ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<addi_eeprom_readw_93c76>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %ax,-0x1c(%rbp)
  movw   $0x0,-0xe(%rbp)
  movzwl -0x1c(%rbp),%eax
  shr    %ax
  or     $0x18,%ah
  movzwl %ax,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0xd,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <addi_eeprom_cmd_93c76@plt>
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L14:
  mov    -0x8(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <addi_eeprom_clk_93c76@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <inl@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0x64,%edi
  call   <udelay@plt>
  shlw   -0xe(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  je     L19
  orw    $0x1,-0xe(%rbp)
L19:
  addl   $0x1,-0xc(%rbp)
L2:
  cmpl   $0xf,-0xc(%rbp)
  jle    L14
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <outl@plt>
  mov    $0x64,%edi
  call   <udelay@plt>
  movzwl -0xe(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned short addi_eeprom_readw_93c76(unsigned long iobase, unsigned short addr) {
  unsigned short val = 0;
  unsigned int cmd;
  unsigned int tmp;
  int i;
  cmd = (0x0180 << 4) | (addr / 2);
  cmd = addi_eeprom_cmd_93c76(iobase, cmd, 13);
  for (i = 0; i < 16; i++) {
    addi_eeprom_clk_93c76(iobase, cmd);
    tmp = inl(iobase);
    udelay(100);
    val <<= 1;
    if (tmp & (1 << 3))
      val |= 0x1;
  }
  outl(0, iobase);
  udelay(100);
  return val;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<addi_eeprom_readw_93c76>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  shr    %si
  or     $0x1800,%si
  movzwl %si,%esi
  mov    $0xd,%edx
  call   <addi_eeprom_cmd_93c76@plt>
  mov    %eax,%r14d
  mov    $0x10,%r12d
  mov    $0x0,%ebx
L5:
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <addi_eeprom_clk_93c76@plt>
  mov    %r13,%rdi
  call   <inl@plt>
  mov    %eax,%ebp
  mov    $0x64,%edi
  call   <udelay@plt>
  add    %ebx,%ebx
  mov    %ebx,%eax
  or     $0x1,%eax
  test   $0x8,%bpl
  cmovne %eax,%ebx
  sub    $0x1,%r12d
  jne    L5
  mov    %r13,%rsi
  mov    $0x0,%edi
  call   <outl@plt>
  mov    $0x64,%edi
  call   <udelay@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
unsigned short addi_eeprom_readw_93c76(unsigned long iobase, unsigned short addr) {
  unsigned short val = 0;
  unsigned int cmd;
  unsigned int tmp;
  int i;
  cmd = (0x0180 << 4) | (addr / 2);
  cmd = addi_eeprom_cmd_93c76(iobase, cmd, 13);
  for (i = 0; i < 16; i++) {
    addi_eeprom_clk_93c76(iobase, cmd);
    tmp = inl(iobase);
    udelay(100);
    val <<= 1;
    if (tmp & (1 << 3))
      val |= 0x1;
  }
  outl(0, iobase);
  udelay(100);
  return val;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<addi_eeprom_readw_93c76>:
  endbr64
  push   %r14
  shr    %si
  mov    $0xd,%edx
  push   %r13
  or     $0x1800,%si
  mov    %rdi,%r13
  push   %r12
  movzwl %si,%esi
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x10,%ebp
  push   %rbx
  call   <addi_eeprom_cmd_93c76@plt>
  mov    %eax,%r14d
  nop
L5:
  mov    %r14d,%esi
  mov    %r13,%rdi
  add    %r12d,%r12d
  call   <addi_eeprom_clk_93c76@plt>
  mov    %r13,%rdi
  call   <inl@plt>
  mov    $0x64,%edi
  mov    %eax,%ebx
  call   <udelay@plt>
  mov    %r12d,%eax
  or     $0x1,%eax
  and    $0x8,%ebx
  cmovne %eax,%r12d
  sub    $0x1,%ebp
  jne    L5
  mov    %r13,%rsi
  xor    %edi,%edi
  call   <outl@plt>
  mov    $0x64,%edi
  call   <udelay@plt>
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
unsigned short addi_eeprom_readw_93c76(unsigned long iobase, unsigned short addr) {
  unsigned short val = 0;
  unsigned int cmd;
  unsigned int tmp;
  int i;
  cmd = (0x0180 << 4) | (addr / 2);
  cmd = addi_eeprom_cmd_93c76(iobase, cmd, 13);
