ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<linear_a2i>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x2f,-0x4(%rbp)
  jle    L73
  cmpb   $0x39,-0x4(%rbp)
  jg     L73
  movzbl -0x4(%rbp),%eax
  sub    $0x30,%eax
  jmp    L7
L73:
  cmpb   $0x40,-0x4(%rbp)
  jle    L33
  cmpb   $0x46,-0x4(%rbp)
  jg     L33
  movzbl -0x4(%rbp),%eax
  sub    $0x37,%eax
  jmp    L7
L33:
  cmpb   $0x60,-0x4(%rbp)
  jle    L32
  cmpb   $0x66,-0x4(%rbp)
  jg     L32
  movzbl -0x4(%rbp),%eax
  sub    $0x57,%eax
  jmp    L7
L32:
  mov    $0x0,%eax
L7:
  pop    %rbp
  ret
```

assistant: ```c
uint32_t linear_a2i(char c) {
  if ((c >= '0') && (c <= '9'))
    return c - '0';
  if ((c >= 'A') && (c <= 'F'))
    return c - 'A' + 10;
  if ((c >= 'a') && (c <= 'f'))
    return c - 'a' + 10;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<linear_a2i>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L6
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L13
  lea    -0x61(%rdi),%eax
  sub    $0x57,%edi
  cmp    $0x6,%al
  mov    $0x0,%eax
  cmovb  %edi,%eax
L6:
  ret
L13:
  lea    -0x37(%rdi),%eax
  ret
```

assistant: ```c
uint32_t linear_a2i(char c) {
  if ((c >= '0') && (c <= '9'))
    return c - '0';
  if ((c >= 'A') && (c <= 'F'))
    return c - 'A' + 10;
  if ((c >= 'a') && (c <= 'f'))
    return c - 'a' + 10;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<linear_a2i>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L17
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L4
  lea    -0x61(%rdi),%eax
  sub    $0x57,%edi
  cmp    $0x6,%al
  mov    $0x0,%eax
  cmovb  %edi,%eax
L17:
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  lea    -0x37(%rdi),%eax
  ret
```

assistant: ```c
uint32_t linear_a2i(char c) {
  if ((c >= '0') && (c <= '9'))
    return c - '0';
  if ((c >= 'A') && (c <= 'F'))
    return c - 'A' + 10;
  if ((c >= 'a') && (c <= 'f'))
    return c - 'a' + 10;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<linear_a2i>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L5
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L19
  lea    -0x61(%rdi),%eax
  sub    $0x57,%edi
  cmp    $0x6,%al
  mov    $0x0,%eax
  cmovb  %edi,%eax
L5:
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  lea    -0x37(%rdi),%eax
  ret
```

assistant: ```c
uint32_t linear_a2i(char c) {
  if ((c >= '0') && (c <= '9'))
    return c - '0';
  if ((c >= 'A') && (c <= 'F'))
    return c - 'A' + 10;
  if ((c >= 'a') && (c <= 'f'))
    return c - 'a' + 10;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<doubles_mat44_rotate_y>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  mov    %rdi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <doubles_mat44_identity@plt>
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <sin@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  movsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x40,%rax
  movsd  -0x10(%rbp),%xmm0
  movq   D6(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x50,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D6
datatype: byte[8]

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

assistant: ```c
void doubles_mat44_rotate_y(double rad, double out[16]) {
  doubles_mat44_identity(out);
  double s = (double)sin(rad), c = (double)cos(rad);
  out[0 * 4 + 0] = c;
  out[0 * 4 + 2] = s;
  out[2 * 4 + 0] = -s;
  out[2 * 4 + 2] = c;
}
```
user: What is the c source code of the assembly code below?

```asm
<doubles_mat44_rotate_y>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  movq   %xmm0,%r12
  mov    %rdi,%rbx
  call   <doubles_mat44_identity@plt>
  movq   %r12,%xmm0
  call   <sin@plt>
  pxor   %xmm2,%xmm2
  cvtsi2sd %rax,%xmm2
  movq   %xmm2,%rbp
  movq   %r12,%xmm0
  call   <cos@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  %xmm0,(%rbx)
  mov    %rbp,0x10(%rbx)
  movq   %rbp,%xmm1
  xorpd  D4(%rip),%xmm1
  movsd  %xmm1,0x40(%rbx)
  movsd  %xmm0,0x50(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
data_type: byte[16]

