ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x70,%rsp
  movdqa D48(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movl   $0x19,0x60(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D40(%rip),%xmm0
  mov    %rbp,%rdi
  movaps %xmm0,0x10(%rsp)
  movdqa D12(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movdqa D42(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  movdqa D5(%rip),%xmm0
  movaps %xmm0,0x40(%rsp)
  movdqa D4(%rip),%xmm0
  movaps %xmm0,0x50(%rsp)
  call   <Print@plt>
  mov    %rbp,%rdi
  call   <TLBR@plt>
  mov    %rbp,%rdi
  call   <BLTR@plt>
  mov    %rbp,%rdi
  call   <TLBR@plt>
  mov    %rbp,%rdi
  call   <BLTR@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x70,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D48
datatype: byte[16]
name: read_data
arguments: data_label: D40
datatype: byte[16]
name: read_data
arguments: data_label: D12
datatype: byte[16]
name: read_data
arguments: data_label: D42
datatype: byte[16]
name: read_data
arguments: data_label: D5
datatype: byte[16]
name: read_data
arguments: data_label: D4
datatype: byte[16]

tool: {"D48": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D40": [0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D12": [0x09,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x0b,0x00,0x00,0x00,0x0c,0x00,0x00,0x00]}

tool: {"D42": [0x0d,0x00,0x00,0x00,0x0e,0x00,0x00,0x00,0x0f,0x00,0x00,0x00,0x10,0x00,0x00,0x00]}

tool: {"D5": [0x11,0x00,0x00,0x00,0x12,0x00,0x00,0x00,0x13,0x00,0x00,0x00,0x14,0x00,0x00,0x00]}

tool: {"D4": [0x15,0x00,0x00,0x00,0x16,0x00,0x00,0x00,0x17,0x00,0x00,0x00,0x18,0x00,0x00,0x00]}

assistant: ```c
void main() {
  int array[5][5], i, j, n = 1;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      array[i][j] = j + i * 5 + 1;
  Print(array);
  TLBR(array);
  BLTR(array);
  TLBR(array);
  BLTR(array);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vdief>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <verrorf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void vdief(int status, const char *msg, va_list ap) {
  verrorf(msg, ap);
  exit(status);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vdief>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%rdi
  mov    %edx,%esi
  call   <verrorf@plt>
  mov    %ebx,%edi
  call   <exit@plt>
  pop    %rbx
  ret
```

assistant: ```c
void vdief(int status, const char *msg, va_list ap) {
  verrorf(msg, ap);
  exit(status);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<vdief>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  mov    %rsi,%rdi
  mov    %edx,%esi
  call   <verrorf@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <exit@plt>
```

