ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  movabs $0x966656409636261,%rax
  movabs $0x96c6b6a0a696867,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movabs $0xa727170096f6e6d,%rax
  mov    %rax,0x10(%rsp)
  movb   $0x0,0x18(%rsp)
  lea    0x20(%rsp),%rbx
  mov    %rsp,%rdi
  mov    %rbx,%rsi
  call   <escape@plt>
  mov    %rbx,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x410,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D4
data_type: string

tool: D4: %s


assistant: ```c
int main() {
  char s[] = "abc\tdef\tghi\njkl\tmno\tpqr\n";
  char t[1000];
  escape(s, t);
  printf("%s\n", t);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x410,%rsp
  movdqa D12(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rbp
  mov    %rsp,%rdi
  movb   $0x0,0x18(%rsp)
  movabs $0xa727170096f6e6d,%rax
  mov    %rbp,%rsi
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  call   <escape@plt>
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D12
data_type: byte[16]
name: retrieve_data
arguments: data_label: D2
data_type: string

tool: {"D12": [0x61,0x62,0x63,0x09,0x64,0x65,0x66,0x09,0x67,0x68,0x69,0x0a,0x6a,0x6b,0x6c,0x09]}

tool: D2: %s


assistant: ```c
int main() {
  char s[] = "abc\tdef\tghi\njkl\tmno\tpqr\n";
  char t[1000];
  escape(s, t);
  printf("%s\n", t);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x410,%rsp
  movdqa D19(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rbp
  mov    %rsp,%rdi
  movb   $0x0,0x18(%rsp)
  movabs $0xa727170096f6e6d,%rax
  mov    %rbp,%rsi
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  call   <escape@plt>
  mov    %rbp,%rsi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D19
type: byte[16]
name: retrieve_data
arguments: label: D12
type: string

tool: {"D19": [0x61,0x62,0x63,0x09,0x64,0x65,0x66,0x09,0x67,0x68,0x69,0x0a,0x6a,0x6b,0x6c,0x09]}

tool: D12: %s


assistant: ```c
int main() {
  char s[] = "abc\tdef\tghi\njkl\tmno\tpqr\n";
  char t[1000];
  escape(s, t);
  printf("%s\n", t);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mod_with_sub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  jmp    L0
L11:
  mov    -0x8(%rbp),%eax
  sub    %eax,-0x4(%rbp)
L0:
  cmpl   $0x0,-0x4(%rbp)
  jg     L11
  cmpl   $0x0,-0x4(%rbp)
  je     L22
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  jmp    L5
L22:
  mov    $0x0,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int mod_with_sub(int a, int b) {
  while (a > 0)
    a -= b;
  return a ? a + b : 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mod_with_sub>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  jle    L17
L4:
  sub    %esi,%eax
  test   %eax,%eax
  jg     L4
L17:
  add    %eax,%esi
  test   %eax,%eax
  cmovne %esi,%eax
  ret
```

assistant: ```c
int mod_with_sub(int a, int b) {
  while (a > 0)
    a -= b;
  return a ? a + b : 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mod_with_sub>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  jle    L9
  nopw   0x0(%rax,%rax,1)
L1:
  sub    %esi,%eax
  test   %eax,%eax
  jg     L1
L9:
  add    %eax,%esi
  test   %eax,%eax
  cmovne %esi,%eax
  ret
```

