ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x53202e412064655a,%rax
  mov    %rax,0xc(%rsp)
  movl   $0x776168,0x14(%rsp)
  movl   $0x64655a,0x8(%rsp)
  movb   $0x4e,0x8(%rsp)
  mov    $0x4,%esi
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x14,%esi
  lea    D54(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x5,%esi
  lea    D59(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xc,%edx
  mov    $0x42,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xc,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xc,%esi
  lea    D63(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D58(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D32
type: string
name: read_data
arguments: tag: D54
type: string
name: read_data
arguments: tag: D59
type: string
name: read_data
arguments: tag: D10
type: string
name: read_data
arguments: tag: D6
type: string
name: read_data
arguments: tag: D40
type: string
name: read_data
arguments: tag: D1
type: string
name: read_data
arguments: tag: D63
type: string
name: read_data
arguments: tag: D58
type: string

tool: D32: The size of an int: %ld


tool: D54: The size of areas (int[]): %ld


tool: D59: The number of ints in areas: %ld.


tool: D10: The first area is %d, the 2nd is %d.


tool: D6: The size of a char: %ld


tool: D40: The size of name (char[]): %ld


tool: D1: The size of full_name (char[]): %ld


tool: D63: The number of chars: %ld


tool: D58: name="%s" and full_name="%s"


assistant: ```c
int main(int argc, char *argv[]) {
  char full_name[] = {'Z', 'e', 'd', ' ', 'A', '.', ' ', 'S', 'h', 'a', 'w', '\0'};
  int areas[] = {'B', 12, 13, 14, 20};
  char name[] = "Zed";
  areas[10] = 100;
  areas[11] = 'A';
  name[0] = 'N';
  printf("The size of an int: %ld\n", sizeof(int));
  printf("The size of areas (int[]): %ld\n", sizeof(areas));
  printf("The number of ints in areas: %ld.\n", sizeof(areas) / sizeof(int));
  printf("The first area is %d, the 2nd is %d.\n", areas[0], areas[1]);
  printf("The size of a char: %ld\n", sizeof(char));
  printf("The size of name (char[]): %ld\n", sizeof(name) / sizeof(char));
  printf("The size of full_name (char[]): %ld\n", sizeof(full_name));
  printf("The number of chars: %ld\n", sizeof(full_name) / sizeof(char));
  printf("name=\"%s\" and full_name=\"%s\"\n", name, full_name);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x4,%esi
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x53202e412064655a,%rax
  movl   $0x64655a,0x8(%rsp)
  movl   $0x776168,0x14(%rsp)
  movb   $0x4e,0x8(%rsp)
  mov    %rax,0xc(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x14,%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x5,%esi
  lea    D43(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%edx
  mov    $0x42,%esi
  xor    %eax,%eax
  lea    D58(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D80(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D83(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%esi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%esi
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D87(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D14
data_type: string
name: disassemble_data
arguments: tag: D6
data_type: string
name: disassemble_data
arguments: tag: D43
data_type: string
name: disassemble_data
arguments: tag: D58
data_type: string
name: disassemble_data
arguments: tag: D80
data_type: string
name: disassemble_data
arguments: tag: D83
data_type: string
name: disassemble_data
arguments: tag: D29
data_type: string
name: disassemble_data
arguments: tag: D44
data_type: string
name: disassemble_data
arguments: tag: D87
data_type: string

tool: D14: The size of an int: %ld


tool: D6: The size of areas (int[]): %ld


tool: D43: The number of ints in areas: %ld.


tool: D58: The first area is %d, the 2nd is %d.