  for (i = 0; i < 16; i++) {
    addi_eeprom_clk_93c76(iobase, cmd);
    tmp = inl(iobase);
    udelay(100);
    val <<= 1;
    if (tmp & (1 << 3))
      val |= 0x1;
  }
  outl(0, iobase);
  udelay(100);
  return val;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<addi_eeprom_readw_93c76>:
  endbr64
  push   %r14
  shr    %si
  mov    $0xd,%edx
  push   %r13
  or     $0x1800,%si
  mov    %rdi,%r13
  push   %r12
  movzwl %si,%esi
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x10,%ebp
  push   %rbx
  call   <addi_eeprom_cmd_93c76@plt>
  mov    %eax,%r14d
  nop
L8:
  mov    %r14d,%esi
  mov    %r13,%rdi
  add    %r12d,%r12d
  call   <addi_eeprom_clk_93c76@plt>
  mov    %r13,%rdi
  call   <inl@plt>
  mov    $0x64,%edi
  mov    %eax,%ebx
  call   <udelay@plt>
  mov    %r12d,%eax
  or     $0x1,%eax
  and    $0x8,%ebx
  cmovne %eax,%r12d
  sub    $0x1,%ebp
  jne    L8
  mov    %r13,%rsi
  xor    %edi,%edi
  call   <outl@plt>
  mov    $0x64,%edi
  call   <udelay@plt>
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
unsigned short addi_eeprom_readw_93c76(unsigned long iobase, unsigned short addr) {
  unsigned short val = 0;
  unsigned int cmd;
  unsigned int tmp;
  int i;
  cmd = (0x0180 << 4) | (addr / 2);
  cmd = addi_eeprom_cmd_93c76(iobase, cmd, 13);
  for (i = 0; i < 16; i++) {
    addi_eeprom_clk_93c76(iobase, cmd);
    tmp = inl(iobase);
    udelay(100);
    val <<= 1;
    if (tmp & (1 << 3))
      val |= 0x1;
  }
  outl(0, iobase);
  udelay(100);
  return val;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x120,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L20:
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x110(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x110(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x114(%rbp)
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x114(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x118(%rbp)
  jmp    L33
L42:
  mov    -0x118(%rbp),%eax
  cltq
  movzbl -0x110(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  subl   $0x1,-0x118(%rbp)
L33:
  cmpl   $0x0,-0x118(%rbp)
  jns    L42
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x118(%rbp)
  jmp    L47
L21:
  mov    $0x2a,%edi
  call   <putchar@plt>
  addl   $0x1,-0x118(%rbp)
L47:
  mov    -0x114(%rbp),%eax
  add    $0xf,%eax
  cmp    %eax,-0x118(%rbp)
  jle    L21
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L20
```

tool_calls: name: unpack_data
arguments: label: D20
data_type: string
name: unpack_data
arguments: label: D26
data_type: string
name: unpack_data
arguments: label: D37
data_type: string
name: unpack_data
arguments: label: D31
data_type: string

tool: D20: Enter string:

Your statement:

tool: D26: %s

tool: D37: Reverse:       

tool: D31: 


assistant: ```c
int main(void) {
  char arr[256];
  int x, i;
  while (1) {
    printf("Enter string:\n\nYour statement:");
    scanf("%s", arr);
    x = strlen(arr);
    printf("Reverse:       ");
    for (i = x - 1; i >= 0; i--)
      printf("%c", arr[i]);
    printf("\n");
    for (i = 0; i <= x + 15; i++)
      printf("*");
    printf("\n\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  lea    D23(%rip),%r15
  lea    D16(%rip),%r14
  lea    D21(%rip),%r13
  jmp    L37
L9:
  lea    D17(%rip),%rdi
  call   <puts@plt>
L37:
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%ebx
  sub    $0x1,%ebx
  js     L52
  movslq %ebx,%rbx
  mov    %rsp,%r12
L32:
  movsbl (%rbx,%r12,1),%edi
  call   <putchar@plt>
  sub    $0x1,%rbx
  test   %ebx,%ebx
  jns    L32
L52:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0xf,%ebp
  js     L9
  mov    $0x0,%ebx
L6:
  mov    $0x2a,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jle    L6
  jmp    L9
```

