ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  sub    $0x8,%rsp
  lea    D14(%rip),%rsi
  lea    D7(%rip),%rdi
  call   <compress@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D14
type_guessed: string
name: disassemble_data
arguments: data_label: D7
type_guessed: string

tool: D14: Opdracht3_compressed.c

tool: D7: Opdracht3.c

assistant: ```c
int main(int argc, char **argv) {
  compress("Opdracht3.c", "Opdracht3_compressed.c");
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rsi
  lea    D0(%rip),%rdi
  call   <compress@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D14
type_guessed: string
name: retrieve_data
arguments: tag: D0
type_guessed: string

tool: D14: Opdracht3_compressed.c

tool: D0: Opdracht3.c

assistant: ```c
int main(int argc, char **argv) {
  compress("Opdracht3.c", "Opdracht3_compressed.c");
  return (0);
}
```
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    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x40(%rbp)
  movl   $0x1e,-0x3c(%rbp)
  movl   $0x3c,-0x38(%rbp)
  movl   $0x64,-0x34(%rbp)
  movl   $0x1,-0x30(%rbp)
  movl   $0x2,-0x2c(%rbp)
  movl   $0x7,-0x28(%rbp)
  movl   $0x9,-0x24(%rbp)
  movl   $0x50,-0x20(%rbp)
  movl   $0x28,-0x1c(%rbp)
  movl   $0x37,-0x18(%rbp)
  movl   $0x16,-0x14(%rbp)
  movl   $0x63,-0x10(%rbp)
  movl   $0xd,-0x48(%rbp)
  movl   $0xa,-0x44(%rbp)
  mov    -0x48(%rbp),%eax
  lea    -0x1(%rax),%esi
  mov    -0x44(%rbp),%edx
  lea    -0x40(%rbp),%rax
  mov    %edx,%ecx
  mov    %esi,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <findRank@plt>
  mov    -0x40(%rbp,%rax,4),%edx
  mov    -0x44(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x4c(%rbp)
  jmp    L14
L3:
  mov    -0x4c(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4c(%rbp)
L14:
  mov    -0x4c(%rbp),%eax
  cmp    -0x48(%rbp),%eax
  jl     L3
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
data_type: string
name: unpack_data
arguments: tag: D11
data_type: string

tool: D3: 
 The element with rank %d is %d

tool: D11: %d 

assistant: ```c
int main(int argc, char const *argv[]) {
  int a[] = {10, 30, 60, 100, 1, 2, 7, 9, 80, 40, 55, 22, 99};
  int size = sizeof(a) / sizeof(a[0]);
  int rank = 10;
  printf("\n The element with rank %d is %d", rank, a[findRank(a, 0, size - 1, rank)]);
  printf("\n");
  for (int i = 0; i < size; ++i) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0xa,(%rsp)
  movl   $0x1e,0x4(%rsp)
  movl   $0x3c,0x8(%rsp)
  movl   $0x64,0xc(%rsp)
  movl   $0x1,0x10(%rsp)
  movl   $0x2,0x14(%rsp)
  movl   $0x7,0x18(%rsp)
  movl   $0x9,0x1c(%rsp)
  movl   $0x50,0x20(%rsp)
  movl   $0x28,0x24(%rsp)
  movl   $0x37,0x28(%rsp)
  movl   $0x16,0x2c(%rsp)
  movl   $0x63,0x30(%rsp)
  mov    %rsp,%rbx
  mov    $0xa,%ecx
  mov    $0xc,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <findRank@plt>
  mov    (%rsp,%rax,4),%edx
  mov    $0xa,%esi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    0x34(%rsp),%r12
  lea    D3(%rip),%rbp
L8:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D0
type_guessed: string
name: disassemble_data
arguments: label: D3
type_guessed: string

tool: D0: 
 The element with rank %d is %d

tool: D3: %d 

assistant: ```c
int main(int argc, char const *argv[]) {
  int a[] = {10, 30, 60, 100, 1, 2, 7, 9, 80, 40, 55, 22, 99};
  int size = sizeof(a) / sizeof(a[0]);
  int rank = 10;
  printf("\n The element with rank %d is %d", rank, a[findRank(a, 0, size - 1, rank)]);
  printf("\n");
  for (int i = 0; i < size; ++i) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xa,%ecx
  mov    $0xc,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x1e0000000a,%rax
  mov    %rsp,%rbx
  movl   $0x63,0x30(%rsp)
  lea    0x34(%rsp),%r12
  mov    %rbx,%rdi
  mov    %rax,(%rsp)
  movabs $0x640000003c,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x200000001,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x900000007,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x2800000050,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x1600000037,%rax
  mov    %rax,0x28(%rsp)
  call   <findRank@plt>
  mov    $0xa,%esi
