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

```asm
<mvp_itoh>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x8,%ebp
  cmp    $0xffff,%edi
  ja     L29
  cmp    $0x100,%edi
  sbb    %ebp,%ebp
  and    $0xfffffffe,%ebp
  add    $0x4,%ebp
L29:
  lea    0x4(%rbp),%edi
  movslq %edi,%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L11
  movb   $0x30,(%rax)
  movb   $0x78,0x1(%rax)
  lea    -0x4(,%rbp,4),%ecx
  lea    0x3(%rax),%rdi
  mov    $0xf,%r9d
L3:
  mov    %r9d,%edx
  shl    %cl,%edx
  and    %ebx,%edx
  shr    %cl,%edx
  and    $0xf,%edx
  lea    0x30(%rdx),%r8d
  lea    0x57(%rdx),%esi
  cmp    $0x9,%dl
  mov    %r8d,%edx
  cmova  %esi,%edx
  mov    %dl,-0x1(%rdi)
  sub    $0x4,%ecx
  add    $0x1,%rdi
  cmp    $0xfffffffc,%ecx
  jne    L3
  mov    %ebp,%ebp
  movb   $0x0,0x2(%rax,%rbp,1)
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *mvp_itoh(int integer) {
  char *str_hex = ((void *)0);
  char *phex;
  unsigned int ui = (unsigned int)integer;
  char half_byte;
  int length;
  int pos;
  int i;
  if (ui <= 0xffffffff) {
    if (ui > 0xffff) {
      length = 8;
    } else if (ui > 0xff) {
      length = 4;
    } else {
      length = 2;
    }
    str_hex = (char *)malloc(length + 4);
    if (((void *)0) != str_hex) {
      phex = str_hex;
      *phex++ = '0';
      *phex++ = 'x';
      for (i = 0; i < length; i++) {
        pos = ((length - i) - 1) * 4;
        half_byte = ((ui & (0x0000000f << pos)) >> pos) & 0x0f;
        if ((half_byte >= 0) && (half_byte <= 9)) {
          *phex++ = '0' + half_byte;
        } else if (half_byte > 9) {
          *phex++ = 'a' + (half_byte - 10);
        }
      }
      *phex = 0;
    }
  }
  return str_hex;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mvp_itoh>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  cmp    $0xffff,%edi
  ja     L30
  cmp    $0x100,%edi
  sbb    %rdi,%rdi
  and    $0xfffffffffffffffe,%rdi
  add    $0x8,%rdi
  cmp    $0x100,%ebx
  sbb    %ebp,%ebp
  and    $0xfffffffe,%ebp
  add    $0x4,%ebp
L22:
  call   <malloc@plt>
  test   %rax,%rax
  je     L15
  mov    $0x7830,%edx
  lea    0x2(%rax),%rdi
  lea    -0x4(,%rbp,4),%ecx
  mov    $0xf,%r9d
  mov    %dx,(%rax)
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r9d,%edx
  add    $0x1,%rdi
  shl    %cl,%edx
  and    %ebx,%edx
  shr    %cl,%edx
  and    $0xf,%edx
  lea    0x30(%rdx),%r8d
  cmp    $0x9,%dl
  lea    0x57(%rdx),%esi
  mov    %r8d,%edx
  cmovg  %esi,%edx
  sub    $0x4,%ecx
  mov    %dl,-0x1(%rdi)
  cmp    $0xfffffffc,%ecx
  jne    L6
  movslq %ebp,%rbp
  movb   $0x0,0x2(%rax,%rbp,1)
L15:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  mov    $0xc,%edi
  mov    $0x8,%ebp
  jmp    L22
```

assistant: ```c
char *mvp_itoh(int integer) {
  char *str_hex = ((void *)0);
  char *phex;
  unsigned int ui = (unsigned int)integer;
  char half_byte;
  int length;
  int pos;
  int i;
  if (ui <= 0xffffffff) {
    if (ui > 0xffff) {
      length = 8;
    } else if (ui > 0xff) {
      length = 4;
    } else {
      length = 2;
    }
    str_hex = (char *)malloc(length + 4);
    if (((void *)0) != str_hex) {
      phex = str_hex;
      *phex++ = '0';
      *phex++ = 'x';
      for (i = 0; i < length; i++) {
        pos = ((length - i) - 1) * 4;
        half_byte = ((ui & (0x0000000f << pos)) >> pos) & 0x0f;
        if ((half_byte >= 0) && (half_byte <= 9)) {
          *phex++ = '0' + half_byte;
        } else if (half_byte > 9) {
          *phex++ = 'a' + (half_byte - 10);
        }
      }
      *phex = 0;
    }
  }
  return str_hex;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<GetLeftRotatedString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x2c(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L34
L25:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x10(%rbp)
  addl   $0x1,-0x20(%rbp)
L34:
  mov    -0x20(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L25
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  jmp    L19
L33:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x10(%rbp)
  subl   $0x1,-0x20(%rbp)
L19:
  cmpl   $0x0,-0x20(%rbp)
  jg     L33
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *GetLeftRotatedString(char *original, int count) {
  int length, i;
  char *new;
  char *source;
  char *destination;
  length = strlen(original);
  new = (char *)malloc((length + 1) * sizeof(char));
  count = count % length;
  source = original;
  destination = new + (length - count);
  for (i = 0; i < count; i++) {
    *destination = *source;
    source++;
    destination++;
  }
  *destination = '\0';
  destination = new;
  for (i = length - count; i > 0; i--) {
    *destination = *source;
    source++;
    destination++;
  }
  return new;
}
```
user: What is the c source code of the assembly code below?