tool_calls: name: parse_data
arguments: label: D23
type_guessed: string
name: parse_data
arguments: label: D16
type_guessed: string
name: parse_data
arguments: label: D21
type_guessed: string
name: parse_data
arguments: label: D17
type_guessed: string

tool: D23: Enter string:

Your statement:

tool: D16: %s

tool: D21: Reverse:       

tool: D17: 


assistant: ```c
int main(void) {
  char arr[256];
  int x, i;
  while (1) {
    printf("Enter string:\n\nYour statement:");
    scanf("%s", arr);
    x = strlen(arr);
    printf("Reverse:       ");
    for (i = x - 1; i >= 0; i--)
      printf("%c", arr[i]);
    printf("\n");
    for (i = 0; i <= x + 15; i++)
      printf("*");
    printf("\n\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D1(%rip),%r15
  push   %r14
  lea    D4(%rip),%r14
  push   %r13
  lea    D6(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
L25:
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edx
  sub    $0x1,%edx
  js     L32
  movslq %edx,%r12
  nopl   0x0(%rax,%rax,1)
L30:
  movsbl (%rbx,%r12,1),%edi
  sub    $0x1,%r12
  call   <putchar@plt>
  test   %r12d,%r12d
  jns    L30
L32:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0xf,%ebp
  js     L8
  xor    %r12d,%r12d
L15:
  mov    $0x2a,%edi
  add    $0x1,%r12d
  call   <putchar@plt>
  cmp    %ebp,%r12d
  jle    L15
L8:
  lea    D2(%rip),%rdi
  call   <puts@plt>
  jmp    L25
```

tool_calls: name: read_data
arguments: data_label: D1
type_guessed: string
name: read_data
arguments: data_label: D4
type_guessed: string
name: read_data
arguments: data_label: D6
type_guessed: string
name: read_data
arguments: data_label: D2
type_guessed: string

tool: D1: Enter string:

Your statement:

tool: D4: %s

tool: D6: Reverse:       

tool: D2: 


assistant: ```c
int main(void) {
  char arr[256];
  int x, i;
  while (1) {
    printf("Enter string:\n\nYour statement:");
    scanf("%s", arr);
    x = strlen(arr);
    printf("Reverse:       ");
    for (i = x - 1; i >= 0; i--)
      printf("%c", arr[i]);
    printf("\n");
    for (i = 0; i <= x + 15; i++)
      printf("*");
    printf("\n\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  lea    D18(%rip),%r15
  push   %r14
  lea    D15(%rip),%r14
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
L44:
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edx
  sub    $0x1,%edx
  js     L40
  movslq %edx,%r12
  nopl   0x0(%rax,%rax,1)
L24:
  movsbl (%rbx,%r12,1),%edi
  sub    $0x1,%r12
  call   <putchar@plt>
  test   %r12d,%r12d
  jns    L24
L40:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0xf,%ebp
  js     L7
  xor    %r12d,%r12d
L35:
  mov    $0x2a,%edi
  add    $0x1,%r12d
  call   <putchar@plt>
  cmp    %ebp,%r12d
  jle    L35
L7:
  lea    D31(%rip),%rdi
  call   <puts@plt>
  jmp    L44
```

tool_calls: name: read_data
arguments: label: D18
datatype: string
name: read_data
arguments: label: D15
datatype: string
name: read_data
arguments: label: D14
datatype: string
name: read_data
arguments: label: D31
datatype: string

tool: D18: Enter string:

