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

```asm
<choldc>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%r13
  mov    %esi,%r15d
  mov    %rdx,0x10(%rsp)
  mov    $0x1,%r12d
  mov    $0x0,%r14d
  test   %esi,%esi
  jg     L34
L66:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L39:
  pxor   %xmm2,%xmm2
  movq   %rbx,%xmm3
  comisd %xmm3,%xmm2
  jae    L9
L97:
  pxor   %xmm4,%xmm4
  movq   %rbx,%xmm5
  ucomisd %xmm5,%xmm4
  ja     L85
  movq   %rbx,%xmm0
  sqrtsd %xmm0,%xmm0
L5:
  mov    0x10(%rsp),%rax
  movsd  %xmm0,(%rax,%r12,8)
L81:
  add    $0x1,%rbp
  cmp    %ebp,%r15d
  jl     L102
L53:
  mov    0x0(%r13,%r12,8),%rdx
  mov    (%rdx,%rbp,8),%rbx
  test   %r14d,%r14d
  jle    L95
  mov    0x0(%r13,%rbp,8),%rcx
  mov    %r14,%rax
L124:
  movsd  (%rdx,%rax,8),%xmm0
  mulsd  (%rcx,%rax,8),%xmm0
  movq   %rbx,%xmm1
  subsd  %xmm0,%xmm1
  movq   %xmm1,%rbx
  sub    $0x1,%rax
  test   %eax,%eax
  jg     L124
L95:
  cmp    %ebp,0xc(%rsp)
  je     L39
  mov    0x0(%r13,%rbp,8),%rax
  mov    0x10(%rsp),%rsi
  movq   %rbx,%xmm0
  divsd  (%rsi,%r12,8),%xmm0
  mov    0x18(%rsp),%rdi
  movsd  %xmm0,(%rax,%rdi,1)
  jmp    L81
L9:
  lea    D4(%rip),%rdi
  call   <nrerror@plt>
  jmp    L97
L85:
  movq   %rbx,%xmm0
  call   <sqrt@plt>
  jmp    L5
L102:
  add    $0x1,%r14
  add    $0x1,%r12
  cmp    %r12d,%r15d
  jl     L66
L34:
  mov    %r12d,0xc(%rsp)
  lea    0x0(,%r12,8),%rax
  mov    %rax,0x18(%rsp)
  mov    %r12,%rbp
  cmp    %r12d,%r15d
  jge    L53
  jmp    L102
```

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

tool: D4: choldc failed

assistant: ```c
void choldc(double **a, int n, double p[]) {
  void nrerror(char error_text[]);
  int i, j, k;
  double sum;
  for (i = 1; i <= n; i++) {
    for (j = i; j <= n; j++) {
      for (sum = a[i][j], k = i - 1; k >= 1; k--)
        sum -= a[i][k] * a[j][k];
      if (i == j) {
        if (sum <= 0.0)
          nrerror("choldc failed");
        p[i] = sqrt(sum);
      } else
        a[j][i] = sum / p[i];
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<choldc>:
  endbr64
  test   %esi,%esi
  jle    L40
  push   %r15
  pxor   %xmm2,%xmm2
  mov    %esi,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
L42:
  lea    0x0(,%r12,8),%r8
  mov    %r12,%rbp
  nopl   0x0(%rax,%rax,1)
L44:
  mov    0x0(%r13,%r12,8),%rdx
  movsd  (%rdx,%rbp,8),%xmm0
  test   %r14,%r14
  je     L16
  mov    0x0(%r13,%rbp,8),%rcx
  mov    %r14,%rax
  nopw   0x0(%rax,%rax,1)
L53:
  movsd  (%rdx,%rax,8),%xmm1
  mulsd  (%rcx,%rax,8),%xmm1
  sub    $0x1,%rax
  subsd  %xmm1,%xmm0
  test   %eax,%eax
  jne    L53
L16:
  cmp    %ebp,%r12d
  je     L47
  mov    0x0(%r13,%rbp,8),%rax
  divsd  (%rbx,%r12,8),%xmm0
  movsd  %xmm0,(%rax,%r8,1)
L74:
  add    $0x1,%rbp
  cmp    %ebp,%r15d
  jge    L44
  add    $0x1,%r12
  add    $0x1,%r14
  cmp    %r12d,%r15d
  jge    L42
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L47:
  comisd %xmm0,%xmm2
  jae    L82
L0:
  ucomisd %xmm0,%xmm2
  ja     L49
  sqrtsd %xmm0,%xmm0
L51:
  movsd  %xmm0,(%rbx,%r12,8)
  jmp    L74
L82:
  lea    D3(%rip),%rdi
  mov    %r8,0x8(%rsp)
  movsd  %xmm0,(%rsp)
  call   <nrerror@plt>
  mov    0x8(%rsp),%r8
  movsd  (%rsp),%xmm0
  pxor   %xmm2,%xmm2
  jmp    L0
L40:
  ret
L49:
  mov    %r8,(%rsp)
  call   <sqrt@plt>
  mov    (%rsp),%r8
  pxor   %xmm2,%xmm2
  jmp    L51
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
datatype: string

