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
<fescape>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x74,-0x4(%rbp)
  je     L83
  cmpl   $0x74,-0x4(%rbp)
  jg     L11
  cmpl   $0x72,-0x4(%rbp)
  je     L96
  cmpl   $0x72,-0x4(%rbp)
  jg     L11
  cmpl   $0x62,-0x4(%rbp)
  je     L6
  cmpl   $0x6e,-0x4(%rbp)
  jne    L11
  mov    $0xa,%eax
  jmp    L182
L6:
  mov    $0x8,%eax
  jmp    L182
L83:
  mov    $0x9,%eax
  jmp    L182
L96:
  mov    $0xd,%eax
  jmp    L182
L11:
  cmpl   $0x2f,-0x4(%rbp)
  jle    L112
  cmpl   $0x37,-0x4(%rbp)
  jle    L73
L112:
  mov    -0x4(%rbp),%eax
  jmp    L182
L73:
  mov    -0x4(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x2,-0x8(%rbp)
  jmp    L33
L26:
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x4(%rbp),%eax
  sub    $0x30,%eax
  or     %edx,%eax
  mov    %eax,-0xc(%rbp)
  subl   $0x1,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jg     L33
  mov    -0xc(%rbp),%eax
  jmp    L182
L33:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x2f,-0x4(%rbp)
  jle    L129
  cmpl   $0x37,-0x4(%rbp)
  jle    L26
L129:
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L17
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
L17:
  mov    -0xc(%rbp),%eax
L182:
  leave
  ret
```

assistant: ```c
int fescape(FILE *f) {
  int n, k, c = fgetc(f);
  switch (c) {
  case 'n':
    return '\n';
  case 'b':
    return '\b';
  case 't':
    return '\t';
  case 'r':
    return '\r';
  }
  if (c < '0' || c > '7')
    return c;
  n = c - '0';
  k = 2;
  while ((c = fgetc(f)) >= '0' && c <= '7') {
    n = n * 8 | c - '0';
    if (--k <= 0)
      return n;
  }
  if (c != EOF)
    ungetc(c, f);
  return n;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fescape>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <fgetc@plt>
  cmp    $0x72,%eax
  je     L105
  mov    %eax,%ebx
  cmp    $0x72,%eax
  jg     L82
  cmp    $0x62,%eax
  je     L22
  cmp    $0x6e,%eax
  jne    L41
  mov    $0xa,%ebx
  jmp    L11
L82:
  cmp    $0x74,%eax
  mov    $0x9,%eax
  cmove  %eax,%ebx
L11:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L41:
  lea    -0x30(%rax),%eax
  cmp    $0x7,%eax
  ja     L11
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <fgetc@plt>
  lea    -0x30(%rax),%edx
  cmp    $0x7,%edx
  ja     L4
  shl    $0x3,%ebx
  or     %edx,%ebx
  mov    %rbp,%rdi
  call   <fgetc@plt>
  lea    -0x30(%rax),%edx
  cmp    $0x7,%edx
  ja     L4
  shl    $0x3,%ebx
  or     %edx,%ebx
  jmp    L11
L4:
  mov    D9(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L11
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
  jmp    L11
L105:
  mov    $0xd,%ebx
  jmp    L11
L22:
  mov    $0x8,%ebx
  jmp    L11
```

assistant: ```c
int fescape(FILE *f) {
  int n, k, c = fgetc(f);
  switch (c) {
  case 'n':
    return '\n';
  case 'b':
    return '\b';
  case 't':
    return '\t';
  case 'r':
    return '\r';
  }
  if (c < '0' || c > '7')
    return c;
  n = c - '0';
  k = 2;
  while ((c = fgetc(f)) >= '0' && c <= '7') {
    n = n * 8 | c - '0';
    if (--k <= 0)
      return n;
  }
  if (c != EOF)
    ungetc(c, f);
  return n;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fescape>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <fgetc@plt>
  cmp    $0x72,%eax
  je     L66
  mov    %eax,%r12d
  jg     L48
  cmp    $0x62,%eax
  je     L103
  cmp    $0x6e,%eax
  jne    L61
  mov    $0xa,%r12d
L52:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L48:
  cmp    $0x74,%eax
  mov    $0x9,%eax
  cmove  %eax,%r12d
  add    $0x8,%rsp
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L103:
  mov    $0x8,%r12d
  jmp    L52
  nopl   0x0(%rax,%rax,1)
L66:
  mov    $0xd,%r12d
  jmp    L52
L61:
  lea    -0x30(%rax),%eax
  cmp    $0x7,%eax
  ja     L52
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <fgetc@plt>
  mov    %eax,%edi
  lea    -0x30(%rax),%eax
  cmp    $0x7,%eax
  ja     L30
  mov    %rbp,%rdi
  shl    $0x3,%r12d
  or     %eax,%r12d
  call   <fgetc@plt>
  mov    %eax,%edi
  lea    -0x30(%rax),%eax
  cmp    $0x7,%eax
  ja     L30
  shl    $0x3,%r12d
  or     %eax,%r12d
  jmp    L52
  nopw   0x0(%rax,%rax,1)
L30:
  mov    D7(%rip),%rax
  cmp    %edi,(%rax)
  je     L52
  mov    %rbp,%rsi
  call   <ungetc@plt>
  jmp    L52
```

assistant: ```c
int fescape(FILE *f) {
  int n, k, c = fgetc(f);
  switch (c) {
  case 'n':
    return '\n';
  case 'b':
    return '\b';
  case 't':
    return '\t';
  case 'r':
    return '\r';
  }
  if (c < '0' || c > '7')
    return c;
  n = c - '0';
  k = 2;
  while ((c = fgetc(f)) >= '0' && c <= '7') {
    n = n * 8 | c - '0';
    if (--k <= 0)
      return n;
  }
  if (c != EOF)
    ungetc(c, f);
  return n;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fescape>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <fgetc@plt>
  cmp    $0x72,%eax
  je     L73
  mov    %eax,%r12d
  jg     L30
  cmp    $0x62,%eax
  je     L93
  cmp    $0x6e,%eax
  jne    L23
  mov    $0xa,%r12d
L48:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L30:
  cmp    $0x74,%eax
  mov    $0x9,%eax
  cmove  %eax,%r12d
  add    $0x8,%rsp
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L93:
  mov    $0x8,%r12d
  jmp    L48
  nopl   0x0(%rax,%rax,1)
L73:
  mov    $0xd,%r12d
  jmp    L48
L23:
  lea    -0x30(%rax),%eax
  cmp    $0x7,%eax
  ja     L48
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <fgetc@plt>
  mov    %eax,%edi
  lea    -0x30(%rax),%eax
  cmp    $0x7,%eax
  ja     L14
  mov    %rbp,%rdi
  shl    $0x3,%r12d
  or     %eax,%r12d
  call   <fgetc@plt>
  mov    %eax,%edi
  lea    -0x30(%rax),%eax
  cmp    $0x7,%eax
  ja     L14
  shl    $0x3,%r12d
  or     %eax,%r12d
  jmp    L48
  nopw   0x0(%rax,%rax,1)
L14:
  mov    D1(%rip),%rax
  cmp    %edi,(%rax)
  je     L48
  mov    %rbp,%rsi
  call   <ungetc@plt>
  jmp    L48
```

