ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x28,%ecx
  lea    D43(%rip),%rax
  mov    %rax,%rdx
  lea    D58(%rip),%rax
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <createAdress@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x25,%ecx
  lea    D41(%rip),%rax
  mov    %rax,%rdx
  lea    D58(%rip),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <createAdress@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <printAddress@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <printAddress@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <destroyAddress@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <destroyAddress@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D43
datatype: string
name: read_data
arguments: data_tag: D58
datatype: string
name: read_data
arguments: data_tag: D30
datatype: string
name: read_data
arguments: data_tag: D41
datatype: string
name: read_data
arguments: data_tag: D57
datatype: string

tool: D43: sam@example.com

tool: D58: Example

tool: D30: Sam

tool: D41: mike@example.com

tool: D57: Mike

assistant: ```c
int main(int argc, char **argv) {
  struct Address *Sam = createAdress("Sam", "Example", "sam@example.com", 40);
  struct Address *Mike = createAdress("Mike", "Example", "mike@example.com", 37);
  printAddress(Sam);
  printAddress(Mike);
  destroyAddress(Sam);
  destroyAddress(Mike);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x28,%ecx
  lea    D37(%rip),%rdx
  lea    D46(%rip),%rbx
  mov    %rbx,%rsi
  lea    D47(%rip),%rdi
  call   <createAdress@plt>
  mov    %rax,%rbp
  mov    $0x25,%ecx
  lea    D12(%rip),%rdx
  mov    %rbx,%rsi
  lea    D40(%rip),%rdi
  call   <createAdress@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <printAddress@plt>
  mov    %rbx,%rdi
  call   <printAddress@plt>
  mov    %rbp,%rdi
  call   <destroyAddress@plt>
  mov    %rbx,%rdi
  call   <destroyAddress@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D37
type_guessed: string
name: read_data
arguments: label: D46
type_guessed: string
name: read_data
arguments: label: D47
type_guessed: string
name: read_data
arguments: label: D12
type_guessed: string
name: read_data
arguments: label: D40
type_guessed: string

tool: D37: sam@example.com

tool: D46: Example

tool: D47: Sam

tool: D12: mike@example.com

tool: D40: Mike

assistant: ```c
int main(int argc, char **argv) {
  struct Address *Sam = createAdress("Sam", "Example", "sam@example.com", 40);
  struct Address *Mike = createAdress("Mike", "Example", "mike@example.com", 37);
  printAddress(Sam);
  printAddress(Mike);
  destroyAddress(Sam);
  destroyAddress(Mike);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x28,%ecx
  lea    D23(%rip),%rdx
  lea    D3(%rip),%rdi
  push   %rbp
  lea    D49(%rip),%rbp
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <createAdress@plt>
  mov    %rbp,%rsi
  mov    $0x25,%ecx
  lea    D12(%rip),%rdx
  lea    D2(%rip),%rdi
  mov    %rax,%r12
  call   <createAdress@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <printAddress@plt>
  mov    %rbp,%rdi
  call   <printAddress@plt>
  mov    %r12,%rdi
  call   <destroyAddress@plt>
  mov    %rbp,%rdi
  call   <destroyAddress@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D23
type_infered: string
name: extract_symbolic_data
arguments: label: D3
type_infered: string
name: extract_symbolic_data
arguments: label: D49
type_infered: string
name: extract_symbolic_data
arguments: label: D12
type_infered: string
name: extract_symbolic_data
arguments: label: D2
type_infered: string

tool: D23: sam@example.com

tool: D3: Sam

tool: D49: Example

tool: D12: mike@example.com

tool: D2: Mike

assistant: ```c
int main(int argc, char **argv) {
  struct Address *Sam = createAdress("Sam", "Example", "sam@example.com", 40);
  struct Address *Mike = createAdress("Mike", "Example", "mike@example.com", 37);
  printAddress(Sam);
  printAddress(Mike);
  destroyAddress(Sam);
  destroyAddress(Mike);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x28,%ecx
  lea    D40(%rip),%rdx
  lea    D17(%rip),%rdi
  push   %rbp
  lea    D31(%rip),%rbp
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <createAdress@plt>
  mov    %rbp,%rsi
  mov    $0x25,%ecx
  lea    D5(%rip),%rdx
  lea    D35(%rip),%rdi
  mov    %rax,%r12
  call   <createAdress@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <printAddress@plt>
  mov    %rbp,%rdi
  call   <printAddress@plt>
  mov    %r12,%rdi
  call   <destroyAddress@plt>
  mov    %rbp,%rdi
  call   <destroyAddress@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D40
type: string
name: extract_symbolic_data
arguments: tag: D17
type: string
name: extract_symbolic_data
arguments: tag: D31
type: string
name: extract_symbolic_data
arguments: tag: D5
type: string
name: extract_symbolic_data
arguments: tag: D35
type: string