tool: D80: The size of a char: %ld


tool: D83: The size of name (char[]): %ld


tool: D29: The size of full_name (char[]): %ld


tool: D44: The number of chars: %ld


tool: D87: name="%s" and full_name="%s"


assistant: ```c
int main(int argc, char *argv[]) {
  char full_name[] = {'Z', 'e', 'd', ' ', 'A', '.', ' ', 'S', 'h', 'a', 'w', '\0'};
  int areas[] = {'B', 12, 13, 14, 20};
  char name[] = "Zed";
  areas[10] = 100;
  areas[11] = 'A';
  name[0] = 'N';
  printf("The size of an int: %ld\n", sizeof(int));
  printf("The size of areas (int[]): %ld\n", sizeof(areas));
  printf("The number of ints in areas: %ld.\n", sizeof(areas) / sizeof(int));
  printf("The first area is %d, the 2nd is %d.\n", areas[0], areas[1]);
  printf("The size of a char: %ld\n", sizeof(char));
  printf("The size of name (char[]): %ld\n", sizeof(name) / sizeof(char));
  printf("The size of full_name (char[]): %ld\n", sizeof(full_name));
  printf("The number of chars: %ld\n", sizeof(full_name) / sizeof(char));
  printf("name=\"%s\" and full_name=\"%s\"\n", name, full_name);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x4,%esi
  lea    D6(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x53202e412064655a,%rax
  movl   $0x64655a,0x8(%rsp)
  movl   $0x776168,0x14(%rsp)
  movb   $0x4e,0x8(%rsp)
  mov    %rax,0xc(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x14,%esi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x5,%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%edx
  mov    $0x42,%esi
  xor    %eax,%eax
  lea    D57(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%esi
  lea    D83(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%esi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D71(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
datatype: string
name: retrieve_data
arguments: label: D22
datatype: string
name: retrieve_data
arguments: label: D11
datatype: string
name: retrieve_data
arguments: label: D57
datatype: string
name: retrieve_data
arguments: label: D20
datatype: string
name: retrieve_data
arguments: label: D58
datatype: string
name: retrieve_data
arguments: label: D83
datatype: string
name: retrieve_data
arguments: label: D27
datatype: string
name: retrieve_data
arguments: label: D71
datatype: string

tool: D6: The size of an int: %ld


tool: D22: The size of areas (int[]): %ld


tool: D11: The number of ints in areas: %ld.


tool: D57: The first area is %d, the 2nd is %d.


tool: D20: The size of a char: %ld


tool: D58: The size of name (char[]): %ld


tool: D83: The size of full_name (char[]): %ld


tool: D27: The number of chars: %ld


tool: D71: name="%s" and full_name="%s"


