  call   *%rbx
  movss  0x8(%rsp),%xmm1
  movss  0x4(%rsp),%xmm2
  addss  0xc(%rsp),%xmm2
  addss  %xmm0,%xmm1
  movaps %xmm2,%xmm0
  movss  %xmm1,0x8(%rsp)
  call   *%rbx
  movss  0x8(%rsp),%xmm1
  movss  %xmm0,0x4(%rsp)
  movaps %xmm1,%xmm0
  call   *%rbx
  movss  %xmm0,0x8(%rsp)
  sub    $0x1,%ebp
  jne    L5
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
type_infered: byte[4]

tool: {"D5": [0x00,0x00,0x80,0x37]}

assistant: ```c
float interchop(float min, float max, float (*fcn)(float x), float (*chop)(float x)) {
  float f, g, h;
  float x, s, y;
  int i, n;
  n = 65536;
  s = (max - min) / n;
  x = min;
  s = chop(s);
  y = 0;
  for (i = 0; i < n; i++) {
    f = fcn(x);
    g = chop(f);
    x = x + s;
    y = y + g;
    x = chop(x);
    y = chop(y);
  }
  return (y);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<stdlib_ndarray_iget_int8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <stdlib_ndarray_iget_ptr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L10
  mov    $0xffffffff,%eax
  jmp    L5
L10:
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stdlib_ndarray_get_ptr_int8@plt>
L5:
  leave
  ret
```

assistant: ```c
int8_t stdlib_ndarray_iget_int8(const struct ndarray *arr, const int64_t idx, int8_t *out) {
  uint8_t *ptr = stdlib_ndarray_iget_ptr(arr, idx);
  if (ptr == NULL) {
    return -1;
  }
  return stdlib_ndarray_get_ptr_int8(ptr, out);
}
```
user: What is the c source code of the assembly code below?

```asm
<stdlib_ndarray_iget_int8>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  call   <stdlib_ndarray_iget_ptr@plt>
  test   %rax,%rax
  je     L17
  mov    %rax,%rdi
  mov    %rbx,%rsi
  call   <stdlib_ndarray_get_ptr_int8@plt>
L19:
  pop    %rbx
  ret
L17:
  mov    $0xffffffff,%eax
  jmp    L19
```

assistant: ```c
int8_t stdlib_ndarray_iget_int8(const struct ndarray *arr, const int64_t idx, int8_t *out) {
  uint8_t *ptr = stdlib_ndarray_iget_ptr(arr, idx);
  if (ptr == NULL) {
    return -1;
  }
  return stdlib_ndarray_get_ptr_int8(ptr, out);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stdlib_ndarray_iget_int8>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  call   <stdlib_ndarray_iget_ptr@plt>
  test   %rax,%rax
  je     L7
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  jmp    1060 <stdlib_ndarray_get_ptr_int8@plt>
L7:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int8_t stdlib_ndarray_iget_int8(const struct ndarray *arr, const int64_t idx, int8_t *out) {
  uint8_t *ptr = stdlib_ndarray_iget_ptr(arr, idx);
  if (ptr == NULL) {
    return -1;
  }
  return stdlib_ndarray_get_ptr_int8(ptr, out);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<stdlib_ndarray_iget_int8>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  call   <stdlib_ndarray_iget_ptr@plt>
  test   %rax,%rax
  je     L1
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  jmp    1060 <stdlib_ndarray_get_ptr_int8@plt>
L1:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int8_t stdlib_ndarray_iget_int8(const struct ndarray *arr, const int64_t idx, int8_t *out) {
  uint8_t *ptr = stdlib_ndarray_iget_ptr(arr, idx);
  if (ptr == NULL) {
    return -1;
  }
  return stdlib_ndarray_get_ptr_int8(ptr, out);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<square>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  mov    %al,-0x2(%rbp)