tool: D3: choldc failed

assistant: ```c
void choldc(double **a, int n, double p[]) {
  void nrerror(char error_text[]);
  int i, j, k;
  double sum;
  for (i = 1; i <= n; i++) {
    for (j = i; j <= n; j++) {
      for (sum = a[i][j], k = i - 1; k >= 1; k--)
        sum -= a[i][k] * a[j][k];
      if (i == j) {
        if (sum <= 0.0)
          nrerror("choldc failed");
        p[i] = sqrt(sum);
      } else
        a[j][i] = sum / p[i];
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_fun>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L40
L1:
  cmpl   $0x0,-0x14(%rbp)
  jle    L39
  subl   $0x1,-0x14(%rbp)
  jmp    L37
L39:
  cmpl   $0x0,-0x18(%rbp)
  jle    L37
  subl   $0x1,-0x18(%rbp)
L37:
  addl   $0x1,-0x4(%rbp)
L40:
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  test   %eax,%eax
  jg     L1
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int test_fun(int x, int y) {
  int c = 0;
  while (x + y > 0) {
    if (x > 0) {
      x = x - 1;
    } else {
      if (y > 0) {
        y = y - 1;
      } else {
      }
    }
    c = c + 1;
  }
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<test_fun>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  test   %eax,%eax
  jle    L4
  mov    $0x0,%eax
  jmp    L21
L35:
  sub    $0x1,%edi
L1:
  add    $0x1,%eax
  lea    (%rdi,%rsi,1),%edx
  test   %edx,%edx
  jle    L34
L21:
  test   %edi,%edi
  jg     L35
  test   %esi,%esi
  setg   %dl
  movzbl %dl,%edx
  sub    %edx,%esi
  jmp    L1
L34:
  ret
L4:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int test_fun(int x, int y) {
  int c = 0;
  while (x + y > 0) {
    if (x > 0) {
      x = x - 1;
    } else {
      if (y > 0) {
        y = y - 1;
      } else {
      }
    }
    c = c + 1;
  }
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test_fun>:
  endbr64
  lea    (%rdi,%rsi,1),%edx
  xor    %eax,%eax
  test   %edx,%edx
  jg     L7
  jmp    L5
  nop
L40:
  sub    $0x1,%edi
  lea    (%rdi,%rsi,1),%edx
L58:
  add    $0x1,%eax
  test   %edx,%edx
  jle    L0
L7:
  test   %edi,%edi
  jg     L40
  test   %esi,%esi
  jle    L58
  sub    $0x1,%esi
  add    $0x1,%eax
  lea    (%rdi,%rsi,1),%edx
  test   %edx,%edx
  jg     L7
L0:
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  ret
```

assistant: ```c
int test_fun(int x, int y) {
  int c = 0;
  while (x + y > 0) {
    if (x > 0) {
      x = x - 1;
    } else {
      if (y > 0) {
        y = y - 1;
      } else {
      }
    }
    c = c + 1;
  }
  return c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_fun>:
  endbr64
  lea    (%rdi,%rsi,1),%edx
  xor    %eax,%eax
  test   %edx,%edx
  jle    L61
  nopl   (%rax)
L60:
  test   %edi,%edi
  jle    L40
  sub    $0x1,%edi
  add    $0x1,%eax
  lea    (%rsi,%rdi,1),%edx
  test   %edx,%edx
  jg     L60
  ret
  nopw   0x0(%rax,%rax,1)
L40:
  test   %esi,%esi
  jle    L6
  sub    $0x1,%esi
  lea    (%rdi,%rsi,1),%edx
L6:
  add    $0x1,%eax
  test   %edx,%edx
  jle    L58
L11:
  test   %esi,%esi
  jle    L55
  sub    $0x1,%esi
  lea    (%rdi,%rsi,1),%edx
L55:
  add    $0x1,%eax
  test   %edx,%edx
  jg     L11
  ret
  nopl   0x0(%rax,%rax,1)
L58:
  ret
  nopl   0x0(%rax)
L61:
  ret
```

