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

```asm
<pmod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  %xmm0,-0x8(%rbp)
  movsd  %xmm1,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <fmod@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L3
  movsd  -0x8(%rbp),%xmm0
  addsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
L3:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double pmod(double x, double y) {
  x = fmod(x, y);
  if (x < 0)
    x += y;
  return x;
}
```
user: What is the c source code of the assembly code below:

```asm
<pmod>:
  endbr64
  push   %rbx
  movq   %xmm1,%rbx
  call   <fmod@plt>
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  ja     L13
L6:
  pop    %rbx
  ret
L13:
  movq   %rbx,%xmm2
  addsd  %xmm2,%xmm0
  jmp    L6
```

assistant: ```c
double pmod(double x, double y) {
  x = fmod(x, y);
  if (x < 0)
    x += y;
  return x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pmod>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  pxor   %xmm2,%xmm2
  comisd %xmm0,%xmm2
  jbe    L3
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm1,%xmm0
L3:
  add    $0x18,%rsp
  ret
```

assistant: ```c
double pmod(double x, double y) {
  x = fmod(x, y);
  if (x < 0)
    x += y;
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pmod>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  pxor   %xmm2,%xmm2
  comisd %xmm0,%xmm2
  jbe    L8
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm1,%xmm0
L8:
  add    $0x18,%rsp
  ret
```

