ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D90(%rip),%rdi
  lea    D54(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  D75(%rip),%xmm0
  mulss  (%rsp),%xmm0
  lea    D64(%rip),%rdi
  addss  D72(%rip),%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  lea    D93(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  mov    $0x1,%eax
  movss  D31(%rip),%xmm1
  lea    D53(%rip),%rdi
  mulss  %xmm0,%xmm1
  mulss  %xmm1,%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  lea    D25(%rip),%rdi
  movss  D9(%rip),%xmm0
  mulss  0x4(%rsp),%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D90
type_guessed: string
name: unpack_data
arguments: label: D54
type_guessed: string
name: unpack_data
arguments: label: D75
type_guessed: f32
name: unpack_data
arguments: label: D64
type_guessed: string
name: unpack_data
arguments: label: D72
type_guessed: f32
name: unpack_data
arguments: label: D93
type_guessed: string
name: unpack_data
arguments: label: D31
type_guessed: f32
name: unpack_data
arguments: label: D53
type_guessed: string
name: unpack_data
arguments: label: D25
type_guessed: string
name: unpack_data
arguments: label: D9
type_guessed: byte[4]

tool: D90: what is x ? 

tool: D54: %f

tool: D75: 6

tool: D64: y = %.2f


tool: D72: 8

tool: D93: what is the radius ?

tool: D31: 10.13

tool: D53: the area = %.2f


tool: D25: the circumference = %.2f


tool: {"D9": [0x7b,0x14,0xa2,0x41]}

assistant: ```c
int main() {
  float x;
  float y;
  float radius, area, circumference;
  printf("what is x ? \n");
  scanf("%f", &x);
  y = 6.0;
  y *= x;
  y += 8.0;
  printf("y = %.2f\n", y);
  printf("what is the radius ?\n");
  scanf("%f", &radius);
  area = 10.13;
  area *= radius;
  area *= radius;
  printf("the area = %.2f\n", area);
  circumference = 2;
  circumference *= 10.13;
  circumference *= radius;
  printf("the circumference = %.2f\n", circumference);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<addBinary>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %rax,%r13
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%r12
  cmp    %rax,%r13
  mov    %rax,%rdi
  cmovae %r13,%rdi
  add    $0x2,%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L58
  lea    (%rbx,%r13,1),%rdx
  lea    0x0(%rbp,%r12,1),%rcx
  mov    %rax,%rsi
  mov    $0x0,%r10d
  mov    $0x0,%edi
  lea    D13(%rip),%r9
  jmp    L66
L64:
  mov    %rax,%rsi
L66:
  mov    %edi,%eax
  cmp    %rbx,%rdx
  jbe    L14
  movsbl -0x1(%rdx),%eax
  sub    $0x30,%eax
  lea    -0x1(%rdx),%rdx
L14:
  mov    %edi,%r11d
  cmp    %rbp,%rcx
  jbe    L27
  movsbl -0x1(%rcx),%r11d
  sub    $0x30,%r11d
  lea    -0x1(%rcx),%rcx
L27:
  add    %r10d,%eax
  add    %r11d,%eax
  mov    %eax,%r10d
  shr    $0x1f,%r10d
  add    %eax,%r10d
  sar    %r10d
  mov    %eax,%r11d
  shr    $0x1f,%r11d
  add    %r11d,%eax
  and    $0x1,%eax
  sub    %r11d,%eax
  cltq
  movzbl (%r9,%rax,1),%eax
  mov    %al,(%rsi)
  lea    0x1(%rsi),%rax
  cmp    %rbx,%rdx
  jne    L64
  cmp    %rbp,%rcx
  jne    L64
  test   %r10d,%r10d
  je     L10
  movslq %r10d,%r10
  lea    D13(%rip),%rdx
  movzbl (%rdx,%r10,1),%edx
  mov    %dl,(%rax)
  lea    0x2(%rsi),%rax
L10:
  movb   $0x0,(%rax)
  cmp    %r8,%rax
  jbe    L58
  mov    %r8,%rdx
L38:
  movzbl (%rdx),%ecx
  movzbl -0x1(%rax),%esi
  mov    %sil,(%rdx)
  mov    %cl,-0x1(%rax)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %rax,%rdx
  jb     L38
L58:
  mov    %r8,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D13
datatype: string