assistant: ```c
int mod_with_sub(int a, int b) {
  while (a > 0)
    a -= b;
  return a ? a + b : 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mod_with_sub>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  jle    L14
  nopw   0x0(%rax,%rax,1)
L9:
  sub    %esi,%eax
  test   %eax,%eax
  jg     L9
L14:
  add    %eax,%esi
  test   %eax,%eax
  cmovne %esi,%eax
  ret
```

assistant: ```c
int mod_with_sub(int a, int b) {
  while (a > 0)
    a -= b;
  return a ? a + b : 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<TriShapeAndDrv>:
  endbr64
  mov    %rcx,%rax
  mov    $0x0,%ecx
  cmp    $0x2,%edi
  jg     L8
  movsd  (%rsi),%xmm0
  movsd  0x8(%rsi),%xmm2
  movsd  D44(%rip),%xmm3
  movapd %xmm3,%xmm1
  subsd  %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  movsd  %xmm0,(%rdx)
  movsd  %xmm3,(%rax)
  movq   $0x0,0x8(%rax)
  movsd  %xmm2,0x8(%rdx)
  movq   $0x0,0x10(%rax)
  movsd  %xmm3,0x18(%rax)
  movsd  %xmm1,0x10(%rdx)
  movsd  D14(%rip),%xmm3
  movsd  %xmm3,0x28(%rax)
  movsd  %xmm3,0x20(%rax)
  mov    $0x3,%ecx
  jne    L8
  movsd  D39(%rip),%xmm4
  movapd %xmm0,%xmm3
  mulsd  %xmm4,%xmm3
  movapd %xmm2,%xmm5
  mulsd  %xmm3,%xmm5
  movsd  %xmm5,0x18(%rdx)
  mulsd  %xmm2,%xmm4
  movapd %xmm1,%xmm5
  mulsd  %xmm4,%xmm5
  movsd  %xmm5,0x20(%rdx)
  mulsd  %xmm3,%xmm1
  movsd  %xmm1,0x28(%rdx)
  movsd  %xmm4,0x30(%rax)
  movsd  %xmm3,0x38(%rax)
  movq   D51(%rip),%xmm1
  movapd %xmm4,%xmm5
  xorpd  %xmm1,%xmm5
  movsd  %xmm5,0x40(%rax)
  xorpd  %xmm1,%xmm3
  movsd  D23(%rip),%xmm5
  movapd %xmm3,%xmm1
  subsd  %xmm5,%xmm1
  movsd  D58(%rip),%xmm6
  mulsd  %xmm6,%xmm2
  addsd  %xmm2,%xmm1
  movsd  %xmm1,0x48(%rax)
  mulsd  %xmm6,%xmm0
  subsd  %xmm5,%xmm0
  subsd  %xmm4,%xmm0
  movsd  %xmm0,0x50(%rax)
  movsd  %xmm3,0x58(%rax)
  mov    $0x6,%ecx
L8:
  mov    %ecx,%eax
  ret
```

tool_calls: name: read_data
arguments: data_label: D44
datatype: f64
name: read_data
arguments: data_label: D14
datatype: f32[2]
name: read_data
arguments: data_label: D39
datatype: f32[2]
name: read_data
arguments: data_label: D51
datatype: f32[2]
name: read_data
arguments: data_label: D23
datatype: f64
name: read_data
arguments: data_label: D58
datatype: f64