assistant: ```c
double pmod(double x, double y) {
  x = fmod(x, y);
  if (x < 0)
    x += y;
  return x;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<yule_distance_bool>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L101
L102:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L90
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L90
  mov    $0x1,%eax
  jmp    L116
L90:
  mov    $0x0,%eax
L116:
  add    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L88
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L88
  mov    $0x1,%eax
  jmp    L57
L88:
  mov    $0x0,%eax
L57:
  add    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L60
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L60
  mov    $0x1,%eax
  jmp    L118
L60:
  mov    $0x0,%eax
L118:
  add    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L40
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L40
  mov    $0x1,%eax
  jmp    L132
L40:
  mov    $0x0,%eax
L132:
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x14(%rbp)
L101:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L102
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x4(%rbp),%xmm0
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x8(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x10(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  imul   -0x8(%rbp),%eax
  add    %edx,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double yule_distance_bool(const char *u, const char *v, int n) {
  int i = 0;
  int ntt = 0, nff = 0, nft = 0, ntf = 0;
  for (i = 0; i < n; i++) {
    ntt += (u[i] && v[i]);
    ntf += (u[i] && !v[i]);
    nft += (!u[i] && v[i]);
    nff += (!u[i] && !v[i]);
  }
  return (2.0 * ntf * nft) / (double)(ntt * nff + ntf * nft);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<yule_distance_bool>:
  endbr64
  push   %rbp
  push   %rbx
  test   %edx,%edx
  jle    L119
  mov    %rdi,%r8
  mov    %edx,%edx
  mov    $0x0,%eax
  mov    $0x0,%r11d
  mov    $0x0,%r10d
  mov    $0x0,%edi
  mov    $0x0,%r9d
  mov    $0x1,%ebp
  mov    $0x0,%ebx
  jmp    L82
L119:
  mov    $0x0,%r11d
  mov    $0x0,%r10d
  mov    $0x0,%edi
  mov    $0x0,%r9d
L67:
  pxor   %xmm0,%xmm0
  cvtsi2sd %r11d,%xmm0
  addsd  %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %r10d,%xmm1
  mulsd  %xmm1,%xmm0
  imul   %edi,%r9d
  imul   %r11d,%r10d
  add    %r10d,%r9d
  pxor   %xmm1,%xmm1
  cvtsi2sd %r9d,%xmm1
  divsd  %xmm1,%xmm0
  pop    %rbx
  pop    %rbp
  ret
L83:
  add    $0x1,%r9d
L30:
  test   %cl,%cl
  jne    L8
L55:
  add    $0x1,%r10d
L8:
  mov    %ebx,%ecx
L113:
  add    %ecx,%edi
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L67
L82:
  movzbl (%r8,%rax,1),%ecx
  test   %cl,%cl
  je     L0
  cmpb   $0x0,(%rsi,%rax,1)
  jne    L83
  jne    L30
  add    $0x1,%r11d
  test   %cl,%cl
  jne    L8
  mov    %ebp,%ecx
  jmp    L113
L0:
  cmpb   $0x0,(%rsi,%rax,1)
  jne    L55
  cmpb   $0x0,(%rsi,%rax,1)
  jne    L8
  mov    %ebp,%ecx
  jmp    L113
```

assistant: ```c
double yule_distance_bool(const char *u, const char *v, int n) {
  int i = 0;
  int ntt = 0, nff = 0, nft = 0, ntf = 0;
  for (i = 0; i < n; i++) {
    ntt += (u[i] && v[i]);
    ntf += (u[i] && !v[i]);
    nft += (!u[i] && v[i]);
    nff += (!u[i] && !v[i]);
  }
  return (2.0 * ntf * nft) / (double)(ntt * nff + ntf * nft);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<yule_distance_bool>:
  endbr64
  test   %edx,%edx
  jle    L27
  movslq %edx,%rdx
  xor    %eax,%eax
  xor    %r11d,%r11d
  xor    %r8d,%r8d
  xor    %r9d,%r9d
  xor    %r10d,%r10d
  jmp    L26
  nop
L55:
  test   %cl,%cl
  je     L23
  add    $0x1,%r10d
L36:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L51
L26:
  cmpb   $0x0,(%rdi,%rax,1)
  movzbl (%rsi,%rax,1),%ecx
  jne    L55
  cmp    $0x1,%cl
  sbb    $0xffffffff,%r8d
  cmp    $0x1,%cl
  adc    $0x0,%r9d
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L26
L51:
  pxor   %xmm0,%xmm0
  imul   %r9d,%r10d
  pxor   %xmm1,%xmm1
  cvtsi2sd %r11d,%xmm0
  cvtsi2sd %r8d,%xmm1
  imul   %r8d,%r11d
  addsd  %xmm0,%xmm0
  add    %r11d,%r10d
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %r10d,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopl   (%rax)
L23:
  add    $0x1,%r11d
  jmp    L36
  xchg   %ax,%ax
L27:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double yule_distance_bool(const char *u, const char *v, int n) {
  int i = 0;
  int ntt = 0, nff = 0, nft = 0, ntf = 0;
  for (i = 0; i < n; i++) {
    ntt += (u[i] && v[i]);
    ntf += (u[i] && !v[i]);
    nft += (!u[i] && v[i]);
    nff += (!u[i] && !v[i]);
  }
  return (2.0 * ntf * nft) / (double)(ntt * nff + ntf * nft);
}
```
user: Translate the below assembly code into C programming language:

```asm
<yule_distance_bool>:
  endbr64
  test   %edx,%edx
  jle    L19
  movslq %edx,%rdx
  xor    %eax,%eax
  xor    %r10d,%r10d
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  xor    %r11d,%r11d
  jmp    L66
  nop
L6:
  test   %cl,%cl
  je     L11
  add    $0x1,%r11d
L30:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L17
L66:
  cmpb   $0x0,(%rdi,%rax,1)
  movzbl (%rsi,%rax,1),%ecx
  jne    L6
  cmp    $0x1,%cl
  sbb    $0xffffffff,%r9d
  cmp    $0x1,%cl
  adc    $0x0,%r8d
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L66
L17:
  pxor   %xmm0,%xmm0
  imul   %r11d,%r8d
  pxor   %xmm1,%xmm1
  cvtsi2sd %r10d,%xmm0
  cvtsi2sd %r9d,%xmm1
  imul   %r10d,%r9d
  addsd  %xmm0,%xmm0
  add    %r9d,%r8d
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %r8d,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopl   (%rax)
L11:
  add    $0x1,%r10d
  jmp    L30
  xchg   %ax,%ax
L19:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double yule_distance_bool(const char *u, const char *v, int n) {
  int i = 0;
  int ntt = 0, nff = 0, nft = 0, ntf = 0;
  for (i = 0; i < n; i++) {
    ntt += (u[i] && v[i]);
    ntf += (u[i] && !v[i]);
    nft += (!u[i] && v[i]);
    nff += (!u[i] && !v[i]);
  }
  return (2.0 * ntf * nft) / (double)(ntt * nff + ntf * nft);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sqrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L10
L13:
  movsd  -0x8(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movapd %xmm0,%xmm1
  addsd  -0x18(%rbp),%xmm1
  movsd  -0x8(%rbp),%xmm0
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  divsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L10:
  cmpl   $0x9,-0xc(%rbp)
  jle    L13
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double sqrt(double x) {
  double ret = x;
  int i;
  for (i = 0; i < 10; i++)
    ret = (ret * ret + x) / (ret * 2.0);
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sqrt>:
  endbr64
  movapd %xmm0,%xmm3
  mov    $0xa,%eax
L8:
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  addsd  %xmm3,%xmm1
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  movapd %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  sub    $0x1,%eax
  jne    L8
  ret
```