  lea    D7(%rip),%rdi
  mov    (%rsp,%rax,4),%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  xchg   %ax,%ax
L10:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L10
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D1
datatype: string
name: parse_data
arguments: data_tag: D7
datatype: string

tool: D1: %d 

tool: D7: 
 The element with rank %d is %d

assistant: ```c
int main(int argc, char const *argv[]) {
  int a[] = {10, 30, 60, 100, 1, 2, 7, 9, 80, 40, 55, 22, 99};
  int size = sizeof(a) / sizeof(a[0]);
  int rank = 10;
  printf("\n The element with rank %d is %d", rank, a[findRank(a, 0, size - 1, rank)]);
  printf("\n");
  for (int i = 0; i < size; ++i) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xa,%ecx
  mov    $0xc,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  movdqa D11(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  movl   $0x63,0x30(%rsp)
  lea    0x34(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D37(%rip),%xmm0
  mov    %rbx,%rdi
  movaps %xmm0,0x10(%rsp)
  movdqa D30(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <findRank@plt>
  mov    $0xa,%esi
  lea    D24(%rip),%rdi
  mov    (%rsp,%rax,4),%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  nopl   (%rax)
L1:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L1
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D6
type_infered: string
name: read_data
arguments: data_tag: D11
type_infered: byte[16]
name: read_data
arguments: data_tag: D37
type_infered: byte[16]
name: read_data
arguments: data_tag: D30
type_infered: byte[16]
name: read_data
arguments: data_tag: D24
type_infered: string

tool: D6: %d 

tool: {"D11": [0x0a,0x00,0x00,0x00,0x1e,0x00,0x00,0x00,0x3c,0x00,0x00,0x00,0x64,0x00,0x00,0x00]}

tool: {"D37": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x09,0x00,0x00,0x00]}

tool: {"D30": [0x50,0x00,0x00,0x00,0x28,0x00,0x00,0x00,0x37,0x00,0x00,0x00,0x16,0x00,0x00,0x00]}

tool: D24: 
 The element with rank %d is %d

assistant: ```c
int main(int argc, char const *argv[]) {
  int a[] = {10, 30, 60, 100, 1, 2, 7, 9, 80, 40, 55, 22, 99};
  int size = sizeof(a) / sizeof(a[0]);
  int rank = 10;
  printf("\n The element with rank %d is %d", rank, a[findRank(a, 0, size - 1, rank)]);
  printf("\n");
  for (int i = 0; i < size; ++i) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<christmas_tree>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  movl   $0x1,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x2,-0x18(%rbp)
  jmp    L78
L41:
  addl   $0x2,-0xc(%rbp)
  addl   $0x1,-0x18(%rbp)
L78:
  mov    -0x18(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jle    L41
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L16
L4:
  movl   $0x1,-0x14(%rbp)
  jmp    L70
L20:
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L70:
  mov    -0x14(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L20
  movl   $0x1,-0x10(%rbp)
  jmp    L28
L57:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L28:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L57
  addl   $0x2,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L16:
  mov    -0x18(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jle    L4
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
data_type: string
name: extract_symbolic_data
arguments: data_tag: D1
data_type: string
name: extract_symbolic_data
arguments: data_tag: D6
data_type: string

tool: D17:  

tool: D1: *

tool: D6: 


assistant: ```c
void christmas_tree(int height) {
  int i, j, p;
  int last = 1;
  int chars = 1;
  for (i = 2; i <= height; i++) {
    last = last + 2;
  }
  int spaces = last / 2;
  for (i = 1; i <= height; i++) {
    for (j = 1; j <= spaces; j++) {
      printf(" ");
    }
    for (p = 1; p <= chars; p++) {
      printf("*");
    }
    chars = chars + 2;
    spaces = spaces - 1;
    printf("\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<christmas_tree>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r15d
  cmp    $0x1,%edi
  jle    L71
  lea    0x1(%rdi),%edx
  mov    $0x2,%eax
L75:
  add    $0x1,%eax
  cmp    %edx,%eax
  jne    L75
