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   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x2,-0x14(%rbp)
  je     L66
  mov    $0x1,%eax
  jmp    L32
L66:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  lea    -0xc(%rbp),%rdx
  lea    D30(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L12
  mov    $0x1,%eax
  jmp    L32
L12:
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L5
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L5:
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L33
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L33:
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L32:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D30
datatype: string
name: disassemble_data
arguments: data_tag: D9
datatype: string
name: disassemble_data
arguments: data_tag: D29
datatype: string
name: disassemble_data
arguments: data_tag: D17
datatype: string

tool: D30: %d

tool: D9: fizz

tool: D29: buzz

tool: D17: 


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  if (argc != 2)
    return 1;
  if (sscanf(argv[1], "%d", &i) != 1)
    return 1;
  if (i % 2 == 0) {
    printf("fizz");
  }
  if (i % 3 == 0) {
    printf("buzz");
  }
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x1,%eax
  cmp    $0x2,%edi
  je     L28
L13:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L79
  add    $0x18,%rsp
  ret
L28:
  lea    0x4(%rsp),%rdx
  mov    0x8(%rsi),%rdi
  lea    D21(%rip),%rsi
  call   <sscanf@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  cmp    $0x1,%edx
  jne    L13
  testb  $0x1,0x4(%rsp)
  je     L19
L64:
  mov    0x4(%rsp),%eax
  mov    $0x3,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L58
L12:
  lea    D31(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L13
L19:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L64
L58:
  lea    D39(%rip),%rdi
  call   <printf@plt>
  jmp    L12
L79:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D21
type_guessed: string
name: parse_data
arguments: data_label: D31
type_guessed: string
name: parse_data
arguments: data_label: D4
type_guessed: string
name: parse_data
arguments: data_label: D39
type_guessed: string

tool: D21: %d

tool: D31: 


tool: D4: fizz

tool: D39: buzz

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  if (argc != 2)
    return 1;
  if (sscanf(argv[1], "%d", &i) != 1)
    return 1;
  if (i % 2 == 0) {
    printf("fizz");
  }
  if (i % 3 == 0) {
    printf("buzz");
  }
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L61
L43:
  mov    $0x1,%eax
L11:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x18,%rsp
  ret
L61:
  mov    0x8(%rsi),%rdi
  lea    0x4(%rsp),%rdx
  lea    D38(%rip),%rsi
  call   <sscanf@plt>
  sub    $0x1,%eax
  jne    L43
  mov    0x4(%rsp),%eax
  test   $0x1,%al
  je     L71
L26:
  mov    $0x3,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L47
L75:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L11
L71:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  jmp    L26
L47:
  lea    D27(%rip),%rdi
  call   <printf@plt>
  jmp    L75
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D38
type_guessed: string
name: read_data
arguments: data_tag: D8
type_guessed: string
name: read_data
arguments: data_tag: D1
type_guessed: string
name: read_data
arguments: data_tag: D27
type_guessed: string

tool: D38: %d

tool: D8: 


tool: D1: fizz

tool: D27: buzz

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  if (argc != 2)
    return 1;
  if (sscanf(argv[1], "%d", &i) != 1)
    return 1;
  if (i % 2 == 0) {
    printf("fizz");
  }
  if (i % 3 == 0) {
    printf("buzz");
  }
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L38
L55:
  mov    $0x1,%eax
L45:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L28
  add    $0x18,%rsp
  ret
L38:
  mov    0x8(%rsi),%rdi
  lea    0x4(%rsp),%rdx
  lea    D31(%rip),%rsi
  call   <sscanf@plt>
  sub    $0x1,%eax
  jne    L55
  mov    0x4(%rsp),%eax
  test   $0x1,%al
  je     L54
L5:
  mov    $0x3,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L43
L40:
  lea    D12(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L45
L54:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  jmp    L5
L43:
  lea    D20(%rip),%rdi
  call   <printf@plt>
  jmp    L40
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D31
type_guessed: string
name: parse_data
arguments: data_tag: D12
type_guessed: string
name: parse_data
arguments: data_tag: D6
type_guessed: string
name: parse_data
arguments: data_tag: D20
type_guessed: string