assistant: ```c
int main(int argc, char *argv[]) {
  char full_name[] = {'Z', 'e', 'd', ' ', 'A', '.', ' ', 'S', 'h', 'a', 'w', '\0'};
  int areas[] = {'B', 12, 13, 14, 20};
  char name[] = "Zed";
  areas[10] = 100;
  areas[11] = 'A';
  name[0] = 'N';
  printf("The size of an int: %ld\n", sizeof(int));
  printf("The size of areas (int[]): %ld\n", sizeof(areas));
  printf("The number of ints in areas: %ld.\n", sizeof(areas) / sizeof(int));
  printf("The first area is %d, the 2nd is %d.\n", areas[0], areas[1]);
  printf("The size of a char: %ld\n", sizeof(char));
  printf("The size of name (char[]): %ld\n", sizeof(name) / sizeof(char));
  printf("The size of full_name (char[]): %ld\n", sizeof(full_name));
  printf("The number of chars: %ld\n", sizeof(full_name) / sizeof(char));
  printf("name=\"%s\" and full_name=\"%s\"\n", name, full_name);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<minKey>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L33
L36:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L29
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jle    L29
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L29:
  addl   $0x1,-0x4(%rbp)
L33:
  cmpl   $0x4,-0x4(%rbp)
  jle    L36
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int minKey(int key[], int mstSet[]) {
  int min = INT_MAX, min_index, v;
  for (v = 0; v < 5; v++)
    if (mstSet[v] == 0 && key[v] < min)
      min = key[v], min_index = v;
  return min_index;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<minKey>:
  endbr64
  mov    D4(%rip),%rax
  mov    (%rax),%r8d
  mov    $0x0,%eax
  jmp    L37
L3:
  add    $0x1,%rax
  cmp    $0x5,%rax
  je     L38
L37:
  cmpl   $0x0,(%rsi,%rax,4)
  jne    L3
  mov    (%rdi,%rax,4),%ecx
  cmp    %r8d,%ecx
  jge    L3
  mov    %eax,%r9d
  mov    %ecx,%r8d
  jmp    L3
L38:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int minKey(int key[], int mstSet[]) {
  int min = INT_MAX, min_index, v;
  for (v = 0; v < 5; v++)
    if (mstSet[v] == 0 && key[v] < min)
      min = key[v], min_index = v;
  return min_index;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<minKey>:
  endbr64
  mov    D3(%rip),%rax
  xor    %r8d,%r8d
  mov    (%rax),%ecx
  xor    %eax,%eax
L25:
  mov    (%rsi,%rax,4),%edx
  test   %edx,%edx
  jne    L24
  mov    (%rdi,%rax,4),%edx
  cmp    %ecx,%edx
  jge    L24
  mov    %eax,%r8d
  mov    %edx,%ecx
L24:
  add    $0x1,%rax
  cmp    $0x5,%rax
  jne    L25
  mov    %r8d,%eax
  ret
```

assistant: ```c
int minKey(int key[], int mstSet[]) {
  int min = INT_MAX, min_index, v;
  for (v = 0; v < 5; v++)
    if (mstSet[v] == 0 && key[v] < min)
      min = key[v], min_index = v;