```asm
<GetLeftRotatedString>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  call   <strlen@plt>
  mov    %eax,%ebp
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %r12d,%eax
  cltd
  idiv   %ebp
  sub    %edx,%ebp
  movslq %ebp,%rsi
  add    %rdi,%rsi
  test   %edx,%edx
  jle    L30
  mov    %edx,%edx
  mov    $0x0,%eax
L14:
  movzbl (%rbx,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L14
  add    %rdx,%rbx
  add    %rdx,%rsi
L30:
  movb   $0x0,(%rsi)
  test   %ebp,%ebp
  jle    L33
  lea    -0x1(%rbp),%eax
  lea    0x1(%rdi,%rax,1),%rcx
  mov    %rdi,%rax
  sub    %rdi,%rbx
L28:
  movzbl (%rax,%rbx,1),%edx
  mov    %dl,(%rax)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L28
L33:
  mov    %rdi,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *GetLeftRotatedString(char *original, int count) {
  int length, i;
  char *new;
  char *source;
  char *destination;
  length = strlen(original);
  new = (char *)malloc((length + 1) * sizeof(char));
  count = count % length;
  source = original;
  destination = new + (length - count);
  for (i = 0; i < count; i++) {
    *destination = *source;
    source++;
    destination++;
  }
  *destination = '\0';
  destination = new;
  for (i = length - count; i > 0; i--) {
    *destination = *source;
    source++;
    destination++;
  }
  return new;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<GetLeftRotatedString>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  mov    %eax,%ebp
  call   <malloc@plt>
  mov    %rax,%r8
  mov    %r12d,%eax
  cltd
  idiv   %ebp
  sub    %edx,%ebp
  movslq %ebp,%rdi
  add    %r8,%rdi
  test   %edx,%edx
  jle    L36
  movslq %edx,%rdx
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L1:
  movzbl (%rbx,%rcx,1),%esi
  mov    %sil,(%rdi,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L1
  add    %rcx,%rbx
  add    %rcx,%rdi
L36:
  movb   $0x0,(%rdi)
  test   %ebp,%ebp
  jle    L6
  lea    -0x1(%rbp),%eax
  sub    %r8,%rbx
  lea    0x1(%r8,%rax,1),%rcx
  mov    %r8,%rax
  nopl   0x0(%rax)
L11:
  movzbl (%rax,%rbx,1),%edx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  cmp    %rcx,%rax
  jne    L11
L6:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *GetLeftRotatedString(char *original, int count) {
  int length, i;
  char *new;
  char *source;
  char *destination;
  length = strlen(original);
  new = (char *)malloc((length + 1) * sizeof(char));
  count = count % length;
  source = original;
  destination = new + (length - count);
  for (i = 0; i < count; i++) {
    *destination = *source;
    source++;
    destination++;
  }
  *destination = '\0';
  destination = new;
  for (i = length - count; i > 0; i--) {
    *destination = *source;
    source++;
    destination++;
  }
  return new;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0xa,%eax
  jg     L9
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  js     L9
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L9:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D19
type_infered: string
name: extract_symbolic_data
arguments: label: D23
type_infered: string

tool: D19: Enter an integer 1-10:

tool: D23: %d

assistant: ```c
int main() {
  int input;
  printf("Enter an integer 1-10:");
  scanf("%d", &input);
  if (input <= 10 && input >= 0) {
    printf("%d", input - 1);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  cmp    $0xa,%esi
  jbe    L19
L29:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L19:
  sub    $0x1,%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L29
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D10
data_type: string
name: parse_data
arguments: data_tag: D5
data_type: string

tool: D10: Enter an integer 1-10:

tool: D5: %d

assistant: ```c
int main() {
  int input;
  printf("Enter an integer 1-10:");
  scanf("%d", &input);
  if (input <= 10 && input >= 0) {
    printf("%d", input - 1);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rdi
  lea    D5(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  cmp    $0xa,%esi
  jbe    L29
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L29:
  sub    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L8
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D2
data_type: string
name: retrieve_data
arguments: label: D5
data_type: string