Your statement:

tool: D15: %s

tool: D14: Reverse:       

tool: D31: 


assistant: ```c
int main(void) {
  char arr[256];
  int x, i;
  while (1) {
    printf("Enter string:\n\nYour statement:");
    scanf("%s", arr);
    x = strlen(arr);
    printf("Reverse:       ");
    for (i = x - 1; i >= 0; i--)
      printf("%c", arr[i]);
    printf("\n");
    for (i = 0; i <= x + 15; i++)
      printf("*");
    printf("\n\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<histogram>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    %r8d,-0x30(%rbp)
  mov    %r9,-0x40(%rbp)
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x14(%rbp)
  jmp    L7
L39:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rcx,%rdx
  movl   $0x0,(%rdx)
  addl   $0x1,-0x14(%rbp)
L7:
  mov    -0x14(%rbp),%edx
  cmp    -0x30(%rbp),%edx
  jl     L39
  movl   $0x0,-0x10(%rbp)
  jmp    L16
L13:
  movl   $0x0,-0xc(%rbp)
  jmp    L22
L42:
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  movslq %eax,%rdx
  imul   %rcx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x38(%rbp),%rdx
  add    %rdx,%rcx
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rcx,%rdx,4),%edx
  movslq %edx,%rcx
  lea    0x0(,%rcx,4),%rsi
  mov    -0x40(%rbp),%rcx
  add    %rsi,%rcx
  mov    (%rcx),%ecx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rsi
  mov    -0x40(%rbp),%rdx
  add    %rsi,%rdx
  add    $0x1,%ecx
  mov    %ecx,(%rdx)
  addl   $0x1,-0xc(%rbp)
L22:
  mov    -0xc(%rbp),%edx
  cmp    -0x28(%rbp),%edx
  jl     L42
  addl   $0x1,-0x10(%rbp)
L16:
  mov    -0x10(%rbp),%edx
  cmp    -0x24(%rbp),%edx
  jl     L13
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void histogram(int M, int N, int step, int matrix[M][step], int P, int hist[P]) {
  for (int k = 0; k < P; k++) {
    hist[k] = 0;
  }
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      hist[matrix[i][j]] += 1;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<histogram>:
  endbr64
  mov    %edi,%r10d
  mov    %esi,%r11d
  mov    %rcx,%rsi
  movslq %edx,%rdx
  test   %r8d,%r8d
  jle    L22
  mov    %r9,%rax
  lea    -0x1(%r8),%ecx
  lea    0x4(%r9,%rcx,4),%rcx
L37:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L37
L22:
  test   %r10d,%r10d
  jle    L63
  lea    0x0(,%rdx,4),%r8
  lea    -0x1(%r11),%edi
  lea    0x4(%rsi,%rdi,4),%rcx
  mov    $0x0,%esi
  not    %rdi
  shl    $0x2,%rdi
  jmp    L11
L48:
  movslq (%rax),%rdx
  addl   $0x1,(%r9,%rdx,4)
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L48
L49:
  add    $0x1,%esi
  add    %r8,%rcx
  cmp    %esi,%r10d
  je     L63
L11:
  lea    (%rdi,%rcx,1),%rax
  test   %r11d,%r11d
  jg     L48
  jmp    L49
L63:
  ret
```

assistant: ```c
void histogram(int M, int N, int step, int matrix[M][step], int P, int hist[P]) {
  for (int k = 0; k < P; k++) {
    hist[k] = 0;
  }
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      hist[matrix[i][j]] += 1;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<histogram>:
  endbr64
  push   %r13
  mov    %rcx,%r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %r8d,%r8d
  jle    L46
  movslq %r8d,%rdx
  mov    %r9,%rdi
  xor    %esi,%esi
  shl    $0x2,%rdx
  call   <memset@plt>
  mov    %rax,%r9
L46:
  test   %ebx,%ebx
  jle    L4
  lea    -0x1(%rbp),%eax
  mov    $0xfffffffffffffffc,%r8
  lea    0x0(,%r12,4),%rdi
  xor    %esi,%esi
  shl    $0x2,%rax
  lea    0x4(%r13,%rax,1),%rcx
  sub    %rax,%r8
  nopl   (%rax)
L35:
  lea    (%r8,%rcx,1),%rax
  test   %ebp,%ebp
  jle    L7
L40:
  movslq (%rax),%rdx
  add    $0x4,%rax
  addl   $0x1,(%r9,%rdx,4)
  cmp    %rax,%rcx
  jne    L40
L7:
  add    $0x1,%esi
  add    %rdi,%rcx
  cmp    %esi,%ebx
  jne    L35
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void histogram(int M, int N, int step, int matrix[M][step], int P, int hist[P]) {
  for (int k = 0; k < P; k++) {
    hist[k] = 0;
  }
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      hist[matrix[i][j]] += 1;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<histogram>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  movslq %edx,%rbx
  sub    $0x8,%rsp
  test   %r8d,%r8d
  jle    L23
  movslq %r8d,%rdx
  mov    %r9,%rdi
  xor    %esi,%esi
  shl    $0x2,%rdx
  call   <memset@plt>
  mov    %rax,%r9
L23:
  test   %ebp,%ebp
  jle    L40
  test   %r13d,%r13d
  jle    L40
  lea    -0x1(%r13),%eax
  mov    $0xfffffffffffffffc,%r8
  lea    0x0(,%rbx,4),%rdi
  xor    %esi,%esi
  shl    $0x2,%rax
  lea    0x4(%r12,%rax,1),%rcx
  sub    %rax,%r8
  nopl   0x0(%rax)
L39:
  lea    (%r8,%rcx,1),%rax
  nopl   0x0(%rax)
L26:
  movslq (%rax),%rdx
  add    $0x4,%rax
  addl   $0x1,(%r9,%rdx,4)
  cmp    %rcx,%rax
  jne    L26
  add    $0x1,%esi
  add    %rdi,%rcx
  cmp    %esi,%ebp
  jne    L39
L40:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void histogram(int M, int N, int step, int matrix[M][step], int P, int hist[P]) {
  for (int k = 0; k < P; k++) {
    hist[k] = 0;
  }
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      hist[matrix[i][j]] += 1;
    }
  }
}
```
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    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x2,-0x10(%rbp)
  jmp    L10
L73:
  movl   $0x1,-0xc(%rbp)
  jmp    L41
L46:
  mov    $0x40,%eax
  sub    -0xc(%rbp),%eax
  mov    $0xffffffffffffffff,%rdx
  mov    %eax,%ecx
  shr    %cl,%rdx
  mov    %rdx,%rax
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L37
L58:
  mov    -0x4(%rbp),%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x14(%rbp)
  shll   -0x4(%rbp)
L37:
  cmpl   $0x3f,-0x4(%rbp)
  jbe    L58
  movl   $0x0,-0x8(%rbp)
  jmp    L57
L66:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <uint64_to_binary@plt>
  mov    %eax,%esi
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%edi
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    %edi,%r9d
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %esi,%edx
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3f,%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%ecx
  shr    %cl,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x8(%rbp)
L57:
  mov    -0x8(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jb     L66
  addl   $0x1,-0xc(%rbp)
L41:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jb     L46
  shll   -0x10(%rbp)
L10:
  cmpl   $0x40,-0x10(%rbp)
  jbe    L73
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D0: 0x%016llx %s (size=%u, length=%u, rotation=%u)