  lea    -0x1(%r15,%r15,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
L27:
  mov    %eax,%ebp
  mov    $0x1,%r12d
  lea    D11(%rip),%r14
  lea    D18(%rip),%r13
  add    $0x1,%eax
  mov    %eax,0xc(%rsp)
  jmp    L35
L71:
  je     L14
L0:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  mov    %r14,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jle    L9
L12:
  test   %r12d,%r12d
  jle    L106
  mov    $0x1,%ebx
L8:
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %r12d,%ebx
  jle    L8
L106:
  add    $0x2,%r12d
  sub    $0x1,%ebp
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  sub    %ebp,%eax
  cmp    %eax,%r15d
  jl     L0
L35:
  mov    $0x1,%ebx
  test   %ebp,%ebp
  jg     L9
  jmp    L12
L14:
  mov    $0x0,%eax
  jmp    L27
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
type_guessed: string
name: retrieve_data
arguments: data_tag: D18
type_guessed: string
name: retrieve_data
arguments: data_tag: D22
type_guessed: string

tool: D11:  

tool: D18: *

tool: D22: 


assistant: ```c
void christmas_tree(int height) {
  int i, j, p;
  int last = 1;
  int chars = 1;
  for (i = 2; i <= height; i++) {
    last = last + 2;
  }
  int spaces = last / 2;
  for (i = 1; i <= height; i++) {
    for (j = 1; j <= spaces; j++) {
      printf(" ");
    }
    for (p = 1; p <= chars; p++) {
      printf("*");
    }
    chars = chars + 2;
    spaces = spaces - 1;
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<christmas_tree>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%rdi,%rdi,1),%ebx
  sar    %ebx
  sub    $0x18,%rsp
  mov    %edi,0x8(%rsp)
  cmp    $0x1,%edi
  jle    L29
L6:
  lea    0x1(%rbx),%eax
  mov    $0x1,%ebp
  lea    D0(%rip),%r13
  mov    %eax,0xc(%rsp)
  lea    D11(%rip),%r15
  lea    D28(%rip),%r12
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    $0x1,%r14d
  test   %ebx,%ebx
  jle    L5
  nopw   0x0(%rax,%rax,1)
L43:
  mov    %r12,%rdi
  add    $0x1,%r14d
  call   <printf@plt>
  cmp    %ebx,%r14d
  jle    L43
L5:
  mov    $0x1,%r14d
  nopw   0x0(%rax,%rax,1)
L53:
  mov    %r13,%rdi
  add    $0x1,%r14d
  call   <printf@plt>
  cmp    %ebp,%r14d
  jle    L53
  mov    %r15,%rdi
  sub    $0x1,%ebx
  add    $0x2,%ebp
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  sub    %ebx,%eax
  cmp    %eax,0x8(%rsp)
  jge    L13
L66:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L29:
  jne    L66
  xor    %ebx,%ebx
  jmp    L6
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type_infered: string
name: disassemble_data
arguments: data_label: D11
type_infered: string
name: disassemble_data
arguments: data_label: D28
type_infered: string

tool: D0: *

tool: D11: 


tool: D28:  

assistant: ```c
void christmas_tree(int height) {
  int i, j, p;
  int last = 1;
  int chars = 1;
  for (i = 2; i <= height; i++) {
    last = last + 2;
  }
  int spaces = last / 2;
  for (i = 1; i <= height; i++) {
    for (j = 1; j <= spaces; j++) {
      printf(" ");
    }
    for (p = 1; p <= chars; p++) {
      printf("*");
    }
    chars = chars + 2;
    spaces = spaces - 1;
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<christmas_tree>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%rdi,%rdi,1),%ebx
  sar    %ebx
  sub    $0x18,%rsp
  mov    %edi,0x8(%rsp)
  cmp    $0x1,%edi
  jle    L67
L38:
  lea    0x1(%rbx),%eax
  mov    $0x1,%ebp
  lea    D7(%rip),%r13
  mov    %eax,0xc(%rsp)
  lea    D19(%rip),%r15
  lea    D6(%rip),%r12
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    $0x1,%r14d
  test   %ebx,%ebx
  jle    L58
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %r12,%rdi
  add    $0x1,%r14d
  call   <printf@plt>
  cmp    %ebx,%r14d
  jle    L64
L58:
  mov    $0x1,%r14d
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %r13,%rdi
  add    $0x1,%r14d
  call   <printf@plt>
  cmp    %ebp,%r14d
  jle    L11
  mov    %r15,%rdi
  sub    $0x1,%ebx
  add    $0x2,%ebp
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  sub    %ebx,%eax
  cmp    %eax,0x8(%rsp)
  jge    L15
L45:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L67:
  jne    L45
  xor    %ebx,%ebx
  jmp    L38
```