tool: D31: %d

tool: D12: 


tool: D6: fizz

tool: D20: buzz

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  if (argc != 2)
    return 1;
  if (sscanf(argv[1], "%d", &i) != 1)
    return 1;
  if (i % 2 == 0) {
    printf("fizz");
  }
  if (i % 3 == 0) {
    printf("buzz");
  }
  printf("\n");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<srfft>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x34(%rbp),%edx
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <srrec@plt>
  cmpl   $0x1,-0x34(%rbp)
  jle    L18
  mov    -0x34(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <BR_permute@plt>
  mov    -0x34(%rbp),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <BR_permute@plt>
L18:
  mov    -0x34(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  pxor   %xmm2,%xmm2
  cvtsi2sdl -0x1c(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  movsd  D5(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L23
L4:
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movsd  (%rax),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movsd  (%rax),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x20(%rbp)
L23:
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L4
  nop
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_infered: f64

tool: D5: 1

assistant: ```c
void srfft(double *xr, double *xi, int logm) {
  int i, m;
  double fac, *xrp, *xip;
  srrec(xr, xi, logm);
  if (logm > 1) {
    BR_permute(xr, logm);
    BR_permute(xi, logm);
  }
  m = 1 << logm;
  fac = 1.0 / sqrt((double)m);
  xrp = xr;
  xip = xi;
  for (i = 0; i < m; i++) {
    *xrp++ *= fac;
    *xip++ *= fac;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<srfft>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r13d
  call   <srrec@plt>
  cmp    $0x1,%r13d
  jg     L28
L56:
  mov    $0x1,%r12d
  mov    %r13d,%ecx
  shl    %cl,%r12d
  pxor   %xmm0,%xmm0
  cvtsi2sd %r12d,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L17
  sqrtsd %xmm0,%xmm0
L38:
  movsd  D6(%rip),%xmm1
  divsd  %xmm0,%xmm1
  test   %r12d,%r12d
  jle    L53
  mov    %r12d,%r12d
  lea    (%rbx,%r12,8),%rax
L19:
  add    $0x8,%rbx
  movapd %xmm1,%xmm0
  mulsd  -0x8(%rbx),%xmm0
  movsd  %xmm0,-0x8(%rbx)
  add    $0x8,%rbp
  movapd %xmm1,%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  cmp    %rax,%rbx
  jne    L19
L53:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  mov    %r13d,%esi
  mov    %rbx,%rdi
  call   <BR_permute@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <BR_permute@plt>
  jmp    L56
L17:
  call   <sqrt@plt>
  jmp    L38
```

tool_calls: name: unpack_data
arguments: tag: D6
type_infered: f64

tool: D6: 1

assistant: ```c
void srfft(double *xr, double *xi, int logm) {
  int i, m;
  double fac, *xrp, *xip;
  srrec(xr, xi, logm);
  if (logm > 1) {
    BR_permute(xr, logm);
    BR_permute(xi, logm);
  }
  m = 1 << logm;
  fac = 1.0 / sqrt((double)m);
  xrp = xr;
  xip = xi;
  for (i = 0; i < m; i++) {
    *xrp++ *= fac;
    *xip++ *= fac;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<srfft>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <srrec@plt>
  cmp    $0x1,%r13d
  jle    L45
  mov    %r13d,%esi
  mov    %rbx,%rdi
  call   <BR_permute@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <BR_permute@plt>
L45:
  mov    $0x1,%r12d
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  mov    %r13d,%ecx
  shl    %cl,%r12d
  cvtsi2sd %r12d,%xmm0
  ucomisd %xmm0,%xmm1
  ja     L21
  sqrtsd %xmm0,%xmm0
L40:
  movsd  D2(%rip),%xmm1
  divsd  %xmm0,%xmm1
  test   %r12d,%r12d
  jle    L2
  movslq %r12d,%r12
  lea    (%rbx,%r12,8),%rax
  nopl   0x0(%rax,%rax,1)
L25:
  movsd  (%rbx),%xmm0
  add    $0x8,%rbp
  add    $0x8,%rbx
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbx)
  movsd  -0x8(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  cmp    %rax,%rbx
  jne    L25
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  call   <sqrt@plt>
  jmp    L40
```