assistant: ```c
double sqrt(double x) {
  double ret = x;
  int i;
  for (i = 0; i < 10; i++)
    ret = (ret * ret + x) / (ret * 2.0);
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<sqrt>:
  endbr64
  movapd %xmm0,%xmm3
  mov    $0xa,%eax
  nopl   (%rax)
L4:
  movapd %xmm0,%xmm1
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm1
  addsd  %xmm0,%xmm2
  addsd  %xmm3,%xmm1
  movapd %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  sub    $0x1,%eax
  jne    L4
  ret
```

assistant: ```c
double sqrt(double x) {
  double ret = x;
  int i;
  for (i = 0; i < 10; i++)
    ret = (ret * ret + x) / (ret * 2.0);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sqrt>:
  endbr64
  movapd %xmm0,%xmm2
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  addsd  %xmm0,%xmm0
  divsd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm2,%xmm2
  addsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  addsd  %xmm0,%xmm0
  addsd  %xmm1,%xmm2
  divsd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm2,%xmm2
  addsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  addsd  %xmm0,%xmm0
  addsd  %xmm1,%xmm2
  divsd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm2,%xmm2
  addsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  addsd  %xmm0,%xmm0
  addsd  %xmm1,%xmm2
  divsd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm2,%xmm2
  addsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  addsd  %xmm0,%xmm0
  addsd  %xmm1,%xmm2
  divsd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm2,%xmm2
  addsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  ret
```

assistant: ```c
double sqrt(double x) {
  double ret = x;
  int i;
  for (i = 0; i < 10; i++)
    ret = (ret * ret + x) / (ret * 2.0);
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x1,-0x8(%rbp)
  jmp    L37
L35:
  movl   $0x1,-0x4(%rbp)
  jmp    L2
L26:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L26
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L37:
  cmpl   $0x5,-0x8(%rbp)
  jle    L35
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x1,-0x8(%rbp)
  jmp    L60
L3:
  movl   $0x5,-0x4(%rbp)
  jmp    L85
L57:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  subl   $0x1,-0x4(%rbp)
L85:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L57
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L60:
  cmpl   $0x5,-0x8(%rbp)
  jle    L3
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L96
L48:
  addl   $0x1,-0x8(%rbp)
L96:
  cmpl   $0x5,-0x8(%rbp)
  jle    L48
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
type: string
name: read_data
arguments: data_tag: D11
type: string
name: read_data
arguments: data_tag: D33
type: string
name: read_data
arguments: data_tag: D49
type: string
name: read_data
arguments: data_tag: D22
type: string

tool: D2: Star Pattern


tool: D11: *

tool: D33: 