tool: D13: 01

assistant: ```c
char *addBinary(char *left, char *right) {
  const char BASE = 2;
  const size_t leftCount = strlen(left);
  const size_t rightCount = strlen(right);
  const size_t resultCount = 1 + (((leftCount) > (rightCount)) ? (leftCount) : (rightCount));
  char *buffer = (char *)calloc(1 + resultCount, sizeof(*buffer));
  if (buffer) {
    const char *pleft = &left[leftCount];
    const char *pright = &right[rightCount];
    char *presult = buffer;
    int carry = 0;
    while (1) {
      const int leftBit = pleft > left ? *--pleft - '0' : 0;
      const int rightBit = pright > right ? *--pright - '0' : 0;
      const int total = carry + leftBit + rightBit;
      carry = total / BASE;
      *presult = "01"[total % BASE];
      ++presult;
      if (pleft == left && pright == right) {
        break;
      }
    }
    if (carry) {
      *presult++ = "01"[carry];
    }
    *presult = '\0';
    for (char *begin = buffer, *end = presult; begin < end--; ++begin) {
      const char temp = *begin;
      *begin = *end;
      *end = temp;
    }
  }
  return buffer;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<addBinary>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%r13
  call   <strlen@plt>
  mov    $0x1,%esi
  cmp    %rax,%r13
  mov    %rax,%rdi
  mov    %rax,%r12
  cmovae %r13,%rdi
  add    $0x2,%edi
  call   <calloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L17
  lea    (%rbx,%r13,1),%rsi
  lea    0x0(%rbp,%r12,1),%rdi
  mov    %rax,%rdx
  xor    %r10d,%r10d
  lea    D8(%rip),%r9
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L66:
  mov    %rax,%rdx
L39:
  cmp    %rbx,%rsi
  jbe    L14
  movsbl -0x1(%rsi),%eax
  sub    $0x1,%rsi
  lea    -0x30(%r10,%rax,1),%r10d
L14:
  mov    %r10d,%eax
  cmp    %rbp,%rdi
  jbe    L46
  movsbl -0x1(%rdi),%ecx
  sub    $0x1,%rdi
  lea    -0x30(%r10,%rcx,1),%eax
L46:
  mov    %eax,%ecx
  mov    %eax,%r11d
  shr    $0x1f,%r11d
  shr    $0x1f,%ecx
  add    %eax,%ecx
  add    %r11d,%eax
  and    $0x1,%eax
  sar    %ecx
  sub    %r11d,%eax
  mov    %ecx,%r10d
  cltq
  movzbl (%r9,%rax,1),%eax
  mov    %al,(%rdx)
  lea    0x1(%rdx),%rax
  cmp    %rbx,%rsi
  jne    L66
  cmp    %rbp,%rdi
  jne    L66
  test   %ecx,%ecx
  je     L21
  movslq %ecx,%rcx
  lea    0x2(%rdx),%rsi
  movzbl (%r9,%rcx,1),%edx
  mov    %dl,(%rax)
  mov    %rax,%rdx
  mov    %rsi,%rax
L21:
  movb   $0x0,(%rax)
  cmp    %r8,%rax
  jbe    L17
  mov    %r8,%rax
  nopl   0x0(%rax,%rax,1)
L80:
  movzbl (%rax),%ecx
  movzbl (%rdx),%esi
  sub    $0x1,%rdx
  add    $0x1,%rax
  mov    %sil,-0x1(%rax)
  mov    %cl,0x1(%rdx)
  lea    0x1(%rdx),%rcx
  cmp    %rcx,%rax
  jb     L80
L17:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
data_type: string