tool: D44: 1

tool: {"D14": [0.0,-1.875]}

tool: {"D39": [0.0,-2.0]}

tool: {"D51": [0.0,-0.0]}

tool: D23: 2

tool: D58: 4

assistant: ```c
int TriShapeAndDrv(int p, double par[2], double N[], double dN[][2]) {
  int i, j, nshp = 0;
  double L[3];
  if (p > 2)
    return nshp;
  L[0] = par[0];
  L[1] = par[1];
  L[2] = 1.0e0 - par[0] - par[1];
  for (i = 0; i < 3; i++) {
    N[i] = L[i];
    if (i == 2)
      dN[i][0] = dN[i][1] = -1.0e0;
    else {
      for (j = 0; j < 2; j++) {
        if (i == j)
          dN[i][j] = 1.0e0;
        else
          dN[i][j] = 0.0e0;
      }
    }
  }
  nshp = 3;
  if (p > 1) {
    N[3] = -2.0e0 * L[0] * L[1];
    N[4] = -2.0e0 * L[1] * L[2];
    N[5] = -2.0e0 * L[0] * L[2];
    dN[3][0] = -2.0e0 * L[1];
    dN[3][1] = -2.0e0 * L[0];
    dN[4][0] = 2.0e0 * L[1];
    dN[4][1] = -2.0e0 + 2.0e0 * L[0] + 4.0e0 * L[1];
    dN[5][0] = -2.0e0 + 4.0e0 * L[0] + 2.0e0 * L[1];
    dN[5][1] = 2.0e0 * L[0];
    nshp = 6;
  }
  return nshp;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<TriShapeAndDrv>:
  endbr64
  xor    %eax,%eax
  cmp    $0x2,%edi
  jg     L0
  movsd  (%rsi),%xmm0
  movsd  0x8(%rsi),%xmm2
  mov    $0x3,%eax
  movsd  D1(%rip),%xmm3
  movsd  %xmm0,(%rdx)
  movapd %xmm3,%xmm1
  movsd  %xmm3,(%rcx)
  subsd  %xmm0,%xmm1
  movq   $0x0,0x8(%rcx)
  movsd  %xmm2,0x8(%rdx)
  movsd  %xmm3,0x18(%rcx)
  movsd  D56(%rip),%xmm3
  subsd  %xmm2,%xmm1
  movq   $0x0,0x10(%rcx)
  movsd  %xmm1,0x10(%rdx)
  movsd  %xmm3,0x28(%rcx)
  movsd  %xmm3,0x20(%rcx)
  je     L9
L0:
  ret
  nopl   0x0(%rax)
L9:
  movsd  D22(%rip),%xmm4
  movapd %xmm0,%xmm3
  movapd %xmm2,%xmm5
  movsd  D53(%rip),%xmm6
  mov    $0x6,%eax
  mulsd  %xmm4,%xmm3
  mulsd  %xmm2,%xmm4
  mulsd  %xmm6,%xmm0
  mulsd  %xmm6,%xmm2
  mulsd  %xmm3,%xmm5
  movsd  %xmm5,0x18(%rdx)
  movapd %xmm1,%xmm5
  mulsd  %xmm3,%xmm1
  mulsd  %xmm4,%xmm5
  movsd  %xmm1,0x28(%rdx)
  movq   D55(%rip),%xmm1
  movsd  %xmm5,0x20(%rdx)
  movapd %xmm4,%xmm5
  xorpd  %xmm1,%xmm5
  movsd  %xmm3,0x38(%rcx)
  xorpd  %xmm1,%xmm3
  movsd  %xmm5,0x40(%rcx)
  movsd  D7(%rip),%xmm5
  movapd %xmm3,%xmm1
  movsd  %xmm4,0x30(%rcx)
  subsd  %xmm5,%xmm1
  subsd  %xmm5,%xmm0
  movsd  %xmm3,0x58(%rcx)
  addsd  %xmm2,%xmm1
  subsd  %xmm4,%xmm0
  movsd  %xmm1,0x48(%rcx)
  movsd  %xmm0,0x50(%rcx)
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type_infered: f64
name: retrieve_data
arguments: data_tag: D56
type_infered: f32[2]
name: retrieve_data
arguments: data_tag: D22
type_infered: f32[2]
name: retrieve_data
arguments: data_tag: D53
type_infered: f64
name: retrieve_data
arguments: data_tag: D55
type_infered: f32[2]
name: retrieve_data
arguments: data_tag: D7
type_infered: f64