tool: D49: Star Pattern Inverted


tool: D22: Equilateral Pyramid


assistant: ```c
int main() {
  int row, column;
  printf("Star Pattern\n");
  for (row = 1; row <= 5; row++) {
    for (column = 1; column <= row; column++) {
      printf("*");
    }
    printf("\n");
  }
  printf("Star Pattern Inverted\n");
  for (row = 1; row <= 5; row++) {
    for (column = 5; column >= row; column--) {
      printf("*");
    }
    printf("\n");
  }
  printf("Equilateral Pyramid\n");
  for (row = 0; row <= 5; row++) {
  }
}
```
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    $0x8,%rsp
  lea    D61(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%ebp
  lea    D24(%rip),%r12
  lea    D34(%rip),%r13
  jmp    L23
L57:
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jle    L57
L9:
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebp
  cmp    $0x6,%ebp
  je     L46
L23:
  mov    $0x1,%ebx
  test   %ebp,%ebp
  jg     L57
  jmp    L9
L46:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%ebp
  lea    D24(%rip),%r12
  lea    D34(%rip),%r13
L4:
  mov    $0x5,%ebx
  cmp    $0x5,%ebp
  jg     L44
L60:
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  cmp    %ebp,%ebx
  jge    L60
L44:
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebp
  cmp    $0x6,%ebp
  jne    L4
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: label: D61
datatype: string
name: unpack_data
arguments: label: D24
datatype: string
name: unpack_data
arguments: label: D34
datatype: string
name: unpack_data
arguments: label: D8
datatype: string
name: unpack_data
arguments: label: D30
datatype: string

tool: D61: Star Pattern


tool: D24: *

tool: D34: 


tool: D8: Star Pattern Inverted


tool: D30: Equilateral Pyramid


assistant: ```c
int main() {
  int row, column;
  printf("Star Pattern\n");
  for (row = 1; row <= 5; row++) {
    for (column = 1; column <= row; column++) {
      printf("*");
    }
    printf("\n");
  }
  printf("Star Pattern Inverted\n");
  for (row = 1; row <= 5; row++) {
    for (column = 5; column >= row; column--) {
      printf("*");
    }
    printf("\n");
  }
  printf("Equilateral Pyramid\n");
  for (row = 0; row <= 5; row++) {
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D46(%rip),%rdi
  lea    D11(%rip),%r13
  push   %r12
  mov    $0x2,%r12d
  push   %rbp
  lea    D41(%rip),%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <printf@plt>
L35:
  mov    $0x1,%ebx
L34:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r12d
  jne    L34
  mov    %r13,%rdi
  add    $0x1,%r12d
  call   <printf@plt>
  cmp    $0x7,%r12d
  jne    L35
  lea    D30(%rip),%rdi
  mov    $0x1,%r12d
  call   <printf@plt>
L7:
  mov    $0x5,%ebx
L12:
  mov    %rbp,%rdi
  sub    $0x1,%ebx
  call   <printf@plt>
  cmp    %r12d,%ebx
  jge    L12
  mov    %r13,%rdi
  add    $0x1,%r12d
  call   <printf@plt>
  cmp    $0x6,%r12d
  jne    L7
  lea    D33(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: data_label: D46
data_type: string
name: read_data
arguments: data_label: D11
data_type: string
name: read_data
arguments: data_label: D41
data_type: string
name: read_data
arguments: data_label: D30
data_type: string
name: read_data
arguments: data_label: D33
data_type: string

tool: D46: Star Pattern


tool: D11: 


tool: D41: *

tool: D30: Star Pattern Inverted


tool: D33: Equilateral Pyramid


assistant: ```c
int main() {
  int row, column;
  printf("Star Pattern\n");
  for (row = 1; row <= 5; row++) {
    for (column = 1; column <= row; column++) {
      printf("*");
    }
    printf("\n");
  }
  printf("Star Pattern Inverted\n");
  for (row = 1; row <= 5; row++) {
    for (column = 5; column >= row; column--) {
      printf("*");
    }
    printf("\n");
  }
  printf("Equilateral Pyramid\n");
  for (row = 0; row <= 5; row++) {
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D11(%rip),%rdi
  lea    D48(%rip),%r13
  push   %r12
  mov    $0x2,%r12d
  push   %rbp
  lea    D31(%rip),%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <printf@plt>
L15:
  mov    $0x1,%ebx
L3:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r12d
  jne    L3
  mov    %r13,%rdi
  add    $0x1,%r12d
  call   <printf@plt>
  cmp    $0x7,%r12d
  jne    L15
  lea    D33(%rip),%rdi
  mov    $0x1,%r12d
  call   <printf@plt>
L18:
  mov    $0x5,%ebx
L29:
  mov    %rbp,%rdi
  sub    $0x1,%ebx
  call   <printf@plt>
  cmp    %r12d,%ebx