assistant: ```c
int test_fun(int x, int y) {
  int c = 0;
  while (x + y > 0) {
    if (x > 0) {
      x = x - 1;
    } else {
      if (y > 0) {
        y = y - 1;
      } else {
      }
    }
    c = c + 1;
  }
  return c;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cc_challenge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x81,-0x4(%rbp)
  je     L3
  cmpl   $0x82,-0x4(%rbp)
  je     L25
  jmp    L16
L3:
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L23
L25:
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L23
L16:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <decode_cc@plt>
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
L23:
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
datatype: string
name: disassemble_data
arguments: data_tag: D28
datatype: string
name: disassemble_data
arguments: data_tag: D14
datatype: string

tool: D7: GetSessChallenge: Invalid user name

tool: D28: GetSessChallenge: Null user name not enabled

tool: D14: GetSessChallenge: %s


assistant: ```c
void cc_challenge(int cc) {
  switch (cc) {
  case 0x81:
    printf("GetSessChallenge: Invalid user name\n");
    break;
  case 0x82:
    printf("GetSessChallenge: Null user name not enabled\n");
    break;
  default:
    printf("GetSessChallenge: %s\n", decode_cc((unsigned short)0, cc));
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cc_challenge>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x81,%edi
  je     L8
  cmp    $0x82,%edi
  je     L33
  mov    %edi,%esi
  mov    $0x0,%edi
  call   <decode_cc@plt>
  mov    %rax,%rsi
  lea    D25(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L2
L8:
  lea    D15(%rip),%rdi
  call   <puts@plt>
L2:
  add    $0x8,%rsp
  ret
L33:
  lea    D11(%rip),%rdi
  call   <puts@plt>
  jmp    L2
```

tool_calls: name: read_data
arguments: label: D25
data_type: string
name: read_data
arguments: label: D15
data_type: string
name: read_data
arguments: label: D11
data_type: string

tool: D25: GetSessChallenge: %s


tool: D15: GetSessChallenge: Invalid user name

tool: D11: GetSessChallenge: Null user name not enabled

assistant: ```c
void cc_challenge(int cc) {
  switch (cc) {
  case 0x81:
    printf("GetSessChallenge: Invalid user name\n");
    break;
  case 0x82:
    printf("GetSessChallenge: Null user name not enabled\n");
    break;
  default:
    printf("GetSessChallenge: %s\n", decode_cc((unsigned short)0, cc));
    break;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cc_challenge>:
  endbr64
  cmp    $0x81,%edi
  je     L12
  cmp    $0x82,%edi
  jne    L11
  lea    D16(%rip),%rdi
  jmp    1070 <puts@plt>
L11:
  mov    %edi,%esi
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <decode_cc@plt>
  lea    D11(%rip),%rdi
  add    $0x8,%rsp
  mov    %rax,%rsi
  xor    %eax,%eax
  jmp    1080 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L12:
  lea    D14(%rip),%rdi
  jmp    1070 <puts@plt>
```