tool: D2: Enter an integer 1-10:

tool: D5: %d

assistant: ```c
int main() {
  int input;
  printf("Enter an integer 1-10:");
  scanf("%d", &input);
  if (input <= 10 && input >= 0) {
    printf("%d", input - 1);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D10(%rip),%rdi
  lea    D0(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  cmp    $0xa,%esi
  jbe    L26
L12:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L26:
  sub    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L12
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D10
type_infered: string
name: parse_data
arguments: data_label: D0
type_infered: string

tool: D10: Enter an integer 1-10:

tool: D0: %d

assistant: ```c
int main() {
  int input;
  printf("Enter an integer 1-10:");
  scanf("%d", &input);
  if (input <= 10 && input >= 0) {
    printf("%d", input - 1);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<multiply27>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L8
L36:
  movl   $0x0,-0x3c(%rbp)
  jmp    L35
L23:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L35:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L23
  addl   $0x1,-0x40(%rbp)
L8:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L36
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply27(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<multiply27>:
  endbr64
  test   %edi,%edi
  jle    L17
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L3:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L23:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L23
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  ret
```

assistant: ```c
void multiply27(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<multiply27>:
  endbr64
  test   %edi,%edi
  jle    L26
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L11:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L10:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L10
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L11
  pop    %rbx
  ret
L26:
  ret
```

assistant: ```c
void multiply27(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<multiply27>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L15
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L42:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L80
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L80
  nopl   0x0(%rax)
L36:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L36
  cmp    -0x24(%rsp),%r11d
  je     L40
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L40
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L40
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L40:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L42
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L80:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L62:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L62
  mov    -0x18(%rsp),%r8
  jmp    L40
```

assistant: ```c
void multiply27(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<h_activeratind>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  lea    D13(%rip),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L18
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
datatype: string
name: extract_symbolic_data
arguments: tag: D11
datatype: string

tool: D13: h_activeratind

tool: D11: CALLED %s ->

assistant: ```c
void h_activeratind(char *name, int argc, char **argv, void *user_data) {
  int i;
  printf("CALLED %s ->", __FUNCTION__);
  for (i = 0; i < argc; i++) {
    printf("%s\n", argv[i]);
  }
  printf("\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<h_activeratind>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  mov    %rdx,%r12
  lea    D5(%rip),%rsi
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L5
  mov    %r12,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x8(%r12,%rax,8),%rbp
L0:
  mov    (%rbx),%rdi
  call   <puts@plt>
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L0
L5:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
data_type: string
name: read_data
arguments: data_tag: D12
data_type: string

tool: D5: h_activeratind

tool: D12: CALLED %s ->

assistant: ```c
void h_activeratind(char *name, int argc, char **argv, void *user_data) {
  int i;
  printf("CALLED %s ->", __FUNCTION__);
  for (i = 0; i < argc; i++) {
    printf("%s\n", argv[i]);
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<h_activeratind>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %esi,%ebp
  lea    D6(%rip),%rdi
  push   %rbx
  lea    D19(%rip),%rsi
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L3
  lea    -0x1(%rbp),%eax
  lea    0x8(%rbx,%rax,8),%rbp
L15:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <puts@plt>
  cmp    %rbp,%rbx
  jne    L15
L3:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <putchar@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D6
type_guessed: string
name: retrieve_data
arguments: data_label: D19
type_guessed: string

tool: D6: CALLED %s ->

tool: D19: h_activeratind

assistant: ```c
void h_activeratind(char *name, int argc, char **argv, void *user_data) {
  int i;
  printf("CALLED %s ->", __FUNCTION__);
  for (i = 0; i < argc; i++) {
    printf("%s\n", argv[i]);
  }
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<h_activeratind>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %esi,%ebp
  lea    D18(%rip),%rdi
  push   %rbx
  lea    D4(%rip),%rsi
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L9
  lea    -0x1(%rbp),%eax
  lea    0x8(%rbx,%rax,8),%rbp
L18:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <puts@plt>
  cmp    %rbp,%rbx
  jne    L18
L9:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <putchar@plt>
```

tool_calls: name: parse_data
arguments: data_label: D18
type_guessed: string
name: parse_data
arguments: data_label: D4
type_guessed: string

tool: D18: CALLED %s ->

tool: D4: h_activeratind

assistant: ```c
void h_activeratind(char *name, int argc, char **argv, void *user_data) {
  int i;
  printf("CALLED %s ->", __FUNCTION__);
  for (i = 0; i < argc; i++) {
    printf("%s\n", argv[i]);
  }
  printf("\n");
}
```