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void doubles_mat44_rotate_y(double rad, double out[16]) {
  doubles_mat44_identity(out);
  double s = (double)sin(rad), c = (double)cos(rad);
  out[0 * 4 + 0] = c;
  out[0 * 4 + 2] = s;
  out[2 * 4 + 0] = -s;
  out[2 * 4 + 2] = c;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<doubles_mat44_rotate_y>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  movsd  %xmm0,(%rsp)
  call   <doubles_mat44_identity@plt>
  movsd  (%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <sin@plt>
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm0
  cvtsi2sd %rax,%xmm1
  movsd  %xmm1,(%rsp)
  call   <cos@plt>
  pxor   %xmm0,%xmm0
  movsd  (%rsp),%xmm1
  cvtsi2sd %rax,%xmm0
  movsd  %xmm1,0x10(%rbx)
  xorpd  D4(%rip),%xmm1
  movsd  %xmm1,0x40(%rbx)
  movsd  %xmm0,(%rbx)
  movsd  %xmm0,0x50(%rbx)
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type_infered: byte[16]

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void doubles_mat44_rotate_y(double rad, double out[16]) {
  doubles_mat44_identity(out);
  double s = (double)sin(rad), c = (double)cos(rad);
  out[0 * 4 + 0] = c;
  out[0 * 4 + 2] = s;
  out[2 * 4 + 0] = -s;
  out[2 * 4 + 2] = c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<doubles_mat44_rotate_y>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  movsd  %xmm0,(%rsp)
  call   <doubles_mat44_identity@plt>
  movsd  (%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <sin@plt>
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm0
  cvtsi2sd %rax,%xmm1
  movsd  %xmm1,(%rsp)
  call   <cos@plt>
  pxor   %xmm0,%xmm0
  movsd  (%rsp),%xmm1
  cvtsi2sd %rax,%xmm0
  movsd  %xmm1,0x10(%rbx)
  xorpd  D6(%rip),%xmm1
  movsd  %xmm1,0x40(%rbx)
  movsd  %xmm0,(%rbx)
  movsd  %xmm0,0x50(%rbx)
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type: byte[16]

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

assistant: ```c
void doubles_mat44_rotate_y(double rad, double out[16]) {
  doubles_mat44_identity(out);
  double s = (double)sin(rad), c = (double)cos(rad);
  out[0 * 4 + 0] = c;
  out[0 * 4 + 2] = s;
  out[2 * 4 + 0] = -s;
  out[2 * 4 + 2] = c;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<deleteString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L2
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
L2:
  nop
  leave
  ret
```

assistant: ```c
void deleteString(struct String *string) {
  if (string != NULL)
    free(string);
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<deleteString>:
  endbr64
  test   %rdi,%rdi
  je     L4
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L4:
  ret
```

assistant: ```c
void deleteString(struct String *string) {
  if (string != NULL)
    free(string);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<deleteString>:
  endbr64
  test   %rdi,%rdi
  je     L4
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L4:
  ret
```

assistant: ```c
void deleteString(struct String *string) {
  if (string != NULL)
    free(string);
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<deleteString>:
  endbr64
  test   %rdi,%rdi
  je     L3
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L3:
  ret
```