tool: D8: 01

assistant: ```c
char *addBinary(char *left, char *right) {
  const char BASE = 2;
  const size_t leftCount = strlen(left);
  const size_t rightCount = strlen(right);
  const size_t resultCount = 1 + (((leftCount) > (rightCount)) ? (leftCount) : (rightCount));
  char *buffer = (char *)calloc(1 + resultCount, sizeof(*buffer));
  if (buffer) {
    const char *pleft = &left[leftCount];
    const char *pright = &right[rightCount];
    char *presult = buffer;
    int carry = 0;
    while (1) {
      const int leftBit = pleft > left ? *--pleft - '0' : 0;
      const int rightBit = pright > right ? *--pright - '0' : 0;
      const int total = carry + leftBit + rightBit;
      carry = total / BASE;
      *presult = "01"[total % BASE];
      ++presult;
      if (pleft == left && pright == right) {
        break;
      }
    }
    if (carry) {
      *presult++ = "01"[carry];
    }
    *presult = '\0';
    for (char *begin = buffer, *end = presult; begin < end--; ++begin) {
      const char temp = *begin;
      *begin = *end;
      *end = temp;
    }
  }
  return buffer;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    $0x0,%eax
  jmp    L14
L16:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
L14:
  leave
  ret
```

assistant: ```c
char *Strdup(char *s) {
  char *new;
  if ((new = (char *)malloc(strlen(s) + 1)) == NULL)
    return NULL;
  strcpy(new, s);
  return new;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    $0x0,%ebp
  test   %rax,%rax
  je     L9
  mov    %rax,%rdi
  mov    %rax,%rbp
  mov    %rbx,%rsi
  call   <strcpy@plt>
L9:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *Strdup(char *s) {
  char *new;
  if ((new = (char *)malloc(strlen(s) + 1)) == NULL)
    return NULL;
  strcpy(new, s);
  return new;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<Strdup>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L5
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
L5:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *Strdup(char *s) {
  char *new;
  if ((new = (char *)malloc(strlen(s) + 1)) == NULL)
    return NULL;
  strcpy(new, s);
  return new;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Strdup>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L1
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
L1:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *Strdup(char *s) {
  char *new;
  if ((new = (char *)malloc(strlen(s) + 1)) == NULL)
    return NULL;
  strcpy(new, s);
  return new;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<NextPrime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  jg     L18
  movl   $0x3,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L34
  addl   $0x1,-0x4(%rbp)
  jmp    L34
L19:
  addl   $0x2,-0x4(%rbp)
L34:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <divisor@plt>
  cmp    %eax,-0x4(%rbp)
  jg     L19
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int NextPrime(int n) {
  if (n < 3)
    n = 3;
  if (n % 2 == 0)
    n++;
  while (n > divisor(n))
    n = n + 2;
  return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<NextPrime>:
  endbr64
  push   %rbx
  mov    $0x3,%eax
  cmp    %eax,%edi
  cmovge %edi,%eax
  mov    %eax,%ebx
  and    $0x1,%eax
  cmp    $0x1,%eax
  adc    $0x0,%ebx
  jmp    L5
L15:
  add    $0x2,%ebx
L5:
  mov    %ebx,%edi
  call   <divisor@plt>
  cmp    %ebx,%eax
  jl     L15
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int NextPrime(int n) {
  if (n < 3)
    n = 3;
  if (n % 2 == 0)
    n++;
  while (n > divisor(n))
    n = n + 2;
  return n;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<NextPrime>:
  endbr64
  mov    $0x3,%eax
  push   %r12
  cmp    %eax,%edi
  cmovge %edi,%eax
  mov    %eax,%r12d
  and    $0x1,%eax
  cmp    $0x1,%eax
  adc    $0x0,%r12d
  jmp    L13
  nop
L11:
  add    $0x2,%r12d
L13:
  mov    %r12d,%edi
  call   <divisor@plt>
  cmp    %r12d,%eax
  jl     L11
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int NextPrime(int n) {
  if (n < 3)
    n = 3;
  if (n % 2 == 0)
    n++;
  while (n > divisor(n))
    n = n + 2;
  return n;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<NextPrime>:
  endbr64
  mov    $0x3,%eax
  push   %r12
  cmp    %eax,%edi
  cmovge %edi,%eax
  mov    %eax,%r12d
  and    $0x1,%eax
  cmp    $0x1,%eax
  adc    $0x0,%r12d
  jmp    L14
  nop
L18:
  add    $0x2,%r12d
L14:
  mov    %r12d,%edi
  call   <divisor@plt>
  cmp    %r12d,%eax
  jl     L18
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int NextPrime(int n) {
  if (n < 3)
    n = 3;
  if (n % 2 == 0)
    n++;
  while (n > divisor(n))
    n = n + 2;
  return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<user_write_client_msg>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  mov    %esi,%r14d
  mov    %rdx,%r13
  mov    %rcx,%rbp
  test   %rcx,%rcx
  je     L33
  mov    %rdx,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    %eax,%r12d
  lea    0x14(%r12),%r15d
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %r14d,%r14d
  jne    L86
  lea    0x13(%r12),%esi
  mov    %rbp,%rcx
  mov    %r13,%rdx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L42
L33:
  mov    %rdx,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  lea    0xa(%rax),%r15d
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %r14d,%r14d
  jne    L56
  lea    0x9(%r12),%esi
  mov    %r13,%rdx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L42
L41:
  lea    D36(%rip),%rcx
  jmp    L35
L17:
  lea    D36(%rip),%rcx
  jmp    L100
L56:
  lea    D19(%rip),%rcx
  cmp    $0x1,%r14d
  je     L41
L3:
  lea    -0x1(%r15),%esi
  mov    %r13,%rdx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L42
L86:
  lea    D19(%rip),%rcx
  cmp    $0x1,%r14d
  je     L17
L35:
  test   %rbp,%rbp
  je     L3
L100:
  lea    -0x1(%r15),%esi
  mov    %rbp,%r8
  mov    %r13,%rdx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
L42:
  lea    0x1(%rax),%edx
  mov    %rbx,%rsi
  mov    0xc(%rsp),%edi
  call   <write@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdi
  call   <free@plt>
  test   %rbp,%rbp
  setg   %al
  movzbl %al,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D36
type: string
name: extract_symbolic_data
arguments: tag: D19
type: string

tool: D36: OK

tool: D19: FAIL

assistant: ```c
int user_write_client_msg(int fd, int status, const char *msg, char *add) {
  int len, ret = 0;
  char *st, *buff;
  if (add == NULL) {
    len = strlen(msg) + 10;
  } else {
    len = strlen(msg) + strlen(add) + 20;
  }
  buff = malloc(sizeof(char) * len);
  if (status) {
    st = "FAIL";
    if (status == 1) {
      st = "OK";
    }
    if (add == NULL) {
      len = snprintf(buff, len - 1, msg, st);
    } else {
      len = snprintf(buff, len - 1, msg, st, add);
    }
  } else {
    if (add == NULL) {
      len = snprintf(buff, len - 1, msg);
    } else {
      len = snprintf(buff, len - 1, msg, add);
    }
  }
  if (write(fd, buff, len + 1) > 0) {
    ret = 1;
  }
  free(buff);
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<user_write_client_msg>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  test   %rcx,%rcx
  je     L58
  mov    %rcx,%r13
  mov    %rdx,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  add    %eax,%ebx
  lea    0x14(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %r15d,%r15d
  jne    L30
  mov    %rax,%rdi
  lea    0x13(%rbx),%esi
  mov    %r13,%rcx
  mov    %r14,%rdx
  xor    %eax,%eax
  call   <snprintf@plt>
L46:
  mov    %r12,%rsi
  mov    %ebp,%edi
  lea    0x1(%rax),%edx
  call   <write@plt>
  mov    %r12,%rdi
  mov    %rax,%rbx
  call   <free@plt>
  xor    %eax,%eax
  test   %rbx,%rbx
  setg   %al
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L30:
  cmp    $0x1,%r15d
  lea    0x13(%rbx),%esi
  mov    %r13,%r8
  mov    %r14,%rdx
  lea    D3(%rip),%rax
  lea    D21(%rip),%rcx
  mov    %r12,%rdi
  cmovne %rax,%rcx
  xor    %eax,%eax
  call   <snprintf@plt>
  jmp    L46
  nopl   0x0(%rax,%rax,1)
L58:
  mov    %rdx,%rdi
  call   <strlen@plt>
  lea    0xa(%rax),%edi
  mov    %eax,%ebx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %r15d,%r15d
  je     L50
  cmp    $0x1,%r15d
  lea    0x9(%rbx),%esi
  mov    %r14,%rdx
  mov    %r12,%rdi
  lea    D3(%rip),%rax
  lea    D21(%rip),%rcx
  cmovne %rax,%rcx
  xor    %eax,%eax
  call   <snprintf@plt>
  jmp    L46
  nopl   (%rax)
L50:
  mov    %rax,%rdi
  lea    0x9(%rbx),%esi
  mov    %r14,%rdx
  xor    %eax,%eax
  call   <snprintf@plt>
  jmp    L46
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
data_type: string
name: disassemble_data
arguments: data_tag: D21
data_type: string

tool: D3: FAIL

tool: D21: OK

assistant: ```c
int user_write_client_msg(int fd, int status, const char *msg, char *add) {
  int len, ret = 0;
  char *st, *buff;
  if (add == NULL) {
    len = strlen(msg) + 10;
  } else {
    len = strlen(msg) + strlen(add) + 20;
  }
  buff = malloc(sizeof(char) * len);
  if (status) {
    st = "FAIL";
    if (status == 1) {
      st = "OK";
    }
    if (add == NULL) {
      len = snprintf(buff, len - 1, msg, st);
    } else {
      len = snprintf(buff, len - 1, msg, st, add);
    }
  } else {
    if (add == NULL) {
      len = snprintf(buff, len - 1, msg);
    } else {
      len = snprintf(buff, len - 1, msg, add);
    }
  }
  if (write(fd, buff, len + 1) > 0) {
    ret = 1;
  }
  free(buff);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<user_write_client_msg>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  test   %rcx,%rcx
  je     L6
  mov    %rcx,%r13
  mov    %rdx,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  add    %eax,%ebx
  lea    0x14(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %r15d,%r15d
  jne    L35
  mov    %rax,%rdi
  lea    0x13(%rbx),%esi
  mov    %r13,%rcx
  mov    %r14,%rdx
  xor    %eax,%eax
  call   <snprintf@plt>
L30:
  mov    %r12,%rsi
  mov    %ebp,%edi
  lea    0x1(%rax),%edx
  call   <write@plt>
  mov    %r12,%rdi
  mov    %rax,%rbx
  call   <free@plt>
  xor    %eax,%eax
  test   %rbx,%rbx
  setg   %al
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L35:
  cmp    $0x1,%r15d
  lea    0x13(%rbx),%esi
  mov    %r13,%r8
  mov    %r14,%rdx
  lea    D27(%rip),%rax
  lea    D34(%rip),%rcx
  mov    %r12,%rdi
  cmovne %rax,%rcx
  xor    %eax,%eax
  call   <snprintf@plt>
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %rdx,%rdi
  call   <strlen@plt>
  lea    0xa(%rax),%edi
  mov    %eax,%ebx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %r15d,%r15d
  je     L57
  cmp    $0x1,%r15d
  lea    0x9(%rbx),%esi
  mov    %r14,%rdx
  mov    %r12,%rdi
  lea    D27(%rip),%rax
  lea    D34(%rip),%rcx
  cmovne %rax,%rcx
  xor    %eax,%eax
  call   <snprintf@plt>
  jmp    L30
  nopl   (%rax)
L57:
  mov    %rax,%rdi
  lea    0x9(%rbx),%esi
  mov    %r14,%rdx
  xor    %eax,%eax
  call   <snprintf@plt>
  jmp    L30
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D27
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D34
type_guessed: string

tool: D27: FAIL

tool: D34: OK

assistant: ```c
int user_write_client_msg(int fd, int status, const char *msg, char *add) {
  int len, ret = 0;
  char *st, *buff;
  if (add == NULL) {
    len = strlen(msg) + 10;
  } else {
    len = strlen(msg) + strlen(add) + 20;
  }
  buff = malloc(sizeof(char) * len);
  if (status) {
    st = "FAIL";
    if (status == 1) {
      st = "OK";
    }
    if (add == NULL) {
      len = snprintf(buff, len - 1, msg, st);
    } else {
      len = snprintf(buff, len - 1, msg, st, add);
    }
  } else {
    if (add == NULL) {
      len = snprintf(buff, len - 1, msg);
    } else {
      len = snprintf(buff, len - 1, msg, add);
    }
  }
  if (write(fd, buff, len + 1) > 0) {
    ret = 1;
  }
  free(buff);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<find_previous_sibling>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L6:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jne    L5
  mov    -0x4(%rbp),%eax
  jmp    L16
L5:
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L6
  mov    $0xffffffff,%eax
L16:
  pop    %rbp
  ret
```

