ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D45(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D35(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  cmpl   $0x0,0x4(%rsp)
  jle    L46
  mov    %rax,%rbp
  mov    $0x0,%ebx
  lea    D35(%rip),%r13
L19:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0x4(%rsp)
  jg     L19
L46:
  lea    D36(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%eax
  mov    %eax,%ebp
  sub    $0x1,%ebp
  js     L25
  movslq %ebp,%rbp
  shl    $0x2,%rbp
  lea    0x0(%rbp,%r12,1),%rbx
  sub    $0x4,%rbp
  lea    -0x1(%rax),%eax
  shl    $0x2,%rax
  sub    %rax,%rbp
  add    %r12,%rbp
  lea    D33(%rip),%r12
L21:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L21
L25:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D45
datatype: string
name: read_data
arguments: data_label: D35
datatype: string
name: read_data
arguments: data_label: D36
datatype: string
name: read_data
arguments: data_label: D33
datatype: string

tool: D45: Enter the size of the array: 

tool: D35: %d

tool: D36: Here are the same numbers in reverse order: 

tool: D33: %d 


assistant: ```c
int main() {
  int n;
  int i;
  int *a;
  printf("Enter the size of the array: ");
  scanf("%d", &n);
  a = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    scanf("%d", (a + i));
  }
  printf("Here are the same numbers in reverse order: \n");
  for (i = n - 1; i >= 0; i--) {
    printf("%d \n", a[i]);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D10(%rip),%rdi
  lea    D38(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L37
  mov    %r12,%rbp
  xor    %ebx,%ebx
  xchg   %ax,%ax
L44:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L44
L37:
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  js     L3
  movslq %eax,%rbp
  mov    %eax,%eax
  shl    $0x2,%rbp
  shl    $0x2,%rax
  lea    (%r12,%rbp,1),%rbx
  sub    $0x4,%rbp
  sub    %rax,%rbp
  add    %r12,%rbp
  lea    D12(%rip),%r12
  nop
L21:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  sub    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L21
L3:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D10
type: string
name: unpack_data
arguments: data_tag: D38
type: string
name: unpack_data
arguments: data_tag: D11
type: string
name: unpack_data
arguments: data_tag: D12
type: string

tool: D10: Enter the size of the array: 

tool: D38: %d

tool: D11: Here are the same numbers in reverse order: 

tool: D12: %d 


assistant: ```c
int main() {
  int n;
  int i;
  int *a;
  printf("Enter the size of the array: ");
  scanf("%d", &n);
  a = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    scanf("%d", (a + i));
  }
  printf("Here are the same numbers in reverse order: \n");
  for (i = n - 1; i >= 0; i--) {
    printf("%d \n", a[i]);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D9(%rip),%rdi
  lea    D32(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L49
  mov    %r12,%rbp
  xor    %ebx,%ebx
  xchg   %ax,%ax
L25:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L25
L49:
  lea    D20(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  js     L20
  movslq %eax,%rbp
  mov    %eax,%eax
  shl    $0x2,%rbp
  shl    $0x2,%rax
  lea    (%r12,%rbp,1),%rbx
  sub    $0x4,%rbp
  sub    %rax,%rbp
  add    %r12,%rbp
  lea    D3(%rip),%r12
  nop
L2:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  sub    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L2
L20:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
datatype: string
name: disassemble_data
arguments: data_tag: D32
datatype: string
name: disassemble_data
arguments: data_tag: D20
datatype: string
name: disassemble_data
arguments: data_tag: D3
datatype: string

tool: D9: Enter the size of the array: 

tool: D32: %d

tool: D20: Here are the same numbers in reverse order: 

tool: D3: %d 


assistant: ```c
int main() {
  int n;
  int i;
  int *a;
  printf("Enter the size of the array: ");
  scanf("%d", &n);
  a = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    scanf("%d", (a + i));
  }
  printf("Here are the same numbers in reverse order: \n");
  for (i = n - 1; i >= 0; i--) {
    printf("%d \n", a[i]);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D3(%rip),%rax
  mov    %rax,-0x78(%rbp)
  movq   $0x0,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x78(%rbp),%rdx
  lea    -0x70(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sub_string_v2@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_guessed: string

tool: D3: abcd

assistant: ```c
int main() {
  char *s = "abcd";
  char new[100] = {0};
  sub_string_v2(new, s);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movl   $0x0,0x60(%rsp)
  mov    %rsp,%rdi
  lea    D4(%rip),%rsi
  call   <sub_string_v2@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x78,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D4
type_guessed: string