assistant: ```c
int main() {
  uint64_t result;
  unsigned size, length, rotation, e;
  for (size = 2; size <= 64; size *= 2)
    for (length = 1; length < size; ++length) {
      result = 0xffffffffffffffffULL >> (64 - length);
      for (e = size; e < 64; e *= 2)
        result |= result << e;
      for (rotation = 0; rotation < size; ++rotation) {
        printf("0x%016llx %s (size=%u, length=%u, rotation=%u)\n", (unsigned long long)result, uint64_to_binary(result), size, length, rotation);
        result = (result >> 63) | (result << 1);
      }
    }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x6,%r15d
  mov    $0x2,%r12d
  lea    D8(%rip),%r14
L5:
  mov    $0x1,%r13d
  cmp    $0x1,%r12d
  jbe    L3
L39:
  mov    $0x40,%ecx
  sub    %r13d,%ecx
  mov    $0xffffffffffffffff,%rbx
  shr    %cl,%rbx
  cmp    $0x3f,%r12d
  ja     L15
  mov    %r12d,%ecx
L52:
  mov    %ebx,%eax
  shl    %cl,%eax
  or     %eax,%ebx
  add    %ecx,%ecx
  cmp    $0x3f,%ecx
  jbe    L52
L15:
  mov    $0x0,%ebp
L58:
  mov    %ebx,%edi
  call   <uint64_to_binary@plt>
  mov    %eax,%edx
  mov    %ebx,%esi
  mov    %ebp,%r9d
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %r14,%rdi
  call   <printf@plt>
  add    %ebx,%ebx
  add    $0x1,%ebp
  cmp    %r12d,%ebp
  jne    L58
  add    $0x1,%r13d
  cmp    %r12d,%r13d
  jne    L39
L3:
  add    %r12d,%r12d
  sub    $0x1,%r15d
  jne    L5
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
type_infered: string