  jmp    L26
L29:
  mov    -0x18(%rbp),%eax
  mov    $0x0,%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  mov    %al,-0x1(%rbp)
  jmp    L25
L34:
  movsbl -0x2(%rbp),%edx
  movsbl -0x1(%rbp),%eax
  imul   $0xf0,%eax,%eax
  add    %edx,%eax
  cltq
  add    %rax,%rax
  add    $0x6000000,%rax
  mov    -0x20(%rbp),%edx
  mov    %dx,(%rax)
  movzbl -0x1(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1(%rbp)
L25:
  movsbl -0x1(%rbp),%eax
  mov    -0x18(%rbp),%ecx
  mov    -0x1c(%rbp),%edx
  add    %ecx,%edx
  cmp    %edx,%eax
  jl     L34
  movzbl -0x2(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x2(%rbp)
L26:
  movsbl -0x2(%rbp),%eax
  mov    -0x14(%rbp),%ecx
  mov    -0x1c(%rbp),%edx
  add    %ecx,%edx
  cmp    %edx,%eax
  jl     L29
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void square(int x, int y, int side, int color) {
  char _x, _y;
  for (_x = (x > 0 ? x : 0); _x < x + side; ++_x)
    for (_y = (y > 0 ? y : 0); _y < y + side; ++_y)
      ((unsigned short *)0x6000000)[_x + _y * 240] = color;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<square>:
  endbr64
  mov    %ecx,%r8d
  test   %edi,%edi
  mov    $0x0,%ecx
  cmovns %edi,%ecx
  mov    %ecx,%r9d
  movsbl %cl,%ecx
  lea    (%rdi,%rdx,1),%r10d
  cmp    %r10d,%ecx
  jge    L17
  test   %esi,%esi
  mov    $0x0,%r11d
  cmovns %esi,%r11d
  movsbl %r11b,%edi
  lea    (%rdx,%rsi,1),%esi
L28:
  mov    %r11d,%edx
  mov    %edi,%eax
  cmp    %esi,%edi
  jge    L31
L1:
  imul   $0xf0,%eax,%eax
  add    %ecx,%eax
  cltq
  mov    %r8w,0x6000000(%rax,%rax,1)
  add    $0x1,%edx
  movsbl %dl,%eax
  cmp    %esi,%eax
  jl     L1
L31:
  add    $0x1,%r9d
  movsbl %r9b,%ecx
  cmp    %r10d,%ecx
  jl     L28
L17:
  ret
```

assistant: ```c
void square(int x, int y, int side, int color) {
  char _x, _y;
  for (_x = (x > 0 ? x : 0); _x < x + side; ++_x)
    for (_y = (y > 0 ? y : 0); _y < y + side; ++_y)
      ((unsigned short *)0x6000000)[_x + _y * 240] = color;
}
```
user: Translate the below assembly code into C programming language:

```asm
<square>:
  endbr64
  mov    %esi,%eax
  xor    %esi,%esi
  lea    (%rdi,%rdx,1),%r10d
  test   %edi,%edi
  cmovns %edi,%esi
  mov    %esi,%r8d
  movsbl %sil,%esi
  cmp    %r10d,%esi
  jge    L24
  xor    %r9d,%r9d
  test   %eax,%eax
  lea    (%rdx,%rax,1),%edi
  cmovns %eax,%r9d
  mov    %r9d,%r11d
  movsbl %r9b,%r9d
L32:
  mov    %r11d,%edx
  mov    %r9d,%eax
  cmp    %edi,%r9d
  jge    L31
  nopl   0x0(%rax,%rax,1)
L9:
  imul   $0xf0,%eax,%eax
  add    $0x1,%edx
  add    %esi,%eax
  cltq
  mov    %cx,0x6000000(%rax,%rax,1)
  movsbl %dl,%eax
  cmp    %edi,%eax
  jl     L9
L31:
  add    $0x1,%r8d
  movsbl %r8b,%esi
  cmp    %r10d,%esi
  jl     L32
  ret
L24:
  ret
```

assistant: ```c
void square(int x, int y, int side, int color) {
  char _x, _y;
  for (_x = (x > 0 ? x : 0); _x < x + side; ++_x)
    for (_y = (y > 0 ? y : 0); _y < y + side; ++_y)
      ((unsigned short *)0x6000000)[_x + _y * 240] = color;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<square>:
  endbr64
  mov    %esi,%eax
  xor    %esi,%esi
  lea    (%rdi,%rdx,1),%r9d
  test   %edi,%edi
  cmovns %edi,%esi
  mov    %esi,%r8d
  movsbl %sil,%esi
  cmp    %r9d,%esi
  jge    L35
  xor    %r10d,%r10d
  test   %eax,%eax
  lea    (%rdx,%rax,1),%edi