tool_calls: name: unpack_data
arguments: data_tag: D2
type_infered: f64

tool: D2: 1

assistant: ```c
void srfft(double *xr, double *xi, int logm) {
  int i, m;
  double fac, *xrp, *xip;
  srrec(xr, xi, logm);
  if (logm > 1) {
    BR_permute(xr, logm);
    BR_permute(xi, logm);
  }
  m = 1 << logm;
  fac = 1.0 / sqrt((double)m);
  xrp = xr;
  xip = xi;
  for (i = 0; i < m; i++) {
    *xrp++ *= fac;
    *xip++ *= fac;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<srfft>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <srrec@plt>
  cmp    $0x1,%r13d
  jle    L77
  mov    %r13d,%esi
  mov    %rbx,%rdi
  call   <BR_permute@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <BR_permute@plt>
L77:
  mov    $0x1,%r12d
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  mov    %r13d,%ecx
  shl    %cl,%r12d
  cvtsi2sd %r12d,%xmm0
  ucomisd %xmm0,%xmm1
  ja     L69
  sqrtsd %xmm0,%xmm0
L75:
  movsd  D9(%rip),%xmm1
  divsd  %xmm0,%xmm1
  test   %r12d,%r12d
  jle    L22
  lea    0xf(%rbx),%rax
  sub    %rbp,%rax
  cmp    $0x1e,%rax
  jbe    L89
  test   %r13d,%r13d
  je     L89
  mov    %r12d,%edx
  movapd %xmm1,%xmm2
  xor    %eax,%eax
  shr    %edx
  unpcklpd %xmm2,%xmm2
  shl    $0x4,%rdx
  nopl   (%rax)
L63:
  movupd (%rbx,%rax,1),%xmm0
  mulpd  %xmm2,%xmm0
  movups %xmm0,(%rbx,%rax,1)
  movupd 0x0(%rbp,%rax,1),%xmm0
  mulpd  %xmm2,%xmm0
  movups %xmm0,0x0(%rbp,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L63
  mov    %r12d,%edx
  and    $0xfffffffe,%edx
  mov    %edx,%eax
  shl    $0x3,%rax
  add    %rax,%rbx
  add    %rbp,%rax
  cmp    %edx,%r12d
  je     L22
  movsd  (%rbx),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rbx)
  mulsd  (%rax),%xmm1
  movsd  %xmm1,(%rax)
L22:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L89:
  movslq %r12d,%r12
  lea    (%rbx,%r12,8),%rax
  nop
L16:
  movsd  (%rbx),%xmm0
  add    $0x8,%rbp
  add    $0x8,%rbx
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbx)
  movsd  -0x8(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  cmp    %rax,%rbx
  jne    L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L69:
  call   <sqrt@plt>
  jmp    L75
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_infered: f64

tool: D9: 1

assistant: ```c
void srfft(double *xr, double *xi, int logm) {
  int i, m;
  double fac, *xrp, *xip;
  srrec(xr, xi, logm);
  if (logm > 1) {
    BR_permute(xr, logm);
    BR_permute(xi, logm);
  }
  m = 1 << logm;
  fac = 1.0 / sqrt((double)m);
  xrp = xr;
  xip = xi;
  for (i = 0; i < m; i++) {
    *xrp++ *= fac;
    *xip++ *= fac;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<printmatw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L20
L17:
  movl   $0x0,-0x8(%rbp)
  jmp    L51
L2:
  mov    -0xc(%rbp),%eax
  imul   -0x20(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  cltd
  idivl  -0x24(%rbp)
  mov    %edx,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L27
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x8(%rbp)
  jge    L27
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L27:
  addl   $0x1,-0x8(%rbp)
L51:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L2
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0xc(%rbp)
L20:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L17
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D0
datatype: string
name: unpack_data
arguments: tag: D17
datatype: string

tool: D0: %9.3f 

tool: D17:   ...