tool: D40: sam@example.com

tool: D17: Sam

tool: D31: Example

tool: D5: mike@example.com

tool: D35: Mike

assistant: ```c
int main(int argc, char **argv) {
  struct Address *Sam = createAdress("Sam", "Example", "sam@example.com", 40);
  struct Address *Mike = createAdress("Mike", "Example", "mike@example.com", 37);
  printAddress(Sam);
  printAddress(Mike);
  destroyAddress(Sam);
  destroyAddress(Mike);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2,%esi
  mov    $0x3,%edi
  call   <pow@plt>
  cvttsd2si %xmm0,%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%esi
  mov    $0x77359400,%edi
  call   <pow@plt>
  cvttsd2si %xmm0,%esi
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x64,%edx
  mov    $0x1,%esi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edx
  mov    $0x64,%esi
  lea    D10(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edx
  mov    $0x64,%esi
  lea    D35(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x64,%edx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x7d2b7500,%edx
  mov    $0x77359400,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x77359400,%edx
  mov    $0x7d2b7500,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type_guessed: string
name: unpack_data
arguments: data_tag: D36
type_guessed: string
name: unpack_data
arguments: data_tag: D15
type_guessed: string
name: unpack_data
arguments: data_tag: D10
type_guessed: string
name: unpack_data
arguments: data_tag: D35
type_guessed: string
name: unpack_data
arguments: data_tag: D4
type_guessed: string

tool: D5: 3 to the 2nd power is %d


tool: D36: 2e9 to the 10th power is %.0f


tool: D15: X is %d and Y is %d
Performing swap now.


tool: D10: After the swap X is %d and Y is %d!



tool: D35: Problem #19 -- X is %d and Y is %d
Performing swap now.


tool: D4: Problem #19 continued
 -- X is %d and Y is %d
Performing swap now.


assistant: ```c
int main(int argc, char *argv[]) {
  int power = pow(3, 2);
  printf("3 to the 2nd power is %d\n", power);
  printf("2e9 to the 10th power is %.0f\n", pow(2000000000, 10));
  int x = 1;
  int y = 100;
  printf("X is %d and Y is %d\nPerforming swap now.\n", x, y);
  int temp;
  temp = x;
  x = y;
  y = temp;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
  printf("Problem #19 -- X is %d and Y is %d\nPerforming swap now.\n", x, y);
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
  x = 2000000000;
  y = 2100000000;
  printf("Problem #19 continued\n -- X is %d and Y is %d\nPerforming swap now.\n", x, y);
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x2,%esi
  mov    $0x3,%edi
  lea    D32(%rip),%rbp
  call   <pow@plt>
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    $0xa,%esi
  mov    $0x77359400,%edi
  call   <pow@plt>
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    $0x64,%edx
  mov    $0x1,%esi
  xor    %eax,%eax
  lea    D34(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%edx
  xor    %eax,%eax
  mov    $0x64,%esi
  call   <printf@plt>
  mov    $0x1,%edx
  mov    $0x64,%esi
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x64,%edx
  xor    %eax,%eax
  mov    $0x1,%esi
  call   <printf@plt>
  mov    $0x7d2b7500,%edx
  mov    $0x77359400,%esi
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x77359400,%edx
  xor    %eax,%eax
  mov    $0x7d2b7500,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D32
data_type: string
name: parse_data
arguments: data_label: D29
data_type: string
name: parse_data
arguments: data_label: D22
data_type: string
name: parse_data
arguments: data_label: D34
data_type: string
name: parse_data
arguments: data_label: D0
data_type: string
name: parse_data
arguments: data_label: D10
data_type: string

tool: D32: After the swap X is %d and Y is %d!



tool: D29: 3 to the 2nd power is %d


tool: D22: 2e9 to the 10th power is %.0f


tool: D34: X is %d and Y is %d
Performing swap now.


tool: D0: Problem #19 -- X is %d and Y is %d
Performing swap now.


tool: D10: Problem #19 continued
 -- X is %d and Y is %d
Performing swap now.