assistant: ```c
void vdief(int status, const char *msg, va_list ap) {
  verrorf(msg, ap);
  exit(status);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<vdief>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  mov    %rsi,%rdi
  mov    %edx,%esi
  call   <verrorf@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <exit@plt>
```

assistant: ```c
void vdief(int status, const char *msg, va_list ap) {
  verrorf(msg, ap);
  exit(status);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    $0x5,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D25(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movq   $0x0,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <execvp@plt>
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type_infered: string
name: disassemble_data
arguments: data_label: D25
type_infered: string
name: disassemble_data
arguments: data_label: D11
type_infered: string

tool: D7: cal

tool: D25: 2014

tool: D11: This statement is not executed if execvp succeeds.


assistant: ```c
int main() {
  char *arglist[3];
  arglist[0] = (char *)malloc(4 * sizeof(char));
  strcpy(arglist[0], "cal");
  arglist[1] = (char *)malloc(5 * sizeof(char));
  strcpy(arglist[1], "2014");
  arglist[2] = NULL;
  execvp("cal", arglist);
  printf("This statement is not executed if execvp succeeds.\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  lea    D25(%rip),%rbx
  mov    %rbx,%rsi
  call   <strcpy@plt>
  mov    $0x5,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,0x8(%rsp)
  lea    D9(%rip),%rsi
  call   <strcpy@plt>
  movq   $0x0,0x10(%rsp)
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <execvp@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D25
type: string
name: unpack_data
arguments: label: D9
type: string
name: unpack_data
arguments: label: D1
type: string

tool: D25: cal

tool: D9: 2014

tool: D1: This statement is not executed if execvp succeeds.


assistant: ```c
int main() {
  char *arglist[3];
  arglist[0] = (char *)malloc(4 * sizeof(char));
  strcpy(arglist[0], "cal");
  arglist[1] = (char *)malloc(5 * sizeof(char));
  strcpy(arglist[1], "2014");
  arglist[2] = NULL;
  execvp("cal", arglist);
  printf("This statement is not executed if execvp succeeds.\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x4,%edi
  lea    D18(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <strcpy@plt>
  mov    $0x5,%edi
  call   <malloc@plt>
  lea    D13(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,0x8(%rsp)
  call   <strcpy@plt>
  mov    %rbp,%rdi
  mov    %rsp,%rsi
  movq   $0x0,0x10(%rsp)
  call   <execvp@plt>
  lea    D20(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D18
type_guessed: string
name: unpack_data
arguments: data_label: D13
type_guessed: string
name: unpack_data
arguments: data_label: D20
type_guessed: string

tool: D18: cal

tool: D13: 2014

tool: D20: This statement is not executed if execvp succeeds.


assistant: ```c
int main() {
  char *arglist[3];
  arglist[0] = (char *)malloc(4 * sizeof(char));
  strcpy(arglist[0], "cal");
  arglist[1] = (char *)malloc(5 * sizeof(char));
  strcpy(arglist[1], "2014");
  arglist[2] = NULL;
  execvp("cal", arglist);
  printf("This statement is not executed if execvp succeeds.\n");
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x4,%edi
  lea    D1(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <strcpy@plt>
  mov    $0x5,%edi
  call   <malloc@plt>
  lea    D12(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,0x8(%rsp)
  call   <strcpy@plt>
  mov    %rbp,%rdi
  mov    %rsp,%rsi
  movq   $0x0,0x10(%rsp)
  call   <execvp@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D1
type: string
name: disassemble_data
arguments: tag: D12
type: string
name: disassemble_data
arguments: tag: D11
type: string

tool: D1: cal

tool: D12: 2014

tool: D11: This statement is not executed if execvp succeeds.


assistant: ```c
int main() {
  char *arglist[3];
  arglist[0] = (char *)malloc(4 * sizeof(char));
  strcpy(arglist[0], "cal");
  arglist[1] = (char *)malloc(5 * sizeof(char));
  strcpy(arglist[1], "2014");
  arglist[2] = NULL;
  execvp("cal", arglist);
  printf("This statement is not executed if execvp succeeds.\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ectoc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x100,%eax
  test   %eax,%eax
  je     L16
  andl   $0xfffffebf,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  and    $0x800,%eax
  test   %eax,%eax
  je     L17
  andl   $0xff,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
ectoc(c) int c;
{
  if (c & 0x0100)
    c = c & ~(0x0100 | 0x40);
  if (c & 0x0800)
    c = c & 255;
  return (c);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ectoc>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%edx
  and    $0xfffffebf,%edx
  test   $0x1,%ah
  cmovne %edx,%eax
  movzbl %al,%edx
  test   $0x8,%ah
  cmovne %edx,%eax
  ret
```

assistant: ```c
ectoc(c) int c;
{
  if (c & 0x0100)
    c = c & ~(0x0100 | 0x40);
  if (c & 0x0800)
    c = c & 255;
  return (c);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ectoc>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%edx
  and    $0xfffffebf,%edx
  test   $0x1,%ah
  cmovne %edx,%eax
  movzbl %al,%edx
  test   $0x8,%ah
  cmovne %edx,%eax
  ret
```

assistant: ```c
ectoc(c) int c;
{
  if (c & 0x0100)
    c = c & ~(0x0100 | 0x40);
  if (c & 0x0800)
    c = c & 255;
  return (c);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ectoc>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%edx
  and    $0xfffffebf,%edx
  test   $0x1,%ah
  cmovne %edx,%eax
  movzbl %al,%edx
  test   $0x8,%ah
  cmovne %edx,%eax
  ret
```