  jge    L29
  mov    %r13,%rdi
  add    $0x1,%r12d
  call   <printf@plt>
  cmp    $0x6,%r12d
  jne    L18
  lea    D38(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D11
type_guessed: string
name: disassemble_data
arguments: tag: D48
type_guessed: string
name: disassemble_data
arguments: tag: D31
type_guessed: string
name: disassemble_data
arguments: tag: D33
type_guessed: string
name: disassemble_data
arguments: tag: D38
type_guessed: string

tool: D11: Star Pattern


tool: D48: 


tool: D31: *

tool: D33: Star Pattern Inverted


tool: D38: Equilateral Pyramid


assistant: ```c
int main() {
  int row, column;
  printf("Star Pattern\n");
  for (row = 1; row <= 5; row++) {
    for (column = 1; column <= row; column++) {
      printf("*");
    }
    printf("\n");
  }
  printf("Star Pattern Inverted\n");
  for (row = 1; row <= 5; row++) {
    for (column = 5; column >= row; column--) {
      printf("*");
    }
    printf("\n");
  }
  printf("Equilateral Pyramid\n");
  for (row = 0; row <= 5; row++) {
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0xa,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D66(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  movl   $0x14,(%rax)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <puts@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: parse_data
arguments: data_tag: D40
type: string
name: parse_data
arguments: data_tag: D18
type: string
name: parse_data
arguments: data_tag: D4
type: string
name: parse_data
arguments: data_tag: D66
type: string
name: parse_data
arguments: data_tag: D31
type: string
name: parse_data
arguments: data_tag: D64
type: string
name: parse_data
arguments: data_tag: D5
type: string

tool: D40: 
LEARNING THE BASICS OF POINTERS IN C


tool: D18: Address of variable 'number'.: %x


tool: D4: Value stored in 'int_pointer': %x



tool: D66: Value that variable 'number' holds: %d


tool: D31: *int_pointer holds the same value.: %d



tool: D64: Changed value of *int_pointer, printing 'number': %d


tool: D5: 
That's all folks!

assistant: ```c
int main() {
  printf("\nLEARNING THE BASICS OF POINTERS IN C\n\n");
  int number = 10;
  int *int_pointer;
  printf("Address of variable \'number\'.: %x\n", &number);
  int_pointer = &number;
  printf("Value stored in \'int_pointer\': %x\n\n", int_pointer);
  printf("Value that variable \'number\' holds: %d\n", number);
  printf("*int_pointer holds the same value.: %d\n\n", *int_pointer);
  *int_pointer = 20;
  printf("Changed value of *int_pointer, printing \'number\': %d\n", number);
  printf("\nThat\'s all folks!\n");
  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   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D32(%rip),%rdi
  call   <puts@plt>
  movl   $0xa,0x4(%rsp)
  lea    0x4(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D46(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D53(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x14,0x4(%rsp)
  mov    $0x14,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D54(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D32
type: string
name: extract_symbolic_data
arguments: label: D46
type: string
name: extract_symbolic_data
arguments: label: D53
type: string
name: extract_symbolic_data
arguments: label: D40
type: string
name: extract_symbolic_data
arguments: label: D9
type: string
name: extract_symbolic_data
arguments: label: D28
type: string
name: extract_symbolic_data
arguments: label: D54
type: string

tool: D32: 
LEARNING THE BASICS OF POINTERS IN C


tool: D46: Address of variable 'number'.: %x


tool: D53: Value stored in 'int_pointer': %x



tool: D40: Value that variable 'number' holds: %d


tool: D9: *int_pointer holds the same value.: %d



tool: D28: Changed value of *int_pointer, printing 'number': %d


tool: D54: 
That's all folks!

assistant: ```c
int main() {
  printf("\nLEARNING THE BASICS OF POINTERS IN C\n\n");
  int number = 10;
  int *int_pointer;
  printf("Address of variable \'number\'.: %x\n", &number);
  int_pointer = &number;
  printf("Value stored in \'int_pointer\': %x\n\n", int_pointer);
  printf("Value that variable \'number\' holds: %d\n", number);
  printf("*int_pointer holds the same value.: %d\n\n", *int_pointer);
  *int_pointer = 20;
  printf("Changed value of *int_pointer, printing \'number\': %d\n", number);
  printf("\nThat\'s all folks!\n");
  return 0;
}
```