assistant: ```c
void deleteString(struct String *string) {
  if (string != NULL)
    free(string);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<juliam>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  mov    %eax,-0x1c(%rbp)
  mov    -0x30(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  mov    %eax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  mov    %eax,-0x14(%rbp)
  mov    -0x40(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  mov    %eax,-0x10(%rbp)
  mov    -0x48(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  mov    %eax,-0xc(%rbp)
  lea    -0xc(%rbp),%rdi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rsi
  lea    -0x1c(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <juliam4@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int juliam(yr, mo, dy, hr, mn)
short *yr, *mo, *dy, *hr, *mn;
{
  int yy, mm, dd, hh, nn;
  yy = *yr;
  mm = *mo;
  dd = *dy;
  hh = *hr;
  nn = *mn;
  return (juliam4(&yy, &mm, &dd, &hh, &nn));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<juliam>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movswl (%rdi),%eax
  mov    %eax,0x4(%rsp)
  movswl (%rsi),%eax
  mov    %eax,0x8(%rsp)
  movswl (%rdx),%eax
  mov    %eax,0xc(%rsp)
  movswl (%rcx),%eax
  mov    %eax,0x10(%rsp)
  movswl (%r8),%eax
  mov    %eax,0x14(%rsp)
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  lea    0x14(%rsp),%r8
  mov    $0x0,%eax
  call   <juliam4@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int juliam(yr, mo, dy, hr, mn)
short *yr, *mo, *dy, *hr, *mn;
{
  int yy, mm, dd, hh, nn;
  yy = *yr;
  mm = *mo;
  dd = *dy;
  hh = *hr;
  nn = *mn;
  return (juliam4(&yy, &mm, &dd, &hh, &nn));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<juliam>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movswl (%rdi),%eax
  lea    0x4(%rsp),%rdi
  mov    %eax,0x4(%rsp)
  movswl (%rsi),%eax
  lea    0x8(%rsp),%rsi
  mov    %eax,0x8(%rsp)
  movswl (%rdx),%eax
  lea    0xc(%rsp),%rdx
  mov    %eax,0xc(%rsp)
  movswl (%rcx),%eax
  lea    0x10(%rsp),%rcx
  mov    %eax,0x10(%rsp)
  movswl (%r8),%eax
  lea    0x14(%rsp),%r8
  mov    %eax,0x14(%rsp)
  xor    %eax,%eax
  call   <juliam4@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int juliam(yr, mo, dy, hr, mn)
short *yr, *mo, *dy, *hr, *mn;
{
  int yy, mm, dd, hh, nn;
  yy = *yr;
  mm = *mo;
  dd = *dy;
  hh = *hr;
  nn = *mn;
  return (juliam4(&yy, &mm, &dd, &hh, &nn));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<juliam>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movswl (%rdi),%eax
  lea    0x4(%rsp),%rdi
  mov    %eax,0x4(%rsp)
  movswl (%rsi),%eax
  lea    0x8(%rsp),%rsi
  mov    %eax,0x8(%rsp)
  movswl (%rdx),%eax
  lea    0xc(%rsp),%rdx
  mov    %eax,0xc(%rsp)
  movswl (%rcx),%eax
  lea    0x10(%rsp),%rcx
  mov    %eax,0x10(%rsp)
  movswl (%r8),%eax
  lea    0x14(%rsp),%r8
  mov    %eax,0x14(%rsp)
  xor    %eax,%eax
  call   <juliam4@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int juliam(yr, mo, dy, hr, mn)
short *yr, *mo, *dy, *hr, *mn;
{
  int yy, mm, dd, hh, nn;
  yy = *yr;
  mm = *mo;
  dd = *dy;
  hh = *hr;
  nn = *mn;
  return (juliam4(&yy, &mm, &dd, &hh, &nn));
}
```
user: What is the c source code of the assembly code below:

```asm
<_inverse_rot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%esi
  mov    -0x10(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    $0x100000,%eax
  cltd
  idiv   %esi
  mov    %eax,(%rcx)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edi
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rcx
  mov    $0x100000,%eax
  cltd
  idiv   %edi
  mov    %eax,(%rcx)
  mov    -0x10(%rbp),%rax
  add    $0xc,%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  imul   %edx,%eax
  mov    -0x10(%rbp),%rdx
  add    $0x10,%rdx
  lea    0x3ff(%rax),%ecx
  test   %eax,%eax
  cmovs  %ecx,%eax
  sar    $0xa,%eax
  neg    %eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  imul   %edx,%eax
  mov    -0x10(%rbp),%rdx
  add    $0x14,%rdx
  lea    0x3ff(%rax),%ecx
  test   %eax,%eax
  cmovs  %ecx,%eax
  sar    $0xa,%eax
  neg    %eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void _inverse_rot(const int *m, int *o) {
  o[0] = 0;
  o[1] = (1024 * 1024) / m[2];
  o[2] = (1024 * 1024) / m[1];
  o[3] = 0;
  o[4] = -(m[5] * o[2]) / 1024;
  o[5] = -(m[4] * o[1]) / 1024;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_inverse_rot>:
  endbr64
  movl   $0x0,(%rsi)
  mov    $0x100000,%ecx
  mov    %ecx,%eax
  cltd
  idivl  0x8(%rdi)
  mov    %eax,%r8d
  mov    %eax,0x4(%rsi)
  mov    %ecx,%eax
  cltd
  idivl  0x4(%rdi)
  mov    %eax,0x8(%rsi)
  movl   $0x0,0xc(%rsi)
  imul   0x14(%rdi),%eax
  lea    0x3ff(%rax),%edx
  test   %eax,%eax
  cmovns %eax,%edx
  sar    $0xa,%edx
  neg    %edx
  mov    %edx,0x10(%rsi)
  mov    %r8d,%edx
  imul   0x10(%rdi),%edx
  lea    0x3ff(%rdx),%eax
  test   %edx,%edx
  cmovns %edx,%eax
  sar    $0xa,%eax
  neg    %eax
  mov    %eax,0x14(%rsi)
  ret
```

assistant: ```c
void _inverse_rot(const int *m, int *o) {
  o[0] = 0;
  o[1] = (1024 * 1024) / m[2];
  o[2] = (1024 * 1024) / m[1];
  o[3] = 0;
  o[4] = -(m[5] * o[2]) / 1024;
  o[5] = -(m[4] * o[1]) / 1024;
}
```
user: What is the c source code of the assembly code below?

```asm
<_inverse_rot>:
  endbr64
  movl   $0x0,(%rsi)
  mov    $0x100000,%ecx
  mov    %ecx,%eax
  cltd
  idivl  0x8(%rdi)
  mov    %eax,0x4(%rsi)
  mov    %eax,%r8d
  mov    %ecx,%eax
  cltd
  idivl  0x4(%rdi)
  movl   $0x0,0xc(%rsi)
  mov    %eax,0x8(%rsi)
  imul   0x14(%rdi),%eax
  test   %eax,%eax
  lea    0x3ff(%rax),%edx
  cmovns %eax,%edx
  sar    $0xa,%edx
  neg    %edx
  mov    %edx,0x10(%rsi)
  mov    0x10(%rdi),%edx
  imul   %r8d,%edx
  test   %edx,%edx
  lea    0x3ff(%rdx),%eax
  cmovns %edx,%eax
  sar    $0xa,%eax
  neg    %eax
  mov    %eax,0x14(%rsi)
  ret
```

assistant: ```c
void _inverse_rot(const int *m, int *o) {
  o[0] = 0;
  o[1] = (1024 * 1024) / m[2];
  o[2] = (1024 * 1024) / m[1];
  o[3] = 0;
  o[4] = -(m[5] * o[2]) / 1024;
  o[5] = -(m[4] * o[1]) / 1024;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_inverse_rot>:
  endbr64
  movl   $0x0,(%rsi)
  mov    $0x100000,%ecx
  mov    %ecx,%eax
  cltd
  idivl  0x8(%rdi)
  mov    %eax,0x4(%rsi)
  mov    %eax,%r8d
  mov    %ecx,%eax
  cltd
  idivl  0x4(%rdi)
  movl   $0x0,0xc(%rsi)
  mov    %eax,0x8(%rsi)
  imul   0x14(%rdi),%eax
  test   %eax,%eax
  lea    0x3ff(%rax),%edx
  cmovns %eax,%edx
  sar    $0xa,%edx
  neg    %edx
  mov    %edx,0x10(%rsi)
  mov    0x10(%rdi),%edx
  imul   %r8d,%edx
  test   %edx,%edx
  lea    0x3ff(%rdx),%eax
  cmovns %edx,%eax
  sar    $0xa,%eax
  neg    %eax
  mov    %eax,0x14(%rsi)
  ret
```

assistant: ```c
void _inverse_rot(const int *m, int *o) {
  o[0] = 0;
  o[1] = (1024 * 1024) / m[2];
  o[2] = (1024 * 1024) / m[1];
  o[3] = 0;
  o[4] = -(m[5] * o[2]) / 1024;
  o[5] = -(m[4] * o[1]) / 1024;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mov_preg32pimm32_reg8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    $0x88,%edi
  call   <put8@plt>
  mov    -0x4(%rbp),%eax
  or     $0x80,%al
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  shl    $0x3,%eax
  or     %edx,%eax
  mov    %eax,%edi
  call   <put8@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <put32@plt>
  nop
  leave
  ret
```