  return min_index;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<minKey>:
  endbr64
  mov    D1(%rip),%rax
  mov    (%rsi),%r10d
  mov    (%rax),%edx
  test   %r10d,%r10d
  jne    L9
  mov    (%rdi),%eax
  cmp    %eax,%edx
  cmovg  %eax,%edx
L9:
  mov    0x4(%rsi),%eax
  test   %eax,%eax
  jne    L89
  mov    0x4(%rdi),%ecx
  cmp    %ecx,%edx
  jg     L16
L80:
  mov    0x8(%rsi),%r9d
  test   %r9d,%r9d
  jne    L45
  mov    0x8(%rdi),%ecx
  cmp    %ecx,%edx
  jg     L106
L45:
  mov    0xc(%rsi),%r8d
  test   %r8d,%r8d
  jne    L2
  mov    0xc(%rdi),%ecx
  cmp    %ecx,%edx
  jg     L29
L2:
  mov    0x10(%rsi),%ecx
  test   %ecx,%ecx
  jne    L68
  cmp    %edx,0x10(%rdi)
  mov    $0x4,%edx
  cmovl  %edx,%eax
L68:
  ret
  nopl   (%rax)
L89:
  xor    %eax,%eax
  jmp    L80
  nopl   0x0(%rax)
L29:
  mov    %ecx,%edx
  mov    $0x3,%eax
  jmp    L2
  nopl   0x0(%rax)
L106:
  mov    %ecx,%edx
  mov    $0x2,%eax
  jmp    L45
  nopl   0x0(%rax)
L16:
  mov    %ecx,%edx
  mov    $0x1,%eax
  jmp    L80
```

assistant: ```c
int minKey(int key[], int mstSet[]) {
  int min = INT_MAX, min_index, v;
  for (v = 0; v < 5; v++)
    if (mstSet[v] == 0 && key[v] < min)
      min = key[v], min_index = v;
  return min_index;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<clean_block>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L28
L10:
  movl   $0x0,-0x4(%rbp)
  jmp    L24
L39:
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x7,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  pxor   %xmm0,%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  addl   $0x1,-0x4(%rbp)
L24:
  cmpl   $0x1f,-0x4(%rbp)
  jle    L39
  addl   $0x1,-0x8(%rbp)
L28:
  cmpl   $0x1f,-0x8(%rbp)
  jle    L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void clean_block(float Dst[32][32]) {
  int i, j;
  for (i = 0; i < 32; i++)
    for (j = 0; j < 32; j++)
      Dst[i][j] = 0.0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<clean_block>:
  endbr64
  mov    %rdi,%rcx
  lea    0x80(%rdi),%rdx
  add    $0x1080,%rcx
L18:
  lea    -0x80(%rdx),%rax
L19:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L19
  sub    $0xffffffffffffff80,%rdx
  cmp    %rcx,%rdx
  jne    L18
  ret
```

assistant: ```c
void clean_block(float Dst[32][32]) {
  int i, j;
  for (i = 0; i < 32; i++)
    for (j = 0; j < 32; j++)
      Dst[i][j] = 0.0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<clean_block>:
  endbr64
  movq   $0x0,(%rdi)
  mov    %rdi,%rax
  lea    0x8(%rdi),%rdi
  movq   $0x0,0xff0(%rdi)
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%rax
  mov    %rax,%rcx
  xor    %eax,%eax
  add    $0x1000,%ecx
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  ret
```