assistant: ```c
ectoc(c) int c;
{
  if (c & 0x0100)
    c = c & ~(0x0100 | 0x40);
  if (c & 0x0800)
    c = c & 255;
  return (c);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xor_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L1
L2:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  xor    %eax,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L1:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L2
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xor_hash(void *key, int len) {
  unsigned char *p = key;
  unsigned h = 0;
  int i;
  for (i = 0; i < len; i++)
    h ^= p[i];
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xor_hash>:
  endbr64
  test   %esi,%esi
  jle    L28
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x1(%rdi,%rdx,1),%rsi
  mov    $0x0,%edx
L19:
  movzbl (%rax),%ecx
  xor    %ecx,%edx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L19
L14:
  mov    %edx,%eax
  ret
L28:
  mov    $0x0,%edx
  jmp    L14
```

assistant: ```c
unsigned xor_hash(void *key, int len) {
  unsigned char *p = key;
  unsigned h = 0;
  int i;
  for (i = 0; i < len; i++)
    h ^= p[i];
  return h;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xor_hash>:
  endbr64
  test   %esi,%esi
  jle    L12
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rcx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L3:
  movzbl (%rdi),%edx
  add    $0x1,%rdi
  xor    %edx,%eax
  cmp    %rcx,%rdi
  jne    L3
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned xor_hash(void *key, int len) {
  unsigned char *p = key;
  unsigned h = 0;
  int i;
  for (i = 0; i < len; i++)
    h ^= p[i];
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<xor_hash>:
  endbr64
  mov    %rdi,%rcx
  mov    %esi,%edx
  test   %esi,%esi
  jle    L94
  lea    -0x1(%rsi),%eax
  cmp    $0xe,%eax
  jbe    L184
  shr    $0x4,%esi
  pxor   %xmm3,%xmm3
  pxor   %xmm4,%xmm4
  mov    %rdi,%rax
  sub    $0x1,%esi
  pxor   %xmm2,%xmm2
  shl    $0x4,%rsi
  lea    0x10(%rdi,%rsi,1),%rsi
  nopl   0x0(%rax,%rax,1)
L9:
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  movdqa %xmm0,%xmm1
  punpckhbw %xmm4,%xmm0
  movdqa %xmm0,%xmm5
  punpcklbw %xmm4,%xmm1
  punpckhwd %xmm2,%xmm0
  punpcklwd %xmm2,%xmm5
  pxor   %xmm5,%xmm0
  movdqa %xmm1,%xmm5
  punpckhwd %xmm2,%xmm1
  punpcklwd %xmm2,%xmm5
  pxor   %xmm5,%xmm1
  pxor   %xmm1,%xmm0
  pxor   %xmm0,%xmm3
  cmp    %rsi,%rax
  jne    L9
  movdqa %xmm3,%xmm0
  mov    %edx,%esi
  psrldq $0x8,%xmm0
  and    $0xfffffff0,%esi
  pxor   %xmm0,%xmm3
  movdqa %xmm3,%xmm0
  psrldq $0x4,%xmm0
  pxor   %xmm0,%xmm3
  movd   %xmm3,%eax
  test   $0xf,%dl
  je     L32
L123:
  movslq %esi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x1(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x2(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x3(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x4(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x5(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x6(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x7(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x8(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0x9(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0xa(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0xb(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0xc(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  lea    0xd(%rsi),%edi
  cmp    %edi,%edx
  jle    L101
  movslq %edi,%rdi
  add    $0xe,%esi
  movzbl (%rcx,%rdi,1),%edi
  xor    %edi,%eax
  cmp    %esi,%edx
  jle    L101
  movslq %esi,%rsi
  movzbl (%rcx,%rsi,1),%edx
  xor    %edx,%eax
  ret
  nop
L94:
  xor    %eax,%eax
L101:
  ret
L184:
  xor    %esi,%esi
  xor    %eax,%eax
  jmp    L123
L32:
  ret
```