tool: D1: 1

tool: {"D56": [0.0,-1.875]}

tool: {"D22": [0.0,-2.0]}

tool: D53: 4

tool: {"D55": [0.0,-0.0]}

tool: D7: 2

assistant: ```c
int TriShapeAndDrv(int p, double par[2], double N[], double dN[][2]) {
  int i, j, nshp = 0;
  double L[3];
  if (p > 2)
    return nshp;
  L[0] = par[0];
  L[1] = par[1];
  L[2] = 1.0e0 - par[0] - par[1];
  for (i = 0; i < 3; i++) {
    N[i] = L[i];
    if (i == 2)
      dN[i][0] = dN[i][1] = -1.0e0;
    else {
      for (j = 0; j < 2; j++) {
        if (i == j)
          dN[i][j] = 1.0e0;
        else
          dN[i][j] = 0.0e0;
      }
    }
  }
  nshp = 3;
  if (p > 1) {
    N[3] = -2.0e0 * L[0] * L[1];
    N[4] = -2.0e0 * L[1] * L[2];
    N[5] = -2.0e0 * L[0] * L[2];
    dN[3][0] = -2.0e0 * L[1];
    dN[3][1] = -2.0e0 * L[0];
    dN[4][0] = 2.0e0 * L[1];
    dN[4][1] = -2.0e0 + 2.0e0 * L[0] + 4.0e0 * L[1];
    dN[5][0] = -2.0e0 + 4.0e0 * L[0] + 2.0e0 * L[1];
    dN[5][1] = 2.0e0 * L[0];
    nshp = 6;
  }
  return nshp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<TriShapeAndDrv>:
  endbr64
  xor    %eax,%eax
  cmp    $0x2,%edi
  jg     L7
  movsd  (%rsi),%xmm0
  movsd  0x8(%rsi),%xmm2
  mov    $0x3,%eax
  movsd  D52(%rip),%xmm3
  movsd  %xmm0,(%rdx)
  movapd %xmm3,%xmm1
  movsd  %xmm3,(%rcx)
  subsd  %xmm0,%xmm1
  movq   $0x0,0x8(%rcx)
  movsd  %xmm2,0x8(%rdx)
  movsd  %xmm3,0x18(%rcx)
  movsd  D40(%rip),%xmm3
  subsd  %xmm2,%xmm1
  movq   $0x0,0x10(%rcx)
  movsd  %xmm1,0x10(%rdx)
  movsd  %xmm3,0x28(%rcx)
  movsd  %xmm3,0x20(%rcx)
  je     L8
L7:
  ret
  nopl   0x0(%rax)
L8:
  movsd  D50(%rip),%xmm4
  movapd %xmm0,%xmm3
  movapd %xmm2,%xmm5
  movapd %xmm1,%xmm6
  movsd  D35(%rip),%xmm7
  mov    $0x6,%eax
  mulsd  %xmm4,%xmm3
  mulsd  %xmm2,%xmm4
  mulsd  %xmm7,%xmm0
  mulsd  %xmm7,%xmm2
  mulsd  %xmm3,%xmm1
  mulsd  %xmm4,%xmm6
  mulsd  %xmm3,%xmm5
  movsd  %xmm1,0x28(%rdx)
  movq   D25(%rip),%xmm1
  unpcklpd %xmm6,%xmm5
  movsd  D8(%rip),%xmm6
  movups %xmm5,0x18(%rdx)
  movsd  %xmm3,0x38(%rcx)
  xorpd  %xmm1,%xmm3
  subsd  %xmm6,%xmm0
  xorpd  %xmm4,%xmm1
  movapd %xmm3,%xmm5
  movsd  %xmm4,0x30(%rcx)
  subsd  %xmm6,%xmm5
  subsd  %xmm4,%xmm0
  addsd  %xmm2,%xmm5
  unpcklpd %xmm3,%xmm0
  movups %xmm0,0x50(%rcx)
  unpcklpd %xmm5,%xmm1
  movups %xmm1,0x40(%rcx)
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D52
type: f64
name: unpack_data
arguments: data_label: D40
type: f32[2]
name: unpack_data
arguments: data_label: D50
type: f32[2]
name: unpack_data
arguments: data_label: D35
type: f64
name: unpack_data
arguments: data_label: D25
type: f32[2]
name: unpack_data
arguments: data_label: D8
type: f64