tool_calls: name: unpack_data
arguments: tag: D16
type: string
name: unpack_data
arguments: tag: D11
type: string
name: unpack_data
arguments: tag: D14
type: string

tool: D16: GetSessChallenge: Null user name not enabled

tool: D11: GetSessChallenge: %s


tool: D14: GetSessChallenge: Invalid user name

assistant: ```c
void cc_challenge(int cc) {
  switch (cc) {
  case 0x81:
    printf("GetSessChallenge: Invalid user name\n");
    break;
  case 0x82:
    printf("GetSessChallenge: Null user name not enabled\n");
    break;
  default:
    printf("GetSessChallenge: %s\n", decode_cc((unsigned short)0, cc));
    break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<cc_challenge>:
  endbr64
  cmp    $0x81,%edi
  je     L17
  cmp    $0x82,%edi
  jne    L13
  lea    D23(%rip),%rdi
  jmp    1070 <puts@plt>
L13:
  mov    %edi,%esi
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <decode_cc@plt>
  lea    D6(%rip),%rdi
  add    $0x8,%rsp
  mov    %rax,%rsi
  xor    %eax,%eax
  jmp    1080 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L17:
  lea    D26(%rip),%rdi
  jmp    1070 <puts@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D23
datatype: string
name: retrieve_data
arguments: data_label: D6
datatype: string
name: retrieve_data
arguments: data_label: D26
datatype: string

tool: D23: GetSessChallenge: Null user name not enabled

tool: D6: GetSessChallenge: %s


tool: D26: GetSessChallenge: Invalid user name

assistant: ```c
void cc_challenge(int cc) {
  switch (cc) {
  case 0x81:
    printf("GetSessChallenge: Invalid user name\n");
    break;
  case 0x82:
    printf("GetSessChallenge: Null user name not enabled\n");
    break;
  default:
    printf("GetSessChallenge: %s\n", decode_cc((unsigned short)0, cc));
    break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rop3_1_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  movzbl -0x20(%rbp),%eax
  or     -0x1c(%rbp),%al
  mov    %al,-0x2(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  or     -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  notb   -0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  and    -0x1c(%rbp),%al
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  and    %edx,%eax
  or     %eax,%ecx
  mov    %ecx,%edx
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  and    %eax,%ecx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    %ecx,%eax
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_1_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = T | S;
  stk1 = *D | stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rop3_1_0_1>:
  endbr64
  mov    %esi,%eax
  or     (%rdi),%al
  or     %edx,%eax
  not    %esi
  and    %edx,%esi
  and    (%rdi),%sil
  not    %eax
  or     %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_1_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = T | S;
  stk1 = *D | stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rop3_1_0_1>:
  endbr64
  movzbl (%rdi),%eax
  or     %esi,%eax
  not    %esi
  or     %edx,%eax
  and    %edx,%esi
  and    (%rdi),%sil
  not    %eax
  or     %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_1_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = T | S;
  stk1 = *D | stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rop3_1_0_1>:
  endbr64
  movzbl (%rdi),%eax
  or     %esi,%eax
  not    %esi
  or     %edx,%eax
  and    %edx,%esi
  and    (%rdi),%sil
  not    %eax
  or     %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_1_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = T | S;
  stk1 = *D | stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jns    L25
  mov    -0x18(%rbp),%eax
  neg    %eax
  mov    %eax,-0x18(%rbp)
L25:
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L46
L23:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x14(%rbp)
L46:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L23
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <selection_sort@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L28
L57:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L28:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L57
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type_guessed: string
name: disassemble_data
arguments: label: D62
type_guessed: string
name: disassemble_data
arguments: label: D10
type_guessed: string
name: disassemble_data
arguments: label: D15
type_guessed: string
name: disassemble_data
arguments: label: D22
type_guessed: string

tool: D8: Enter the number of elements: 

tool: D62: %d

tool: D10: 