assistant: ```c
int main(int argc, char *argv[]) {
  int power = pow(3, 2);
  printf("3 to the 2nd power is %d\n", power);
  printf("2e9 to the 10th power is %.0f\n", pow(2000000000, 10));
  int x = 1;
  int y = 100;
  printf("X is %d and Y is %d\nPerforming swap now.\n", x, y);
  int temp;
  temp = x;
  x = y;
  y = temp;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
  printf("Problem #19 -- X is %d and Y is %d\nPerforming swap now.\n", x, y);
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
  x = 2000000000;
  y = 2100000000;
  printf("Problem #19 continued\n -- X is %d and Y is %d\nPerforming swap now.\n", x, y);
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x2,%esi
  mov    $0x3,%edi
  lea    D29(%rip),%rbp
  call   <pow@plt>
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    $0xa,%esi
  mov    $0x77359400,%edi
  call   <pow@plt>
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    $0x64,%edx
  mov    $0x1,%esi
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%edx
  xor    %eax,%eax
  mov    $0x64,%esi
  call   <printf@plt>
  mov    $0x1,%edx
  mov    $0x64,%esi
  xor    %eax,%eax
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x64,%edx
  xor    %eax,%eax
  mov    $0x1,%esi
  call   <printf@plt>
  mov    $0x7d2b7500,%edx
  mov    $0x77359400,%esi
  xor    %eax,%eax
  lea    D47(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x77359400,%edx
  xor    %eax,%eax
  mov    $0x7d2b7500,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D29
type_infered: string
name: disassemble_data
arguments: data_tag: D16
type_infered: string
name: disassemble_data
arguments: data_tag: D26
type_infered: string
name: disassemble_data
arguments: data_tag: D24
type_infered: string
name: disassemble_data
arguments: data_tag: D22
type_infered: string
name: disassemble_data
arguments: data_tag: D47
type_infered: string

tool: D29: After the swap X is %d and Y is %d!



tool: D16: 3 to the 2nd power is %d


tool: D26: 2e9 to the 10th power is %.0f


tool: D24: X is %d and Y is %d
Performing swap now.


tool: D22: Problem #19 -- X is %d and Y is %d
Performing swap now.


tool: D47: Problem #19 continued
 -- X is %d and Y is %d
Performing swap now.