assistant: ```c
unsigned xor_hash(void *key, int len) {
  unsigned char *p = key;
  unsigned h = 0;
  int i;
  for (i = 0; i < len; i++)
    h ^= p[i];
  return h;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Column_Transformation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movsd  %xmm0,-0x30(%rbp)
  mov    %esi,-0x34(%rbp)
  mov    %edx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    %r8d,-0x40(%rbp)
  mov    -0x34(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  je     L34
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L23
L7:
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  movsd  (%rax),%xmm0
  mulsd  -0x30(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x40(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,-0x10(%rbp)
  mov    -0x40(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,-0x8(%rbp)
  addl   $0x1,-0x14(%rbp)
L23:
  mov    -0x14(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L7
  jmp    L2
L34:
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L12
L20:
  mov    -0x10(%rbp),%rax
  movsd  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  movsd  (%rax),%xmm0
  mulsd  -0x30(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  mov    -0x10(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x40(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L12:
  mov    -0x14(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L20
L2:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void Column_Transformation(double *A, double x, int col1, int col2, int nrows, int ncols) {
  int i;
  double *pA1, *pA2;
  if (col1 != col2) {
    pA1 = A + col1;
    pA2 = A + col2;
    for (i = 0; i < nrows; pA1 += ncols, pA2 += ncols, i++)
      *pA2 += x * *pA1;
  } else {
    pA1 = A + col1;
    for (i = 0; i < nrows; pA1 += ncols, i++)
      *pA1 += x * *pA1;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Column_Transformation>:
  endbr64
  mov    %rdi,%rax
  mov    %ecx,%edi
  cmp    %edx,%esi
  je     L5
  movslq %esi,%rsi
  lea    (%rax,%rsi,8),%rcx
  movslq %edx,%rdx
  lea    (%rax,%rdx,8),%rax
  test   %edi,%edi
  jle    L42
  movslq %r8d,%r8
  lea    0x0(,%r8,8),%rdx
  mov    $0x0,%esi
L18:
  movapd %xmm0,%xmm1
  mulsd  (%rcx),%xmm1
  addsd  (%rax),%xmm1
  movsd  %xmm1,(%rax)
  add    %rdx,%rcx
  add    %rdx,%rax
  add    $0x1,%esi
  cmp    %esi,%edi
  jne    L18
  ret
L5:
  movslq %esi,%rsi
  lea    (%rax,%rsi,8),%rdx
  test   %ecx,%ecx
  jle    L42
  movslq %r8d,%r8
  lea    0x0(,%r8,8),%rax
  mov    $0x0,%ecx
L6:
  movsd  (%rdx),%xmm2
  movapd %xmm2,%xmm1
  mulsd  %xmm0,%xmm1
  addsd  %xmm2,%xmm1
  movsd  %xmm1,(%rdx)
  add    %rax,%rdx
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L6
L42:
  ret
```

assistant: ```c
void Column_Transformation(double *A, double x, int col1, int col2, int nrows, int ncols) {
  int i;
  double *pA1, *pA2;
  if (col1 != col2) {
    pA1 = A + col1;
    pA2 = A + col2;
    for (i = 0; i < nrows; pA1 += ncols, pA2 += ncols, i++)
      *pA2 += x * *pA1;
  } else {
    pA1 = A + col1;
    for (i = 0; i < nrows; pA1 += ncols, i++)
      *pA1 += x * *pA1;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Column_Transformation>:
  endbr64
  mov    %rdi,%r9
  movslq %esi,%rax
  mov    %ecx,%edi
  lea    (%r9,%rax,8),%rax
  cmp    %edx,%esi
  je     L15
  movslq %edx,%rdx
  movslq %r8d,%r8
  xor    %esi,%esi
  lea    (%r9,%rdx,8),%rdx
  lea    0x0(,%r8,8),%rcx
  test   %edi,%edi
  jle    L19
  nopl   0x0(%rax)
L7:
  movsd  (%rax),%xmm1
  add    $0x1,%esi
  add    %rcx,%rax
  mulsd  %xmm0,%xmm1
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  add    %rcx,%rdx
  cmp    %esi,%edi
  jne    L7
  ret
  xchg   %ax,%ax
L15:
  test   %ecx,%ecx
  jle    L19
  movslq %r8d,%r8
  xor    %ecx,%ecx
  lea    0x0(,%r8,8),%rdx
  nopl   0x0(%rax)
L13:
  movsd  (%rax),%xmm2
  add    $0x1,%ecx
  movapd %xmm2,%xmm1
  mulsd  %xmm0,%xmm1
  addsd  %xmm2,%xmm1
  movsd  %xmm1,(%rax)
  add    %rdx,%rax
  cmp    %ecx,%edi
  jne    L13
L19:
  ret
```