tool: D52: 1

tool: {"D40": [0.0,-1.875]}

tool: {"D50": [0.0,-2.0]}

tool: D35: 4

tool: {"D25": [0.0,-0.0]}

tool: D8: 2

assistant: ```c
int TriShapeAndDrv(int p, double par[2], double N[], double dN[][2]) {
  int i, j, nshp = 0;
  double L[3];
  if (p > 2)
    return nshp;
  L[0] = par[0];
  L[1] = par[1];
  L[2] = 1.0e0 - par[0] - par[1];
  for (i = 0; i < 3; i++) {
    N[i] = L[i];
    if (i == 2)
      dN[i][0] = dN[i][1] = -1.0e0;
    else {
      for (j = 0; j < 2; j++) {
        if (i == j)
          dN[i][j] = 1.0e0;
        else
          dN[i][j] = 0.0e0;
      }
    }
  }
  nshp = 3;
  if (p > 1) {
    N[3] = -2.0e0 * L[0] * L[1];
    N[4] = -2.0e0 * L[1] * L[2];
    N[5] = -2.0e0 * L[0] * L[2];
    dN[3][0] = -2.0e0 * L[1];
    dN[3][1] = -2.0e0 * L[0];
    dN[4][0] = 2.0e0 * L[1];
    dN[4][1] = -2.0e0 + 2.0e0 * L[0] + 4.0e0 * L[1];
    dN[5][0] = -2.0e0 + 4.0e0 * L[0] + 2.0e0 * L[1];
    dN[5][1] = 2.0e0 * L[0];
    nshp = 6;
  }
  return nshp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movl   $0x1,-0x24(%rbp)
  jmp    L4
L20:
  pxor   %xmm2,%xmm2
  cvtsi2sdl -0x24(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x12(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <getLine@plt>
  lea    -0x12(%rbp),%rax
  mov    %rax,%rdi
  call   <stringToDouble@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  -0x20(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  addl   $0x1,-0x24(%rbp)
L4:
  cmpl   $0x4,-0x24(%rbp)
  jle    L20
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D6
type_infered: string
name: read_data
arguments: data_label: D8
type_infered: string

tool: D6: Enter number #%d: 

tool: D8: The sum of the 4 values is %lf


assistant: ```c
int main() {
  char input[10];
  int i;
  double sum = 0.0;
  for (i = 1; i <= 4; i++) {
    printf("Enter number #%d: ", i);
    getLine(input, 10);
    sum += stringToDouble(input);
  }
  printf("The sum of the 4 values is %lf\n", sum);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  pxor   %xmm2,%xmm2
  movq   %xmm2,%r14
  mov    $0x1,%ebx
  lea    D8(%rip),%r12
L3:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  mov    %r12,%rdi
  call   <printf@plt>
  lea    0xe(%rsp),%rbp
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <getLine@plt>
  mov    %rbp,%rdi
  call   <stringToDouble@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movq   %r14,%xmm1
  addsd  %xmm1,%xmm0
  movq   %xmm0,%r14
  add    $0x1,%ebx
  cmp    $0x5,%ebx
  jne    L3
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type_infered: string
name: retrieve_data
arguments: data_tag: D19
type_infered: string

tool: D8: Enter number #%d: 

tool: D19: The sum of the 4 values is %lf


assistant: ```c
int main() {
  char input[10];
  int i;
  double sum = 0.0;
  for (i = 1; i <= 4; i++) {
    printf("Enter number #%d: ", i);
    getLine(input, 10);
    sum += stringToDouble(input);
  }
  printf("The sum of the 4 values is %lf\n", sum);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  pxor   %xmm1,%xmm1
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x1e(%rsp),%rbp
L5:
  pxor   %xmm0,%xmm0
  mov    %r12,%rdi
  movsd  %xmm1,0x8(%rsp)
  cvtsi2sd %ebx,%xmm0
  add    $0x1,%ebx
  call   <printf@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <getLine@plt>
  mov    %rbp,%rdi
  call   <stringToDouble@plt>
  pxor   %xmm0,%xmm0
  movsd  0x8(%rsp),%xmm1
  cvtsi2sd %rax,%xmm0
  addsd  %xmm0,%xmm1
  cmp    $0x5,%ebx
  jne    L5
  movapd %xmm1,%xmm0
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
type: string
name: parse_data
arguments: tag: D3
type: string

tool: D0: Enter number #%d: 

tool: D3: The sum of the 4 values is %lf


assistant: ```c
int main() {
  char input[10];
  int i;
  double sum = 0.0;
  for (i = 1; i <= 4; i++) {
    printf("Enter number #%d: ", i);
    getLine(input, 10);
    sum += stringToDouble(input);
  }
  printf("The sum of the 4 values is %lf\n", sum);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  pxor   %xmm1,%xmm1
  lea    D17(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x1e(%rsp),%rbp
L16:
  pxor   %xmm0,%xmm0
  mov    %r12,%rdi
  movsd  %xmm1,0x8(%rsp)
  cvtsi2sd %ebx,%xmm0
  add    $0x1,%ebx
  call   <printf@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <getLine@plt>
  mov    %rbp,%rdi
  call   <stringToDouble@plt>
  pxor   %xmm0,%xmm0
  movsd  0x8(%rsp),%xmm1
  cvtsi2sd %rax,%xmm0
  addsd  %xmm0,%xmm1
  cmp    $0x5,%ebx
  jne    L16
  movapd %xmm1,%xmm0
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D17
type: string
name: extract_symbolic_data
arguments: data_label: D14
type: string