assistant: ```c
void printmatw(double *a, int m, int n, int w) {
  int i, j, jmod;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      printf("%9.3f ", a[i * n + j]);
      jmod = (j + 1) % w;
      if ((jmod == 0) && (j < (n - 1))) {
        printf("  ...\n");
      }
    }
    printf("\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<printmatw>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %esi,0x14(%rsp)
  mov    %edx,0xc(%rsp)
  test   %esi,%esi
  jle    L0
  mov    %ecx,%ebp
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x8(%rsp)
  lea    D13(%rip),%r14
  lea    -0x1(%rdx),%r15d
  jmp    L80
L70:
  add    $0x1,%rbx
  cmp    %r13,%rbx
  je     L75
L28:
  movsd  (%r12,%rbx,8),%xmm0
  mov    %r14,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    0x1(%rbx),%eax
  cltd
  idiv   %ebp
  test   %edx,%edx
  jne    L70
  cmp    %ebx,%r15d
  jle    L70
  lea    D14(%rip),%rdi
  call   <puts@plt>
  jmp    L70
L75:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,0x8(%rsp)
  mov    0x8(%rsp),%eax
  mov    0xc(%rsp),%esi
  add    %esi,0x10(%rsp)
  cmp    %eax,0x14(%rsp)
  je     L0
L80:
  cmpl   $0x0,0xc(%rsp)
  jle    L75
  mov    0xc(%rsp),%r13d
  movslq 0x10(%rsp),%rax
  mov    0x18(%rsp),%rcx
  lea    (%rcx,%rax,8),%r12
  mov    $0x0,%ebx
  jmp    L28
L0:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_tag: D13
datatype: string
name: read_data
arguments: data_tag: D14
datatype: string

tool: D13: %9.3f 

tool: D14:   ...

assistant: ```c
void printmatw(double *a, int m, int n, int w) {
  int i, j, jmod;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      printf("%9.3f ", a[i * n + j]);
      jmod = (j + 1) % w;
      if ((jmod == 0) && (j < (n - 1))) {
        printf("  ...\n");
      }
    }
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<printmatw>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %esi,0x14(%rsp)
  mov    %edx,0xc(%rsp)
  test   %esi,%esi
  jle    L7
  movl   $0x0,0x10(%rsp)
  mov    %ecx,%ebx
  lea    -0x1(%rdx),%r14d
  lea    D12(%rip),%r13
  movl   $0x0,0x8(%rsp)
  nopl   0x0(%rax)
L9:
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L32
  movslq 0x10(%rsp),%rax
  mov    0x18(%rsp),%rcx
  mov    %r14d,%r12d
  xor    %r15d,%r15d
  lea    (%rcx,%rax,8),%rbp
  jmp    L15
  cs nopw 0x0(%rax,%rax,1)
L8:
  lea    0x1(%r15),%rax
  cmp    %r12,%r15
  je     L32
L10:
  mov    %rax,%r15
L15:
  movsd  0x0(%rbp,%r15,8),%xmm0
  mov    %r13,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    0x1(%r15),%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L8
  cmp    %r15d,%r14d
  jle    L8
  lea    D10(%rip),%rdi
  call   <puts@plt>
  lea    0x1(%r15),%rax
  cmp    %r12,%r15
  jne    L10
  nopl   (%rax)
L32:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,0x8(%rsp)
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%eax
  add    %esi,0x10(%rsp)
  cmp    %eax,0x14(%rsp)
  jne    L9
L7:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
type_guessed: string
name: disassemble_data
arguments: data_tag: D10
type_guessed: string

tool: D12: %9.3f 

tool: D10:   ...