assistant: ```c
int main(int argc, char *argv[]) {
  int power = pow(3, 2);
  printf("3 to the 2nd power is %d\n", power);
  printf("2e9 to the 10th power is %.0f\n", pow(2000000000, 10));
  int x = 1;
  int y = 100;
  printf("X is %d and Y is %d\nPerforming swap now.\n", x, y);
  int temp;
  temp = x;
  x = y;
  y = temp;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
  printf("Problem #19 -- X is %d and Y is %d\nPerforming swap now.\n", x, y);
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
  x = 2000000000;
  y = 2100000000;
  printf("Problem #19 continued\n -- X is %d and Y is %d\nPerforming swap now.\n", x, y);
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
  printf("After the swap X is %d and Y is %d!\n\n", x, y);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<minimum_column_indices>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L10
L35:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L28
L26:
  mov    -0x18(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x10(%rbp),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L3
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L3:
  addl   $0x1,-0x18(%rbp)
L28:
  mov    -0x18(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L26
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L10:
  mov    -0x1c(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L35
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void minimum_column_indices(double *matrix, int M, int N, double *mininds) {
  int i, j, soFarInd;
  double soFar, nextTry;
  for (i = 0; i < M; i++) {
    soFar = matrix[i];
    soFarInd = 0;
    for (j = 0; j < N; j++) {
      nextTry = matrix[i + j * M];
      if (nextTry < soFar) {
        soFar = nextTry;
        soFarInd = j;
      }
    }
    mininds[i] = soFarInd + 1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<minimum_column_indices>:
  endbr64
  test   %esi,%esi
  jle    L57
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %edx,%edi
  movslq %esi,%r8
  shl    $0x3,%r8
  lea    (%rbx,%r8,1),%r10
  mov    %esi,%r11d
  mov    $0x0,%r9d
  mov    $0x0,%ebp
  jmp    L38
L50:
  movsd  (%rdx),%xmm1
  comisd %xmm1,%xmm0
  minsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  cmova  %eax,%esi
  add    %r8,%rdx
L6:
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L50
L55:
  add    $0x1,%esi
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  movsd  %xmm0,(%rcx,%r9,8)
  add    $0x1,%r9
  add    $0x8,%r10
  cmp    %r11,%r9
  je     L16
L38:
  movsd  (%rbx,%r9,8),%xmm0
  mov    %r10,%rdx
  mov    $0x0,%eax
  mov    %ebp,%esi
  test   %edi,%edi
  jg     L6
  jmp    L55
L16:
  pop    %rbx
  pop    %rbp
  ret
L57:
  ret
```

assistant: ```c
void minimum_column_indices(double *matrix, int M, int N, double *mininds) {
  int i, j, soFarInd;
  double soFar, nextTry;
  for (i = 0; i < M; i++) {
    soFar = matrix[i];
    soFarInd = 0;
    for (j = 0; j < N; j++) {
      nextTry = matrix[i + j * M];
      if (nextTry < soFar) {
        soFar = nextTry;
        soFarInd = j;
      }
    }
    mininds[i] = soFarInd + 1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<minimum_column_indices>:
  endbr64
  test   %esi,%esi
  jle    L14
  movslq %esi,%r11
  push   %rbx
  xor    %r9d,%r9d
  mov    %rdi,%rbx
  lea    0x0(,%r11,8),%r8
  lea    (%rdi,%r8,1),%r10
  xchg   %ax,%ax
L21:
  movsd  (%rbx,%r9,8),%xmm0
  mov    %r10,%rsi
  xor    %eax,%eax
  xor    %edi,%edi
  test   %edx,%edx
  jg     L48
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L41:
  movsd  (%rsi),%xmm1
  comisd %xmm1,%xmm0
  minsd  %xmm0,%xmm1
  cmova  %eax,%edi
  movapd %xmm1,%xmm0
  add    %r8,%rsi
L48:
  add    $0x1,%eax
  cmp    %eax,%edx
  jne    L41
  add    $0x1,%edi
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm0
L52:
  movsd  %xmm0,(%rcx,%r9,8)
  add    $0x1,%r9
  add    $0x8,%r10
  cmp    %r11,%r9
  jne    L21
  pop    %rbx
  ret
  nopl   (%rax)
L44:
  mov    D2(%rip),%rax
  movq   %rax,%xmm0
  jmp    L52
L14:
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
type_guessed: byte[8]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void minimum_column_indices(double *matrix, int M, int N, double *mininds) {
  int i, j, soFarInd;
  double soFar, nextTry;
  for (i = 0; i < M; i++) {
    soFar = matrix[i];
    soFarInd = 0;
    for (j = 0; j < N; j++) {
      nextTry = matrix[i + j * M];
      if (nextTry < soFar) {
        soFar = nextTry;
        soFarInd = j;
      }
    }
    mininds[i] = soFarInd + 1;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<minimum_column_indices>:
  endbr64
  test   %esi,%esi
  jle    L19
  movslq %esi,%r11
  push   %rbx
  xor    %r9d,%r9d
  mov    %rdi,%rbx
  lea    0x0(,%r11,8),%r8
  lea    (%rdi,%r8,1),%r10
  nopw   0x0(%rax,%rax,1)
L10:
  movsd  (%rbx,%r9,8),%xmm0
  test   %edx,%edx
  jle    L49
  cmp    $0x1,%edx
  je     L15
  mov    %r10,%rsi
  mov    $0x1,%eax
  xor    %edi,%edi
  nopl   0x0(%rax)
L7:
  movsd  (%rsi),%xmm1
  comisd %xmm1,%xmm0
  minsd  %xmm0,%xmm1
  cmova  %eax,%edi
  add    $0x1,%eax
  movapd %xmm1,%xmm0
  add    %r8,%rsi
  cmp    %edx,%eax
  jne    L7
L65:
  add    $0x1,%edi
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm0
L25:
  movsd  %xmm0,(%rcx,%r9,8)
  add    $0x1,%r9
  add    $0x8,%r10
  cmp    %r11,%r9
  jne    L10
  pop    %rbx
  ret
  nopl   (%rax)
L49:
  mov    D8(%rip),%rax
  movq   %rax,%xmm0
  jmp    L25
  cs nopw 0x0(%rax,%rax,1)
L15:
  xor    %edi,%edi
  jmp    L65
L19:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type_infered: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void minimum_column_indices(double *matrix, int M, int N, double *mininds) {
  int i, j, soFarInd;
  double soFar, nextTry;
  for (i = 0; i < M; i++) {
    soFar = matrix[i];
    soFarInd = 0;
    for (j = 0; j < N; j++) {
      nextTry = matrix[i + j * M];
      if (nextTry < soFar) {
        soFar = nextTry;
        soFarInd = j;
      }
    }
    mininds[i] = soFarInd + 1;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<split_nr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  movl   $0x2,-0x4(%rbp)
  jmp    L0
L4:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  movzbl -0x1c(%rbp),%edx
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dl
  shr    $0x8,%ax
  sar    $0x2,%al
  mov    %edx,%edi
  sar    $0x7,%dil
  sub    %edi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %cl,(%rsi)
  movzbl -0x1c(%rbp),%edx
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dl
  shr    $0x8,%ax
  sar    $0x2,%al
  sar    $0x7,%dl
  sub    %edx,%eax
  mov    %al,-0x1c(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void split_nr(char parts[], char nr_to_split) {
  int dig = 2;
  while (dig--) {
    parts[dig] = nr_to_split % 10;
    nr_to_split /= 10;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<split_nr>:
  endbr64
  mov    $0x67,%edx
  mov    %esi,%eax
  imul   %dl
  mov    %eax,%ecx
  sar    $0xa,%cx
  mov    %esi,%eax
  sar    $0x7,%al
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %sil,0x1(%rdi)
  mov    %edx,%eax
  imul   %cl
  mov    %eax,%edx
  sar    $0xa,%dx
  mov    %ecx,%eax
  sar    $0x7,%al
  sub    %eax,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %cl,(%rdi)
  ret
```