tool: D4: abcd

assistant: ```c
int main() {
  char *s = "abcd";
  char new[100] = {0};
  sub_string_v2(new, s);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  pxor   %xmm0,%xmm0
  lea    D8(%rip),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movl   $0x0,0x60(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <sub_string_v2@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x78,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D8
type_infered: string

tool: D8: abcd

assistant: ```c
int main() {
  char *s = "abcd";
  char new[100] = {0};
  sub_string_v2(new, s);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  pxor   %xmm0,%xmm0
  lea    D4(%rip),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movl   $0x0,0x60(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <sub_string_v2@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x78,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D4
type: string

tool: D4: abcd

assistant: ```c
int main() {
  char *s = "abcd";
  char new[100] = {0};
  sub_string_v2(new, s);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cmatrix_roty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  movq   %xmm0,%rax
  mov    -0x18(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  call   <sin@plt>
  movq   %xmm0,%rax
  mov    -0x18(%rbp),%rdx
  add    $0x30,%rdx
  movq   D5(%rip),%xmm0
  movq   %rax,%xmm1
  xorpd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  movsd  %xmm0,(%rdx)
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x20,%rax
  movsd  D6(%rip),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x38,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x10(%rax),%rbx
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  call   <sin@plt>
  movq   %xmm0,%rax
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  add    $0x28,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x40(%rax),%rbx
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  movq   %xmm0,%rax
  mov    %rax,(%rbx)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D5
type_infered: f32[2]
name: read_data
arguments: label: D6
type_infered: f64

tool: {"D5": [0.0,-0.0]}

tool: D6: 1

assistant: ```c
void cmatrix_roty(double *m, double angle) {
  m[((((1) - 1) * 3) + ((1) - 1))] = cos(angle);
  m[((((2) - 1) * 3) + ((1) - 1))] = 0.0;
  m[((((3) - 1) * 3) + ((1) - 1))] = -sin(angle);
  m[((((1) - 1) * 3) + ((2) - 1))] = 0.0;
  m[((((2) - 1) * 3) + ((2) - 1))] = 1.0;
  m[((((3) - 1) * 3) + ((2) - 1))] = 0.0;
  ;
  m[((((1) - 1) * 3) + ((3) - 1))] = sin(angle);
  m[((((2) - 1) * 3) + ((3) - 1))] = 0.0;
  m[((((3) - 1) * 3) + ((3) - 1))] = cos(angle);
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmatrix_roty>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  lea    0x8(%rsp),%rdi
  mov    %rsp,%rsi
  call   <sincos@plt>
  movsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  movsd  %xmm0,(%rbx)
  movq   $0x0,0x18(%rbx)
  movapd %xmm1,%xmm2
  xorpd  D9(%rip),%xmm2
  movsd  %xmm2,0x30(%rbx)
  movq   $0x0,0x8(%rbx)
  mov    D2(%rip),%rax
  mov    %rax,0x20(%rbx)
  movq   $0x0,0x38(%rbx)
  movsd  %xmm1,0x10(%rbx)
  movq   $0x0,0x28(%rbx)
  movsd  %xmm0,0x40(%rbx)
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
datatype: f32[4]
name: disassemble_data
arguments: label: D2
datatype: f64

tool: {"D9": [0.0,-0.0,0.0,0.0]}

tool: D2: 1

assistant: ```c
void cmatrix_roty(double *m, double angle) {
  m[((((1) - 1) * 3) + ((1) - 1))] = cos(angle);
  m[((((2) - 1) * 3) + ((1) - 1))] = 0.0;
  m[((((3) - 1) * 3) + ((1) - 1))] = -sin(angle);
  m[((((1) - 1) * 3) + ((2) - 1))] = 0.0;
  m[((((2) - 1) * 3) + ((2) - 1))] = 1.0;
  m[((((3) - 1) * 3) + ((2) - 1))] = 0.0;
  ;
  m[((((1) - 1) * 3) + ((3) - 1))] = sin(angle);
  m[((((2) - 1) * 3) + ((3) - 1))] = 0.0;
  m[((((3) - 1) * 3) + ((3) - 1))] = cos(angle);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cmatrix_roty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  lea    0x8(%rsp),%rdi
  mov    %rsp,%rsi
  call   <sincos@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm0
  movq   $0x0,0x18(%rbx)
  mov    D13(%rip),%rax
  movq   $0x0,0x8(%rbx)
  movapd %xmm1,%xmm2
  xorpd  D8(%rip),%xmm2
  movq   $0x0,0x38(%rbx)
  mov    %rax,0x20(%rbx)
  movq   $0x0,0x28(%rbx)
  movsd  %xmm0,(%rbx)
  movsd  %xmm2,0x30(%rbx)
  movsd  %xmm1,0x10(%rbx)
  movsd  %xmm0,0x40(%rbx)
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D13
type_infered: f64
name: parse_data
arguments: label: D8
type_infered: f32[4]