tool: D17: Enter number #%d: 

tool: D14: The sum of the 4 values is %lf


assistant: ```c
int main() {
  char input[10];
  int i;
  double sum = 0.0;
  for (i = 1; i <= 4; i++) {
    printf("Enter number #%d: ", i);
    getLine(input, 10);
    sum += stringToDouble(input);
  }
  printf("The sum of the 4 values is %lf\n", sum);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<s_swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x78(%rbp)
  mov    %esi,-0x7c(%rbp)
  mov    %edx,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x7c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x78(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  lea    -0x70(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x80(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x78(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x7c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x78(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x80(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x78(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  lea    -0x70(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void s_swap(char **Pair_Name, int i, int j) {
  char temp[100];
  strcpy(temp, Pair_Name[i]);
  strcpy(Pair_Name[i], Pair_Name[j]);
  strcpy(Pair_Name[j], temp);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<s_swap>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %rdi,%rbp
  mov    %edx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movslq %esi,%rsi
  lea    (%rdi,%rsi,8),%r13
  mov    %rsp,%r12
  mov    0x0(%r13),%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  movslq %ebx,%rbx
  lea    0x0(%rbp,%rbx,8),%rbx
  mov    (%rbx),%rsi
  mov    0x0(%r13),%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    (%rbx),%rdi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void s_swap(char **Pair_Name, int i, int j) {
  char temp[100];
  strcpy(temp, Pair_Name[i]);
  strcpy(Pair_Name[i], Pair_Name[j]);
  strcpy(Pair_Name[j], temp);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<s_swap>:
  endbr64
  push   %r13
  movslq %esi,%rsi
  push   %r12
  lea    (%rdi,%rsi,8),%r13
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %edx,%rbx
  lea    0x0(%rbp,%rbx,8),%rbx
  sub    $0x78,%rsp
  mov    0x0(%r13),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    (%rbx),%rsi
  mov    0x0(%r13),%rdi
  call   <strcpy@plt>
  mov    (%rbx),%rdi
  mov    %r12,%rsi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void s_swap(char **Pair_Name, int i, int j) {
  char temp[100];
  strcpy(temp, Pair_Name[i]);
  strcpy(Pair_Name[i], Pair_Name[j]);
  strcpy(Pair_Name[j], temp);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<s_swap>:
  endbr64
  push   %r13
  movslq %esi,%rsi
  push   %r12
  lea    (%rdi,%rsi,8),%r13
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %edx,%rbx
  lea    0x0(%rbp,%rbx,8),%rbx
  sub    $0x78,%rsp
  mov    0x0(%r13),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    (%rbx),%rsi
  mov    0x0(%r13),%rdi
  call   <strcpy@plt>
  mov    (%rbx),%rdi
  mov    %r12,%rsi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void s_swap(char **Pair_Name, int i, int j) {
  char temp[100];
  strcpy(temp, Pair_Name[i]);
  strcpy(Pair_Name[i], Pair_Name[j]);
  strcpy(Pair_Name[j], temp);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <time@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <localtime@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    0x4(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type_infered: string

tool: D4: %d horas, %d minutos


assistant: ```c
main() {
  struct tm *fh;
  time_t segundos;
  time(&segundos);
  fh = localtime(&segundos);
  printf("%d horas, %d minutos\n", fh->tm_hour, fh->tm_min);
}
```
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   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  mov    %rbx,%rdi
  call   <time@plt>
  mov    %rbx,%rdi
  call   <localtime@plt>
  mov    0x4(%rax),%edx
  mov    (%rax),%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
datatype: string

tool: D0: %d horas, %d minutos


assistant: ```c
main() {
  struct tm *fh;
  time_t segundos;
  time(&segundos);
  fh = localtime(&segundos);
  printf("%d horas, %d minutos\n", fh->tm_hour, fh->tm_min);
}
```