assistant: ```c
int find_previous_sibling(int con[500][5], int max, int b) {
  int i = 0;
  for (i = 0; i < max; i++) {
    if (con[i][0] == b)
      return i;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<find_previous_sibling>:
  endbr64
  test   %esi,%esi
  jle    L3
  mov    $0x0,%eax
L5:
  cmp    %edx,(%rdi)
  je     L10
  add    $0x1,%eax
  add    $0x14,%rdi
  cmp    %eax,%esi
  jne    L5
  mov    $0xffffffff,%eax
  ret
L3:
  mov    $0xffffffff,%eax
L10:
  ret
```

assistant: ```c
int find_previous_sibling(int con[500][5], int max, int b) {
  int i = 0;
  for (i = 0; i < max; i++) {
    if (con[i][0] == b)
      return i;
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<find_previous_sibling>:
  endbr64
  test   %esi,%esi
  jle    L0
  xor    %eax,%eax
  jmp    L19
  nopl   0x0(%rax)
L5:
  add    $0x1,%eax
  add    $0x14,%rdi
  cmp    %eax,%esi
  je     L0
L19:
  cmp    %edx,(%rdi)
  jne    L5
  ret
L0:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int find_previous_sibling(int con[500][5], int max, int b) {
  int i = 0;
  for (i = 0; i < max; i++) {
    if (con[i][0] == b)
      return i;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<find_previous_sibling>:
  endbr64
  test   %esi,%esi
  jle    L18
  xor    %eax,%eax
  jmp    L14
  nopl   0x0(%rax)
L7:
  add    $0x1,%eax
  add    $0x14,%rdi
  cmp    %eax,%esi
  je     L18
L14:
  cmp    %edx,(%rdi)
  jne    L7
  ret
L18:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int find_previous_sibling(int con[500][5], int max, int b) {
  int i = 0;
  for (i = 0; i < max; i++) {
    if (con[i][0] == b)
      return i;
  }
  return -1;
}
```