assistant: ```c
void Column_Transformation(double *A, double x, int col1, int col2, int nrows, int ncols) {
  int i;
  double *pA1, *pA2;
  if (col1 != col2) {
    pA1 = A + col1;
    pA2 = A + col2;
    for (i = 0; i < nrows; pA1 += ncols, pA2 += ncols, i++)
      *pA2 += x * *pA1;
  } else {
    pA1 = A + col1;
    for (i = 0; i < nrows; pA1 += ncols, i++)
      *pA1 += x * *pA1;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Column_Transformation>:
  endbr64
  mov    %rdi,%r9
  movslq %esi,%rax
  mov    %ecx,%edi
  lea    (%r9,%rax,8),%rax
  cmp    %edx,%esi
  je     L13
  movslq %edx,%rdx
  lea    (%r9,%rdx,8),%rdx
  test   %ecx,%ecx
  jle    L50
  movslq %r8d,%rsi
  xor    %ecx,%ecx
  shl    $0x3,%rsi
  cmp    $0x1,%r8d
  jne    L66
  nopw   0x0(%rax,%rax,1)
L57:
  movsd  (%rax),%xmm1
  add    $0x1,%ecx
  add    %rsi,%rax
  mulsd  %xmm0,%xmm1
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  add    %rsi,%rdx
  cmp    %edi,%ecx
  jne    L57
  ret
  cs nopw 0x0(%rax,%rax,1)
L66:
  movsd  (%rax),%xmm1
  add    $0x1,%ecx
  add    %rsi,%rax
  mulsd  %xmm0,%xmm1
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  add    %rsi,%rdx
  cmp    %ecx,%edi
  jne    L66
  ret
  xchg   %ax,%ax
L13:
  test   %ecx,%ecx
  jle    L50
  movslq %r8d,%rcx
  xor    %edx,%edx
  shl    $0x3,%rcx
  cmp    $0x1,%r8d
  jne    L9
  nopl   0x0(%rax,%rax,1)
L17:
  movsd  (%rax),%xmm2
  add    $0x1,%edx
  movapd %xmm2,%xmm1
  mulsd  %xmm0,%xmm1
  addsd  %xmm2,%xmm1
  movsd  %xmm1,(%rax)
  add    %rcx,%rax
  cmp    %edx,%edi
  jne    L17
L50:
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  movsd  (%rax),%xmm2
  add    $0x1,%edx
  movapd %xmm2,%xmm1
  mulsd  %xmm0,%xmm1
  addsd  %xmm2,%xmm1
  movsd  %xmm1,(%rax)
  add    %rcx,%rax
  cmp    %edx,%edi
  jne    L9
  ret
```

assistant: ```c
void Column_Transformation(double *A, double x, int col1, int col2, int nrows, int ncols) {
  int i;
  double *pA1, *pA2;
  if (col1 != col2) {
    pA1 = A + col1;
    pA2 = A + col2;
    for (i = 0; i < nrows; pA1 += ncols, pA2 += ncols, i++)
      *pA2 += x * *pA1;
  } else {
    pA1 = A + col1;
    for (i = 0; i < nrows; pA1 += ncols, i++)
      *pA1 += x * *pA1;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hexchar_to_bin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x2f,-0x4(%rbp)
  jle    L67
  cmpb   $0x39,-0x4(%rbp)
  jg     L67
  movzbl -0x4(%rbp),%eax
  sub    $0x30,%eax
  jmp    L62
L67:
  cmpb   $0x40,-0x4(%rbp)
  jle    L38
  cmpb   $0x46,-0x4(%rbp)
  jg     L38
  movzbl -0x4(%rbp),%eax
  sub    $0x37,%eax
  jmp    L62
L38:
  cmpb   $0x60,-0x4(%rbp)
  jle    L53
  cmpb   $0x66,-0x4(%rbp)
  jg     L53
  movzbl -0x4(%rbp),%eax
  sub    $0x57,%eax
  jmp    L62
L53:
  mov    $0xffffffff,%eax
L62:
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((__unused__)) int8_t hexchar_to_bin(char hex) {
  if (hex >= '0' && hex <= '9')
    return hex - '0';
  if (hex >= 'A' && hex <= 'F')
    return hex - 'A' + 10;
  if (hex >= 'a' && hex <= 'f')
    return hex - 'a' + 10;
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hexchar_to_bin>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L15
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L11
  lea    -0x61(%rdi),%eax
  sub    $0x57,%edi
  cmp    $0x6,%al
  mov    $0xffffffff,%eax
  cmovb  %edi,%eax
L15:
  ret
L11:
  lea    -0x37(%rdi),%eax
  ret
```

assistant: ```c
__attribute__((__unused__)) int8_t hexchar_to_bin(char hex) {
  if (hex >= '0' && hex <= '9')
    return hex - '0';
  if (hex >= 'A' && hex <= 'F')
    return hex - 'A' + 10;
  if (hex >= 'a' && hex <= 'f')
    return hex - 'a' + 10;
  return -1;
}
```