tool: D13: 1

tool: {"D8": [0.0,-0.0,0.0,0.0]}

assistant: ```c
void cmatrix_roty(double *m, double angle) {
  m[((((1) - 1) * 3) + ((1) - 1))] = cos(angle);
  m[((((2) - 1) * 3) + ((1) - 1))] = 0.0;
  m[((((3) - 1) * 3) + ((1) - 1))] = -sin(angle);
  m[((((1) - 1) * 3) + ((2) - 1))] = 0.0;
  m[((((2) - 1) * 3) + ((2) - 1))] = 1.0;
  m[((((3) - 1) * 3) + ((2) - 1))] = 0.0;
  ;
  m[((((1) - 1) * 3) + ((3) - 1))] = sin(angle);
  m[((((2) - 1) * 3) + ((3) - 1))] = 0.0;
  m[((((3) - 1) * 3) + ((3) - 1))] = cos(angle);
}
```
user: What is the c source code of the assembly code below?

```asm
<cmatrix_roty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  lea    0x8(%rsp),%rdi
  mov    %rsp,%rsi
  call   <sincos@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm0
  movq   $0x0,0x18(%rbx)
  mov    D17(%rip),%rax
  movq   $0x0,0x8(%rbx)
  movapd %xmm1,%xmm2
  xorpd  D11(%rip),%xmm2
  movq   $0x0,0x38(%rbx)
  mov    %rax,0x20(%rbx)
  movq   $0x0,0x28(%rbx)
  movsd  %xmm0,(%rbx)
  movsd  %xmm2,0x30(%rbx)
  movsd  %xmm1,0x10(%rbx)
  movsd  %xmm0,0x40(%rbx)
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D17
type_guessed: f64
name: parse_data
arguments: data_tag: D11
type_guessed: f32[4]

tool: D17: 1

tool: {"D11": [0.0,-0.0,0.0,0.0]}

assistant: ```c
void cmatrix_roty(double *m, double angle) {
  m[((((1) - 1) * 3) + ((1) - 1))] = cos(angle);
  m[((((2) - 1) * 3) + ((1) - 1))] = 0.0;
  m[((((3) - 1) * 3) + ((1) - 1))] = -sin(angle);
  m[((((1) - 1) * 3) + ((2) - 1))] = 0.0;
  m[((((2) - 1) * 3) + ((2) - 1))] = 1.0;
  m[((((3) - 1) * 3) + ((2) - 1))] = 0.0;
  ;
  m[((((1) - 1) * 3) + ((3) - 1))] = sin(angle);
  m[((((2) - 1) * 3) + ((3) - 1))] = 0.0;
  m[((((3) - 1) * 3) + ((3) - 1))] = cos(angle);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<printMoneyResults>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D26
datatype: string
name: retrieve_data
arguments: data_label: D12
datatype: string
name: retrieve_data
arguments: data_label: D9
datatype: string

tool: D26: total = %i 


tool: D12: dollars = %i 


tool: D9: cents = %i 


assistant: ```c
void printMoneyResults(int total, int *dollars, int *cents) {
  printf("\n");
  printf("total = %i \n", total);
  printf("dollars = %i \n", *dollars);
  printf("cents = %i \n", *cents);
}
```
user: Translate the below assembly code into C programming language:

```asm
<printMoneyResults>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r12d,%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x0(%rbp),%esi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%rbx),%esi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D16
type_infered: string
name: disassemble_data
arguments: data_tag: D15
type_infered: string
name: disassemble_data
arguments: data_tag: D0
type_infered: string

tool: D16: total = %i 


tool: D15: dollars = %i 


tool: D0: cents = %i 


assistant: ```c
void printMoneyResults(int total, int *dollars, int *cents) {
  printf("\n");
  printf("total = %i \n", total);
  printf("dollars = %i \n", *dollars);
  printf("cents = %i \n", *cents);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<printMoneyResults>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  mov    $0xa,%edi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <putchar@plt>
  mov    %r12d,%esi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x0(%rbp),%esi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rbx),%esi
  lea    D16(%rip),%rdi
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1070 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D1
type: string
name: retrieve_data
arguments: label: D27
type: string
name: retrieve_data
arguments: label: D16
type: string