assistant: ```c
void clean_block(float Dst[32][32]) {
  int i, j;
  for (i = 0; i < 32; i++)
    for (j = 0; j < 32; j++)
      Dst[i][j] = 0.0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<clean_block>:
  endbr64
  movq   $0x0,(%rdi)
  mov    %rdi,%rax
  lea    0x8(%rdi),%rdi
  movq   $0x0,0xff0(%rdi)
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%rax
  mov    %rax,%rcx
  xor    %eax,%eax
  add    $0x1000,%ecx
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  ret
```

assistant: ```c
void clean_block(float Dst[32][32]) {
  int i, j;
  for (i = 0; i < 32; i++)
    for (j = 0; j < 32; j++)
      Dst[i][j] = 0.0;
}
```
user: What is the c source code of the assembly code below?

```asm
<ft_atoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  jmp    L155
L146:
  addq   $0x1,-0x18(%rbp)
L155:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L125
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L146
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L146
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L146
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xd,%al
  je     L146
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xb,%al
  je     L146
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xc,%al
  je     L146
L125:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  sete   %al
  mov    %al,-0x5(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2b,%al
  je     L126
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L80
L126:
  addq   $0x1,-0x18(%rbp)
L80:
  movl   $0x0,-0x4(%rbp)
  jmp    L123
L128:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L101
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L101
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %ecx,%eax
  mov    %eax,-0x4(%rbp)
L123:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L128
L101:
  cmpb   $0x0,-0x5(%rbp)
  je     L52
  mov    -0x4(%rbp),%eax
  neg    %eax
  jmp    L23
L52:
  mov    -0x4(%rbp),%eax
L23:
  pop    %rbp
  ret
```

assistant: ```c
int ft_atoi(const char *str) {
  int n;
  char s;
  while (*str && (*str == '\n' || *str == ' ' || *str == '\t' || *str == '\r' || *str == '\v' || *str == '\f'))
    str++;
  s = (*str == 45 ? 1 : 0);
  if (*str == 43 || *str == 45)
    str++;
  n = 0;
  while (*str) {
    if (*str > 47 && *str < 58)
      n = n * 10 + ((int)(*str++) - 48);
    else
      break;
  }
  return (s ? -n : n);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_atoi>:
  endbr64
  movzbl (%rdi),%eax
  mov    $0x0,%edx
  test   %al,%al
  jne    L68
  jmp    L119
L4:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L109
L68:
  mov    %eax,%ecx
  lea    -0x9(%rax),%edx
  cmp    $0x4,%dl
  jbe    L4
  cmp    $0x20,%al
  je     L4
  cmp    $0x2d,%al
  sete   %r8b
  sub    $0x2b,%ecx
  test   $0xfd,%cl
  jne    L94
  lea    0x1(%rdi),%rcx
  movzbl 0x1(%rdi),%eax
  test   %al,%al
  je     L32
L103:
  mov    $0x0,%edx
L2:
  lea    -0x30(%rax),%esi
  cmp    $0x9,%sil
  ja     L11
  lea    (%rdx,%rdx,4),%edx
  add    $0x1,%rcx
  movsbl %al,%eax
  lea    -0x30(%rax,%rdx,2),%edx
  movzbl (%rcx),%eax
  test   %al,%al
  jne    L2
L11:
  mov    %edx,%eax
  neg    %eax
  test   %r8b,%r8b
  cmovne %eax,%edx
L119:
  mov    %edx,%eax
  ret
L32:
  mov    $0x0,%edx
  jmp    L11
L109:
  mov    $0x0,%r8d
L94:
  movzbl (%rdi),%eax
  test   %al,%al
  je     L87
  mov    %rdi,%rcx
  jmp    L103
L87:
  mov    $0x0,%edx
  jmp    L119
```

assistant: ```c
int ft_atoi(const char *str) {
  int n;
  char s;
  while (*str && (*str == '\n' || *str == ' ' || *str == '\t' || *str == '\r' || *str == '\v' || *str == '\f'))
    str++;
  s = (*str == 45 ? 1 : 0);
  if (*str == 43 || *str == 45)
    str++;
  n = 0;
  while (*str) {
    if (*str > 47 && *str < 58)
      n = n * 10 + ((int)(*str++) - 48);
    else
      break;
  }
  return (s ? -n : n);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ft_atoi>:
  endbr64
  jmp    L88
  cs nopw 0x0(%rax,%rax,1)
L103:
  lea    -0x9(%rdx),%eax
  cmp    $0x4,%al
  jbe    L99
  cmp    $0x20,%dl
  jne    L87
L99:
  add    $0x1,%rdi
L88:
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L103
L27:
  xor    %r8d,%r8d
L117:
  mov    %r8d,%eax
  ret
L87:
  lea    -0x2b(%rdx),%eax
  test   $0xfd,%al
  jne    L70
  movzbl 0x1(%rdi),%eax
  lea    0x1(%rdi),%rcx
  test   %al,%al
  je     L118
L11:
  xor    %r8d,%r8d
  jmp    L72
  cs nopw 0x0(%rax,%rax,1)
L19:
  movsbl %al,%eax
  lea    (%r8,%r8,4),%esi
  add    $0x1,%rcx
  lea    (%rax,%rsi,2),%r8d
  movzbl (%rcx),%eax
  test   %al,%al
  je     L47
L72:
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L19
L47:
  cmp    $0x2d,%dl
  jne    L117
  neg    %r8d
  mov    %r8d,%eax
  ret
L118:
  xor    %r8d,%r8d
  jmp    L47
L70:
  movzbl (%rdi),%eax
  mov    %rdi,%rcx
  test   %al,%al
  jne    L11
  jmp    L27
```