tool_calls: name: read_data
arguments: label: D7
type_infered: string
name: read_data
arguments: label: D19
type_infered: string
name: read_data
arguments: label: D6
type_infered: string

tool: D7: *

tool: D19: 


tool: D6:  

assistant: ```c
void christmas_tree(int height) {
  int i, j, p;
  int last = 1;
  int chars = 1;
  for (i = 2; i <= height; i++) {
    last = last + 2;
  }
  int spaces = last / 2;
  for (i = 1; i <= height; i++) {
    for (j = 1; j <= spaces; j++) {
      printf(" ");
    }
    for (p = 1; p <= chars; p++) {
      printf("*");
    }
    chars = chars + 2;
    spaces = spaces - 1;
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_strequ>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L43
  cmpq   $0x0,-0x10(%rbp)
  jne    L43
  mov    $0x1,%eax
  jmp    L3
L43:
  cmpq   $0x0,-0x8(%rbp)
  jne    L12
  cmpq   $0x0,-0x10(%rbp)
  jne    L49
L12:
  cmpq   $0x0,-0x10(%rbp)
  jne    L93
  cmpq   $0x0,-0x8(%rbp)
  je     L93
L49:
  mov    $0x0,%eax
  jmp    L3
L126:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L93:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L21
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L21
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L126
L21:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L26
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L26
  mov    $0x1,%eax
  jmp    L3
L26:
  mov    $0x0,%eax
L3:
  pop    %rbp
  ret
```

assistant: ```c
int ft_strequ(char const *s1, char const *s2) {
  if (s1 == NULL && s2 == NULL)
    return (1);
  if ((s1 == NULL && s2 != NULL) || (s2 == NULL && s1 != NULL))
    return (0);
  while (*s1 != '\0' && *s1 == *s2 && *s2 != '\0') {
    s1++;
    s2++;
  }
  if (*s2 == '\0' && *s1 == '\0')
    return (1);
  else
    return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ft_strequ>:
  endbr64
  mov    $0x1,%eax
  mov    %rdi,%rcx
  or     %rsi,%rcx
  je     L4
  test   %rdi,%rdi
  sete   %cl
  test   %rsi,%rsi
  sete   %dl
  mov    $0x0,%eax
  cmp    %dl,%cl
  jne    L4
  movzbl (%rdi),%eax
  test   %al,%al
  je     L42
L34:
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L42
  cmp    %al,%dl
  jne    L42
  add    $0x1,%rdi
  add    $0x1,%rsi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L34
L42:
  or     (%rsi),%al
  sete   %al
  movzbl %al,%eax
L4:
  ret
```