assistant: ```c
void printmatw(double *a, int m, int n, int w) {
  int i, j, jmod;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      printf("%9.3f ", a[i * n + j]);
      jmod = (j + 1) % w;
      if ((jmod == 0) && (j < (n - 1))) {
        printf("  ...\n");
      }
    }
    printf("\n");
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<printmatw>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,0x20(%rsp)
  mov    %esi,0x28(%rsp)
  test   %esi,%esi
  jle    L68
  lea    -0x1(%rdx),%eax
  movl   $0x0,0x14(%rsp)
  mov    %edx,%r13d
  mov    %ecx,%ebp
  cmp    %eax,%edx
  mov    %eax,0x2c(%rsp)
  lea    D16(%rip),%r12
  lea    D4(%rip),%r15
  cmovle %edx,%eax
  movl   $0x0,0x10(%rsp)
  mov    %eax,0xc(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L67:
  test   %r13d,%r13d
  jle    L78
  mov    0x2c(%rsp),%eax
  test   %eax,%eax
  jle    L61
  movslq 0x14(%rsp),%rax
  mov    0x20(%rsp),%rsi
  xor    %r14d,%r14d
  mov    %rax,0x18(%rsp)
  lea    (%rsi,%rax,8),%rbx
  nopl   0x0(%rax,%rax,1)
L74:
  movsd  (%rbx),%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  add    $0x1,%r14d
  call   <printf@plt>
  mov    %r14d,%eax
  cltd
  idiv   %ebp
  test   %edx,%edx
  jne    L50
  mov    %r15,%rdi
  call   <puts@plt>
L50:
  add    $0x8,%rbx
  cmp    0xc(%rsp),%r14d
  jl     L74
  cmp    %r13d,%r14d
  jge    L78
L69:
  mov    0x20(%rsp),%rax
  mov    0x18(%rsp),%rcx
  movslq %r14d,%rbx
  lea    (%rax,%rcx,8),%r14
  nopl   0x0(%rax,%rax,1)
L82:
  movsd  (%r14,%rbx,8),%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,%r13d
  jg     L82
L78:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,0x10(%rsp)
  mov    0x10(%rsp),%eax
  add    %r13d,0x14(%rsp)
  cmp    %eax,0x28(%rsp)
  jne    L67
L68:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L61:
  movslq 0x14(%rsp),%rax
  xor    %r14d,%r14d
  mov    %rax,0x18(%rsp)
  jmp    L69
```

tool_calls: name: unpack_data
arguments: data_tag: D16
type_guessed: string
name: unpack_data
arguments: data_tag: D4
type_guessed: string

tool: D16: %9.3f 

tool: D4:   ...

assistant: ```c
void printmatw(double *a, int m, int n, int w) {
  int i, j, jmod;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      printf("%9.3f ", a[i * n + j]);
      jmod = (j + 1) % w;
      if ((jmod == 0) && (j < (n - 1))) {
        printf("  ...\n");
      }
    }
    printf("\n");
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L11
L5:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L11:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x8(%rbp)
  jl     L5
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type_infered: string

tool: D8: the reversed string is = %s

assistant: ```c
void reverse(char s[]) {
  char temp;
  int i, len;
  len = strlen(s) - 1;
  for (i = 0; i < strlen(s) / 2; i++) {
    temp = s[i];
    s[i] = s[len];
    s[len] = temp;
    len--;
  }
  printf("the reversed string is = %s", s);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  lea    -0x1(%rax),%ebp
  movslq %ebp,%rbp
  add    %r12,%rbp
  mov    $0x0,%ebx
  jmp    L3
L8:
  movzbl (%r12,%rbx,1),%eax
  movzbl 0x0(%rbp),%edx
  mov    %dl,(%r12,%rbx,1)
  mov    %al,0x0(%rbp)
  add    $0x1,%rbx
  sub    $0x1,%rbp
L3:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rax,1),%eax
  sar    %eax
  cmp    %ebx,%eax
  jg     L8
  mov    %r12,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
type_guessed: string

tool: D7: the reversed string is = %s

assistant: ```c
void reverse(char s[]) {
  char temp;
  int i, len;
  len = strlen(s) - 1;
  for (i = 0; i < strlen(s) / 2; i++) {
    temp = s[i];
    s[i] = s[len];
    s[len] = temp;
    len--;
  }
  printf("the reversed string is = %s", s);
}
```