assistant: ```c
void split_nr(char parts[], char nr_to_split) {
  int dig = 2;
  while (dig--) {
    parts[dig] = nr_to_split % 10;
    nr_to_split /= 10;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<split_nr>:
  endbr64
  mov    $0x67,%edx
  mov    %esi,%eax
  imul   %dl
  mov    %eax,%ecx
  mov    %esi,%eax
  sar    $0x7,%al
  sar    $0xa,%cx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %edx,%eax
  imul   %cl
  mov    %sil,0x1(%rdi)
  mov    %eax,%edx
  mov    %ecx,%eax
  sar    $0xa,%dx
  sar    $0x7,%al
  sub    %eax,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %cl,(%rdi)
  ret
```

assistant: ```c
void split_nr(char parts[], char nr_to_split) {
  int dig = 2;
  while (dig--) {
    parts[dig] = nr_to_split % 10;
    nr_to_split /= 10;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<split_nr>:
  endbr64
  mov    $0x67,%edx
  mov    %esi,%eax
  imul   %dl
  mov    %eax,%ecx
  mov    %esi,%eax
  sar    $0x7,%al
  sar    $0xa,%cx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %edx,%eax
  imul   %cl
  mov    %sil,0x1(%rdi)
  mov    %eax,%edx
  mov    %ecx,%eax
  sar    $0xa,%dx
  sar    $0x7,%al
  sub    %eax,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %cl,(%rdi)
  ret
```

assistant: ```c
void split_nr(char parts[], char nr_to_split) {
  int dig = 2;
  while (dig--) {
    parts[dig] = nr_to_split % 10;
    nr_to_split /= 10;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movb   $0x0,D14(%rip)
  mov    $0x66,%eax
  add    $0x1,%eax
  mov    %al,D16(%rip)
  mov    $0x66,%eax
  sub    $0x1,%eax
  mov    %al,D16(%rip)
  mov    $0x66,%eax
  add    $0x1,%eax
  mov    %al,D16(%rip)
  mov    $0x66,%eax
  sub    $0x1,%eax
  mov    %al,D16(%rip)
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D16
type_infered: byte

tool: D16: 102

assistant: ```c
void f(void) {
  "foo"[0] = 0;
  "foo"[0]++;
  "foo"[0]--;
  ++"foo"[0];
  --"foo"[0];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<f>:
  endbr64
  movb   $0x65,D8(%rip)
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
datatype: byte

tool: D8: 102

assistant: ```c
void f(void) {
  "foo"[0] = 0;
  "foo"[0]++;
  "foo"[0]--;
  ++"foo"[0];
  --"foo"[0];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<f>:
  endbr64
  movb   $0x65,D2(%rip)
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
datatype: byte

tool: D2: 102

assistant: ```c
void f(void) {
  "foo"[0] = 0;
  "foo"[0]++;
  "foo"[0]--;
  ++"foo"[0];
  --"foo"[0];
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<f>:
  endbr64
  movb   $0x65,D0(%rip)
  ret
```

tool_calls: name: read_data
arguments: tag: D0
type_guessed: byte

tool: D0: 102

assistant: ```c
void f(void) {
  "foo"[0] = 0;
  "foo"[0]++;
  "foo"[0]--;
  ++"foo"[0];
  --"foo"[0];
}
```