assistant: ```c
int ft_strequ(char const *s1, char const *s2) {
  if (s1 == NULL && s2 == NULL)
    return (1);
  if ((s1 == NULL && s2 != NULL) || (s2 == NULL && s1 != NULL))
    return (0);
  while (*s1 != '\0' && *s1 == *s2 && *s2 != '\0') {
    s1++;
    s2++;
  }
  if (*s2 == '\0' && *s1 == '\0')
    return (1);
  else
    return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strequ>:
  endbr64
  mov    %rdi,%rdx
  mov    $0x1,%eax
  or     %rsi,%rdx
  je     L3
  test   %rdi,%rdi
  sete   %cl
  test   %rsi,%rsi
  sete   %dl
  xor    %eax,%eax
  cmp    %dl,%cl
  je     L25
L3:
  ret
  nopl   0x0(%rax)
L21:
  cmp    %al,(%rsi)
  jne    L38
  add    $0x1,%rdi
  add    $0x1,%rsi
L25:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L21
  xor    %eax,%eax
  cmpb   $0x0,(%rsi)
  sete   %al
  ret
  nopl   0x0(%rax)
L38:
  xor    %eax,%eax
  ret
```

assistant: ```c
int ft_strequ(char const *s1, char const *s2) {
  if (s1 == NULL && s2 == NULL)
    return (1);
  if ((s1 == NULL && s2 != NULL) || (s2 == NULL && s1 != NULL))
    return (0);
  while (*s1 != '\0' && *s1 == *s2 && *s2 != '\0') {
    s1++;
    s2++;
  }
  if (*s2 == '\0' && *s1 == '\0')
    return (1);
  else
    return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_strequ>:
  endbr64
  mov    %rdi,%rdx
  mov    $0x1,%eax
  or     %rsi,%rdx
  je     L32
  test   %rdi,%rdi
  sete   %cl
  test   %rsi,%rsi
  sete   %dl
  xor    %eax,%eax
  cmp    %dl,%cl
  je     L16
L32:
  ret
  nopl   0x0(%rax)
L24:
  cmp    %al,(%rsi)
  jne    L34
  add    $0x1,%rdi
  add    $0x1,%rsi
L16:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L24
  xor    %eax,%eax
  cmpb   $0x0,(%rsi)
  sete   %al
  ret
  nopl   0x0(%rax)
L34:
  xor    %eax,%eax
  ret
```

assistant: ```c
int ft_strequ(char const *s1, char const *s2) {
  if (s1 == NULL && s2 == NULL)
    return (1);
  if ((s1 == NULL && s2 != NULL) || (s2 == NULL && s1 != NULL))
    return (0);
  while (*s1 != '\0' && *s1 == *s2 && *s2 != '\0') {
    s1++;
    s2++;
  }
  if (*s2 == '\0' && *s1 == '\0')
    return (1);
  else
    return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<num_bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x10(%rbp),%eax
  mov    $0xa,%edx
  lea    D2(%rip),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <mpz_init_set_str@plt>
  mov    -0xc(%rbp),%eax
  mov    $0xa,%edx
  lea    D10(%rip),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <mpz_init_set_str@plt>
  mov    -0x8(%rbp),%eax
  mov    $0xa,%edx
  lea    D1(%rip),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <mpz_init_set_str@plt>
  mov    -0x4(%rbp),%eax
  mov    $0xa,%edx
  lea    D6(%rip),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <mpz_init_set_str@plt>
L0:
  mov    -0x4(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <mpz_powm@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x14(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_add@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <mpz_cmp@plt>
  test   %rax,%rax
  jns    L0
  mov    -0x14(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_sub@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <mpz_clear@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <mpz_clear@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <mpz_clear@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <mpz_clear@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type_infered: string
name: extract_symbolic_data
arguments: tag: D10
type_infered: string
name: extract_symbolic_data
arguments: tag: D1
type_infered: string
name: extract_symbolic_data
arguments: tag: D6
type_infered: string

tool: D2: 100

tool: D10: 0

tool: D1: 2

tool: D6: 1000000000000000000000000000

assistant: ```c
void num_bits(mpz_t bits, mpz_t number) {
  mpz_t pow, ground, counter, module;
  mpz_init_set_str(counter, "100", 10);
  mpz_init_set_str(pow, "0", 10);
  mpz_init_set_str(ground, "2", 10);
  mpz_init_set_str(module, "1000000000000000000000000000", 10);
  do {
    mpz_powm(counter, ground, pow, module);
    mpz_add(pow, bits, pow);
  } while (mpz_cmp(number, counter) >= 0);
  mpz_sub(bits, pow, bits);
  mpz_clear(pow);
  mpz_clear(ground);
  mpz_clear(counter);
  mpz_clear(module);
}
```