tool: D1: total = %i 


tool: D27: dollars = %i 


tool: D16: cents = %i 


assistant: ```c
void printMoneyResults(int total, int *dollars, int *cents) {
  printf("\n");
  printf("total = %i \n", total);
  printf("dollars = %i \n", *dollars);
  printf("cents = %i \n", *cents);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<printMoneyResults>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  mov    $0xa,%edi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <putchar@plt>
  mov    %r12d,%esi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x0(%rbp),%esi
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rbx),%esi
  lea    D18(%rip),%rdi
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1070 <printf@plt>
```

tool_calls: name: read_data
arguments: data_label: D22
datatype: string
name: read_data
arguments: data_label: D25
datatype: string
name: read_data
arguments: data_label: D18
datatype: string

tool: D22: total = %i 


tool: D25: dollars = %i 


tool: D18: cents = %i 


assistant: ```c
void printMoneyResults(int total, int *dollars, int *cents) {
  printf("\n");
  printf("total = %i \n", total);
  printf("dollars = %i \n", *dollars);
  printf("cents = %i \n", *cents);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<coco_create_unique_directory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x1,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <coco_directory_exists@plt>
  test   %eax,%eax
  jne    L1
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <coco_create_directory@plt>
  jmp    L8
L28:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    -0xc(%rbp),%edx
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <coco_strdupf@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <coco_directory_exists@plt>
  test   %eax,%eax
  jne    L4
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <coco_free_memory@plt>
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <coco_create_directory@plt>
  jmp    L8
L4:
  addl   $0x1,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <coco_free_memory@plt>
L1:
  cmpl   $0x3e6,-0xc(%rbp)
  jle    L28
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <coco_error@plt>
  nop
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D18
type_guessed: string
name: parse_data
arguments: tag: D9
type_guessed: string

tool: D18: %s-%03d

tool: D9: coco_create_unique_path(): could not create a unique path with name %s

assistant: ```c
void coco_create_unique_directory(char **path) {
  int counter = 1;
  char *new_path;
  if (!coco_directory_exists(*path)) {
    coco_create_directory(*path);
    return;
  }
  while (counter < 999) {
    new_path = coco_strdupf("%s-%03d", *path, counter);
    if (!coco_directory_exists(new_path)) {
      coco_free_memory(*path);
      *path = new_path;
      coco_create_directory(*path);
      return;
    } else {
      counter++;
      coco_free_memory(new_path);
    }
  }
  coco_error("coco_create_unique_path(): could not create a unique path with name %s", *path);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<coco_create_unique_directory>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    (%rdi),%rdi
  call   <coco_directory_exists@plt>
  mov    $0x1,%ebp
  lea    D15(%rip),%r13
  test   %eax,%eax
  je     L34
L12:
  mov    %ebp,%edx
  mov    (%r12),%rsi
  mov    %r13,%rdi
  call   <coco_strdupf@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <coco_directory_exists@plt>
  test   %eax,%eax
  je     L46
  add    $0x1,%ebp
  mov    %rbx,%rdi
  call   <coco_free_memory@plt>
  cmp    $0x3e7,%ebp
  jne    L12
  mov    (%r12),%rsi
  lea    D6(%rip),%rdi
  call   <coco_error@plt>
  jmp    L37
L34:
  mov    (%r12),%rdi
  call   <coco_create_directory@plt>
  jmp    L37
L46:
  mov    (%r12),%rdi
  call   <coco_free_memory@plt>
  mov    %rbx,(%r12)
  mov    %rbx,%rdi
  call   <coco_create_directory@plt>
L37:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D15
type_guessed: string
name: unpack_data
arguments: data_label: D6
type_guessed: string

tool: D15: %s-%03d

tool: D6: coco_create_unique_path(): could not create a unique path with name %s

assistant: ```c
void coco_create_unique_directory(char **path) {
  int counter = 1;
  char *new_path;
  if (!coco_directory_exists(*path)) {
    coco_create_directory(*path);
    return;
  }
  while (counter < 999) {
    new_path = coco_strdupf("%s-%03d", *path, counter);
    if (!coco_directory_exists(new_path)) {
      coco_free_memory(*path);
      *path = new_path;
      coco_create_directory(*path);
      return;
    } else {
      counter++;
      coco_free_memory(new_path);
    }
  }
  coco_error("coco_create_unique_path(): could not create a unique path with name %s", *path);
  return;
}
```