assistant: ```c
int ft_atoi(const char *str) {
  int n;
  char s;
  while (*str && (*str == '\n' || *str == ' ' || *str == '\t' || *str == '\r' || *str == '\v' || *str == '\f'))
    str++;
  s = (*str == 45 ? 1 : 0);
  if (*str == 43 || *str == 45)
    str++;
  n = 0;
  while (*str) {
    if (*str > 47 && *str < 58)
      n = n * 10 + ((int)(*str++) - 48);
    else
      break;
  }
  return (s ? -n : n);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_atoi>:
  endbr64
  jmp    L57
  cs nopw 0x0(%rax,%rax,1)
L19:
  lea    -0x9(%rdx),%eax
  cmp    $0x4,%al
  jbe    L84
  cmp    $0x20,%dl
  jne    L5
L84:
  add    $0x1,%rdi
L57:
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L19
  xor    %r8d,%r8d
L102:
  mov    %r8d,%eax
  ret
L5:
  lea    -0x2b(%rdx),%eax
  test   $0xfd,%al
  jne    L74
  movzbl 0x1(%rdi),%eax
  lea    0x1(%rdi),%rcx
  test   %al,%al
  je     L83
L91:
  xor    %r8d,%r8d
  jmp    L92
  cs nopw 0x0(%rax,%rax,1)
L10:
  movsbl %al,%eax
  lea    (%r8,%r8,4),%esi
  add    $0x1,%rcx
  lea    (%rax,%rsi,2),%r8d
  movzbl (%rcx),%eax
  test   %al,%al
  je     L118
L92:
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L10
L118:
  cmp    $0x2d,%dl
  jne    L102
  neg    %r8d
  mov    %r8d,%eax
  ret
L83:
  xor    %r8d,%r8d
  jmp    L118
L74:
  mov    %rdi,%rcx
  mov    %edx,%eax
  jmp    L91
```

assistant: ```c
int ft_atoi(const char *str) {
  int n;
  char s;
  while (*str && (*str == '\n' || *str == ' ' || *str == '\t' || *str == '\r' || *str == '\v' || *str == '\f'))
    str++;
  s = (*str == 45 ? 1 : 0);
  if (*str == 43 || *str == 45)
    str++;
  n = 0;
  while (*str) {
    if (*str > 47 && *str < 58)
      n = n * 10 + ((int)(*str++) - 48);
    else
      break;
  }
  return (s ? -n : n);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<base64_encode>:
  endbr64
  mov    %rdx,%r10
  test   %esi,%esi
  jle    L7
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r9d
  mov    $0x2,%ebx
  mov    $0x1,%r8d
  mov    $0x0,%eax
  mov    $0x0,%r13d
  lea    D2(%rip),%r12
  jmp    L41
L92:
  mov    %ebx,-0x4(%rsp)
  mov    %r13d,%r11d
  cmp    %ebx,%r9d
  jle    L46
  movzbl 0x2(%rsi),%r11d
L46:
  movslq %eax,%rbp
  mov    %edx,%esi
  sar    $0x2,%esi
  movslq %esi,%rsi
  movzbl (%r12,%rsi,1),%esi
  mov    %sil,(%r10,%rbp,1)
  lea    0x2(%rax),%esi
  lea    0x1(%rax),%ebp
  movslq %ebp,%rbp
  shl    $0x4,%edx
  and    $0x30,%edx
  mov    %ecx,%r15d
  sar    $0x4,%r15d
  or     %r15d,%edx
  movslq %edx,%rdx
  movzbl (%r12,%rdx,1),%edx
  mov    %dl,(%r10,%rbp,1)
  cmp    %r14d,%r9d
  jle    L35
  movslq %esi,%rsi
  shl    $0x2,%ecx
  and    $0x3c,%ecx
  mov    %r11d,%edx
  sar    $0x6,%edx
  or     %edx,%ecx
  movslq %ecx,%rcx
  movzbl (%r12,%rcx,1),%edx
  mov    %dl,(%r10,%rsi,1)
  lea    0x3(%rax),%esi
L35:
  mov    %esi,%eax
  cmp    -0x4(%rsp),%r9d
  jle    L5
  lea    0x1(%rsi),%eax
  movslq %esi,%rsi
  and    $0x3f,%r11d
  movzbl (%r12,%r11,1),%edx
  mov    %dl,(%r10,%rsi,1)
L5:
  add    $0x3,%rdi
  lea    0x3(%r8),%edx
  add    $0x3,%ebx
  add    $0x2,%r8d
  cmp    %r8d,%r9d
  jle    L13
  mov    %edx,%r8d
L41:
  mov    %rdi,%rsi
  movzbl (%rdi),%edx
  mov    %r8d,%r14d
  mov    %r13d,%ecx
  cmp    %r8d,%r9d
  jle    L92
  movzbl 0x1(%rdi),%ecx
  jmp    L92
L13:
  test   $0x3,%al
  je     L6
  add    $0x1,%eax
  movslq %eax,%rdx
L45:
  mov    %edx,%eax
  movb   $0x3d,-0x1(%r10,%rdx,1)
  mov    %edx,%ecx
  and    $0x3,%ecx
  add    $0x1,%rdx
  test   %ecx,%ecx
  jne    L45
L6:
  cltq
  movb   $0x0,(%r10,%rax,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  mov    $0x0,%eax
  cltq
  movb   $0x0,(%rdx,%rax,1)
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
datatype: string

tool: D2: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/

assistant: ```c
void base64_encode(const unsigned char *src, int src_len, char *dst) {
  static const char *b64 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
  int i, j, a, b, c;
  for (i = j = 0; i < src_len; i += 3) {
    a = src[i];
    b = i + 1 >= src_len ? 0 : src[i + 1];
    c = i + 2 >= src_len ? 0 : src[i + 2];
    dst[j++] = b64[a >> 2];
    dst[j++] = b64[((a & 3) << 4) | (b >> 4)];
    if (i + 1 < src_len) {
      dst[j++] = b64[(b & 15) << 2 | (c >> 6)];
    }
    if (i + 2 < src_len) {
      dst[j++] = b64[c & 63];
    }
  }
  while (j % 4 != 0) {
    dst[j++] = '=';
  }
  dst[j++] = '\0';
}
```