tool: D15: [%d] = 

tool: D22: [%d]: %d


assistant: ```c
int main() {
  int n, i;
  int *array;
  printf("Enter the number of elements: ");
  scanf("%d", &n);
  if (n < 0) {
    n = -n;
  }
  array = (int *)malloc(sizeof(int) * n);
  printf("\n\n");
  for (i = 0; i < n; i++) {
    printf("[%d] = ", i);
    scanf("%d", &array[i]);
    printf("\n");
  }
  printf("\n\n");
  selection_sort(array, n);
  for (i = 0; i < n; i++) {
    printf("[%d]: %d\n", i, array[i]);
  }
  free(array);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D33(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D38(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  js     L5
L43:
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  lea    D4(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L33
  mov    %r13,%rbp
  mov    $0x0,%ebx
  lea    D55(%rip),%r14
  lea    D38(%rip),%r12
L29:
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0x4(%rsp)
  jg     L29
L33:
  lea    D4(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%esi
  mov    %r13,%rdi
  call   <selection_sort@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L27
  mov    %r13,%rbp
  mov    $0x0,%ebx
  lea    D1(%rip),%r12
L20:
  mov    0x0(%rbp),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0x4(%rsp)
  jg     L20
L27:
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  neg    %eax
  mov    %eax,0x4(%rsp)
  jmp    L43
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D33
type_infered: string
name: disassemble_data
arguments: label: D38
type_infered: string
name: disassemble_data
arguments: label: D4
type_infered: string
name: disassemble_data
arguments: label: D55
type_infered: string
name: disassemble_data
arguments: label: D1
type_infered: string

tool: D33: Enter the number of elements: 

tool: D38: %d

tool: D4: 


tool: D55: [%d] = 

tool: D1: [%d]: %d


assistant: ```c
int main() {
  int n, i;
  int *array;
  printf("Enter the number of elements: ");
  scanf("%d", &n);
  if (n < 0) {
    n = -n;
  }
  array = (int *)malloc(sizeof(int) * n);
  printf("\n\n");
  for (i = 0; i < n; i++) {
    printf("[%d] = ", i);
    scanf("%d", &array[i]);
    printf("\n");
  }
  printf("\n\n");
  selection_sort(array, n);
  for (i = 0; i < n; i++) {
    printf("[%d]: %d\n", i, array[i]);
  }
  free(array);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D30(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  lea    D35(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jns    L14
  neg    %edi
  mov    %edi,0x4(%rsp)
L14:
  shl    $0x2,%edi
  lea    D21(%rip),%r15
  call   <malloc@plt>
  mov    %r15,%rdi
  mov    %rax,%r13
  call   <puts@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L2
  mov    %r13,%rbp
  xor    %ebx,%ebx
  lea    D12(%rip),%r14
  xchg   %ax,%ax
L31:
  mov    %ebx,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L31
L2:
  mov    %r15,%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%esi
  mov    %r13,%rdi
  call   <selection_sort@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L51
  mov    %r13,%rbp
  xor    %ebx,%ebx
  lea    D27(%rip),%r12
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    0x0(%rbp),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L16
L51:
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D30
type_guessed: string
name: disassemble_data
arguments: tag: D35
type_guessed: string
name: disassemble_data
arguments: tag: D21
type_guessed: string
name: disassemble_data
arguments: tag: D12
type_guessed: string
name: disassemble_data
arguments: tag: D27
type_guessed: string

tool: D30: Enter the number of elements: 

tool: D35: %d

tool: D21: 


tool: D12: [%d] = 

tool: D27: [%d]: %d


assistant: ```c
int main() {
  int n, i;
  int *array;
  printf("Enter the number of elements: ");
  scanf("%d", &n);
  if (n < 0) {
    n = -n;
  }
  array = (int *)malloc(sizeof(int) * n);
  printf("\n\n");
  for (i = 0; i < n; i++) {
    printf("[%d] = ", i);
    scanf("%d", &array[i]);
    printf("\n");
  }
  printf("\n\n");
  selection_sort(array, n);
  for (i = 0; i < n; i++) {
    printf("[%d]: %d\n", i, array[i]);
  }
  free(array);
  return 0;
}
```