tool: D8: 0x%016llx %s (size=%u, length=%u, rotation=%u)


assistant: ```c
int main() {
  uint64_t result;
  unsigned size, length, rotation, e;
  for (size = 2; size <= 64; size *= 2)
    for (length = 1; length < size; ++length) {
      result = 0xffffffffffffffffULL >> (64 - length);
      for (e = size; e < 64; e *= 2)
        result |= result << e;
      for (rotation = 0; rotation < size; ++rotation) {
        printf("0x%016llx %s (size=%u, length=%u, rotation=%u)\n", (unsigned long long)result, uint64_to_binary(result), size, length, rotation);
        result = (result >> 63) | (result << 1);
      }
    }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  mov    $0x40,%r14d
  push   %r13
  lea    D6(%rip),%r13
  push   %r12
  push   %rbp
  mov    $0x2,%ebp
  push   %rbx
  sub    $0x18,%rsp
  movl   $0x6,0xc(%rsp)
  nopl   0x0(%rax)
L36:
  mov    $0x1,%r12d
  cs nopw 0x0(%rax,%rax,1)
L34:
  mov    %r14d,%ecx
  mov    $0xffffffffffffffff,%rbx
  sub    %r12d,%ecx
  shr    %cl,%rbx
  cmp    $0x3f,%ebp
  ja     L24
  mov    %ebp,%ecx
  nopw   0x0(%rax,%rax,1)
L25:
  mov    %ebx,%eax
  shl    %cl,%eax
  add    %ecx,%ecx
  or     %eax,%ebx
  cmp    $0x3f,%ecx
  jbe    L25
L24:
  xor    %r15d,%r15d
L6:
  mov    %ebx,%edi
  call   <uint64_to_binary@plt>
  mov    %ebx,%esi
  mov    %r15d,%r9d
  mov    %r12d,%r8d
  mov    %eax,%edx
  mov    %ebp,%ecx
  mov    %r13,%rdi
  add    $0x1,%r15d
  call   <printf@plt>
  add    %ebx,%ebx
  cmp    %ebp,%r15d
  jne    L6
  add    $0x1,%r12d
  cmp    %r15d,%r12d
  jne    L34
  subl   $0x1,0xc(%rsp)
  lea    (%r12,%r12,1),%ebp
  jne    L36
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D6
data_type: string

tool: D6: 0x%016llx %s (size=%u, length=%u, rotation=%u)


assistant: ```c
int main() {
  uint64_t result;
  unsigned size, length, rotation, e;
  for (size = 2; size <= 64; size *= 2)
    for (length = 1; length < size; ++length) {
      result = 0xffffffffffffffffULL >> (64 - length);
      for (e = size; e < 64; e *= 2)
        result |= result << e;
      for (rotation = 0; rotation < size; ++rotation) {
        printf("0x%016llx %s (size=%u, length=%u, rotation=%u)\n", (unsigned long long)result, uint64_to_binary(result), size, length, rotation);
        result = (result >> 63) | (result << 1);
      }
    }
  return 0;
}
```