assistant: ```c
void mov_preg32pimm32_reg8(int reg32, unsigned int imm32, int reg8) {
  put8(0x88);
  put8(0x80 | reg32 | (reg8 << 3));
  put32(imm32);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mov_preg32pimm32_reg8>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %esi,%ebp
  mov    %edx,%ebx
  mov    $0x88,%edi
  call   <put8@plt>
  lea    0x0(,%rbx,8),%edi
  or     %r12d,%edi
  or     $0x80,%dil
  call   <put8@plt>
  mov    %ebp,%edi
  call   <put32@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void mov_preg32pimm32_reg8(int reg32, unsigned int imm32, int reg8) {
  put8(0x88);
  put8(0x80 | reg32 | (reg8 << 3));
  put32(imm32);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mov_preg32pimm32_reg8>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  mov    $0x88,%edi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  call   <put8@plt>
  lea    0x0(,%rbx,8),%edi
  or     %r12d,%edi
  or     $0x80,%dil
  call   <put8@plt>
  pop    %rbx
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1070 <put32@plt>
```

assistant: ```c
void mov_preg32pimm32_reg8(int reg32, unsigned int imm32, int reg8) {
  put8(0x88);
  put8(0x80 | reg32 | (reg8 << 3));
  put32(imm32);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mov_preg32pimm32_reg8>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  mov    $0x88,%edi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  call   <put8@plt>
  lea    0x0(,%rbx,8),%edi
  or     %r12d,%edi
  or     $0x80,%dil
  call   <put8@plt>
  pop    %rbx
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1070 <put32@plt>
```

assistant: ```c
void mov_preg32pimm32_reg8(int reg32, unsigned int imm32, int reg8) {
  put8(0x88);
  put8(0x80 | reg32 | (reg8 << 3));
  put32(imm32);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<PairwiseCombs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  imul   %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double PairwiseCombs(int nsam) {
  double combs = (nsam - 1) * (nsam / 2);
  return combs;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<PairwiseCombs>:
  endbr64
  lea    -0x1(%rdi),%edx
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  imul   %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  ret
```

assistant: ```c
double PairwiseCombs(int nsam) {
  double combs = (nsam - 1) * (nsam / 2);
  return combs;
}
```
user: Translate the below assembly code into C programming language:

```asm
<PairwiseCombs>:
  endbr64
  mov    %edi,%eax
  lea    -0x1(%rdi),%edx
  pxor   %xmm0,%xmm0
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  imul   %edx,%eax
  cvtsi2sd %eax,%xmm0
  ret
```

assistant: ```c
double PairwiseCombs(int nsam) {
  double combs = (nsam - 1) * (nsam / 2);
  return combs;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<PairwiseCombs>:
  endbr64
  mov    %edi,%eax
  lea    -0x1(%rdi),%edx
  pxor   %xmm0,%xmm0
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  imul   %edx,%eax
  cvtsi2sd %eax,%xmm0
  ret
```

assistant: ```c
double PairwiseCombs(int nsam) {
  double combs = (nsam - 1) * (nsam / 2);
  return combs;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<native_test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movsd  %xmm0,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L18:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <sprintf@plt>
  addl   $0x1,-0x4(%rbp)
L7:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jb     L18
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int native_test(unsigned int count, char *fmt, double val) {
  static char buf[(1024)];
  unsigned int i;
  for (i = 0; i < count; i++)
    sprintf(buf, fmt, val);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<native_test>:
  endbr64
  test   %edi,%edi
  je     L7
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%r12
  movq   %xmm0,%r13
  mov    $0x0,%ebx
  lea    D6(%rip),%r14
L8:
  movq   %r13,%xmm0
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <sprintf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L8
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int native_test(unsigned int count, char *fmt, double val) {
  static char buf[(1024)];
  unsigned int i;
  for (i = 0; i < count; i++)
    sprintf(buf, fmt, val);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<native_test>:
  endbr64
  test   %edi,%edi
  je     L3
  push   %r14
  movq   %xmm0,%r14
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L7:
  movq   %r14,%xmm0
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <sprintf@plt>
  cmp    %ebx,%ebp
  jne    L7
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  ret
```

assistant: ```c
int native_test(unsigned int count, char *fmt, double val) {
  static char buf[(1024)];
  unsigned int i;
  for (i = 0; i < count; i++)
    sprintf(buf, fmt, val);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<native_test>:
  endbr64
  test   %edi,%edi
  je     L13
  push   %r14
  movq   %xmm0,%r14
  push   %r13
  lea    D6(%rip),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L6:
  movq   %r14,%xmm0
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <sprintf@plt>
  cmp    %ebx,%ebp
  jne    L6
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
int native_test(unsigned int count, char *fmt, double val) {
  static char buf[(1024)];
  unsigned int i;
  for (i = 0; i < count; i++)
    sprintf(buf, fmt, val);
  return 0;
}
```