  cmovns %eax,%r10d
  mov    %r10d,%r11d
  movsbl %r10b,%r10d
  cmp    %edi,%r10d
  jge    L35
  nopl   (%rax)
L13:
  mov    %r11d,%edx
  mov    %r10d,%eax
  xchg   %ax,%ax
L11:
  imul   $0xf0,%eax,%eax
  add    $0x1,%edx
  add    %esi,%eax
  cltq
  mov    %cx,0x6000000(%rax,%rax,1)
  movsbl %dl,%eax
  cmp    %edi,%eax
  jl     L11
  add    $0x1,%r8d
  movsbl %r8b,%esi
  cmp    %r9d,%esi
  jl     L13
L35:
  ret
```

assistant: ```c
void square(int x, int y, int side, int color) {
  char _x, _y;
  for (_x = (x > 0 ? x : 0); _x < x + side; ++_x)
    for (_y = (y > 0 ? y : 0); _y < y + side; ++_y)
      ((unsigned short *)0x6000000)[_x + _y * 240] = color;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ceillog2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x43300000,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movsd  D9(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  shr    $0x14,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  and    -0x14(%rbp),%eax
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  add    %edx,%eax
  sub    $0x3ff,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D9
data_type: f64

tool: D9: 4503599627370496.0

assistant: ```c
int ceillog2(unsigned v) {
  union {
    unsigned int u[2];
    double d;
  } t;
  t.u[1] = 0x43300000;
  t.u[!1] = v;
  t.d -= 4503599627370496.0;
  return ((t.u[1] >> 20) - 0x3FF) + ((v & (v - 1)) != 0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ceillog2>:
  endbr64
  mov    %edi,%edx
  movabs $0x4330000000000000,%rax
  or     %rdx,%rax
  movq   %rax,%xmm0
  subsd  D0(%rip),%xmm0
  movq   %xmm0,%rax
  shr    $0x34,%rax
  lea    -0x1(%rdi),%edx
  and    %edi,%edx
  cmp    $0x1,%edx
  sbb    $0x3fe,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
type_guessed: f64

tool: D0: 4503599627370496.0

assistant: ```c
int ceillog2(unsigned v) {
  union {
    unsigned int u[2];
    double d;
  } t;
  t.u[1] = 0x43300000;
  t.u[!1] = v;
  t.d -= 4503599627370496.0;
  return ((t.u[1] >> 20) - 0x3FF) + ((v & (v - 1)) != 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ceillog2>:
  endbr64
  movabs $0x4330000000000000,%rax
  mov    %edi,%edx
  or     %rdx,%rax
  lea    -0x1(%rdi),%edx
  movq   %rax,%xmm0
  subsd  D1(%rip),%xmm0
  and    %edi,%edx
  movq   %xmm0,%rax
  shr    $0x34,%rax
  cmp    $0x1,%edx
  sbb    $0x3fe,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: label: D1
type: f64

tool: D1: 4503599627370496.0

assistant: ```c
int ceillog2(unsigned v) {
  union {
    unsigned int u[2];
    double d;
  } t;
  t.u[1] = 0x43300000;
  t.u[!1] = v;
  t.d -= 4503599627370496.0;
  return ((t.u[1] >> 20) - 0x3FF) + ((v & (v - 1)) != 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ceillog2>:
  endbr64
  movabs $0x4330000000000000,%rax
  mov    %edi,%edx
  or     %rdx,%rax
  lea    -0x1(%rdi),%edx
  movq   %rax,%xmm0
  subsd  D7(%rip),%xmm0
  and    %edi,%edx
  movq   %xmm0,%rax
  shr    $0x34,%rax
  cmp    $0x1,%edx
  sbb    $0x3fe,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_infered: f64

tool: D7: 4503599627370496.0

assistant: ```c
int ceillog2(unsigned v) {
  union {
    unsigned int u[2];
    double d;
  } t;
  t.u[1] = 0x43300000;
  t.u[!1] = v;
  t.d -= 4503599627370496.0;
  return ((t.u[1] >> 20) - 0x3FF) + ((v & (v - 1)) != 0);
}
```