assistant: ```c
int fescape(FILE *f) {
  int n, k, c = fgetc(f);
  switch (c) {
  case 'n':
    return '\n';
  case 'b':
    return '\b';
  case 't':
    return '\t';
  case 'r':
    return '\r';
  }
  if (c < '0' || c > '7')
    return c;
  n = c - '0';
  k = 2;
  while ((c = fgetc(f)) >= '0' && c <= '7') {
    n = n * 8 | c - '0';
    if (--k <= 0)
      return n;
  }
  if (c != EOF)
    ungetc(c, f);
  return n;
}
```
user: What is the c source code of the assembly code below?

```asm
<stringisprintable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L15:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isprint@plt>
  test   %eax,%eax
  jne    L17
  mov    $0x0,%eax
  jmp    L3
L17:
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L15
  mov    $0x1,%eax
L3:
  leave
  ret
```

assistant: ```c
int stringisprintable(const char *str, int strlength) {
  int i;
  for (i = 0; i < strlength; i++)
    if (!isprint((int)(unsigned char)str[i]))
      return 0;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<stringisprintable>:
  endbr64
  test   %esi,%esi
  jle    L29
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rbp
L18:
  movzbl (%rbx),%edi
  call   <isprint@plt>
  test   %eax,%eax
  je     L24
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L18
  mov    $0x1,%eax
L24:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L29:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int stringisprintable(const char *str, int strlength) {
  int i;
  for (i = 0; i < strlength; i++)
    if (!isprint((int)(unsigned char)str[i]))
      return 0;
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stringisprintable>:
  endbr64
  test   %esi,%esi
  jle    L24
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x1(%rdi,%rax,1),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L9:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L0
L6:
  movzbl (%rbx),%edi
  call   <isprint@plt>
  test   %eax,%eax
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L0:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
L24:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int stringisprintable(const char *str, int strlength) {
  int i;
  for (i = 0; i < strlength; i++)
    if (!isprint((int)(unsigned char)str[i]))
      return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<stringisprintable>:
  endbr64
  test   %esi,%esi
  jle    L3
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x1(%rdi,%rax,1),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L12
  nopl   0x0(%rax,%rax,1)
L9:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L14
L12:
  movzbl (%rbx),%edi
  call   <isprint@plt>
  test   %eax,%eax
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L14:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
L3:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int stringisprintable(const char *str, int strlength) {
  int i;
  for (i = 0; i < strlength; i++)
    if (!isprint((int)(unsigned char)str[i]))
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sign>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L5
  mov    $0x2d,%eax
  jmp    L18
L5:
  mov    $0x2b,%eax
L18:
  pop    %rbp
  ret
```

assistant: ```c
char sign(int a) {
  if (a < 0) {
    return '-';
  }
  return '+';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sign>:
  endbr64
  sar    $0x1f,%edi
  and    $0x2,%edi
  lea    0x2b(%rdi),%eax
  ret
```

assistant: ```c
char sign(int a) {
  if (a < 0) {
    return '-';
  }
  return '+';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sign>:
  endbr64
  sar    $0x1f,%edi
  and    $0x2,%edi
  lea    0x2b(%rdi),%eax
  ret
```

assistant: ```c
char sign(int a) {
  if (a < 0) {
    return '-';
  }
  return '+';
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<wkd_cksum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L18:
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %eax,-0x4(%rbp)
L1:
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  test   %eax,%eax
  jne    L18
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int wkd_cksum(const unsigned char *b, int len) {
  unsigned int ret = 0;
  while (len--)
    ret += b;
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<wkd_cksum>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %esi,%esi
  je     L18
  mov    %edi,%esi
  mov    $0x0,%edx
L3:
  lea    (%rdx,%rdi,1),%ecx
  sub    $0x1,%eax
  add    %esi,%edx
  cmp    $0xffffffff,%eax
  jne    L3
L23:
  mov    %ecx,%eax
  ret
L18:
  mov    $0x0,%ecx
  jmp    L23
```

assistant: ```c
unsigned int wkd_cksum(const unsigned char *b, int len) {
  unsigned int ret = 0;
  while (len--)
    ret += b;
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<wkd_cksum>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %esi,%esi
  je     L19
  mov    %edi,%ecx
  xor    %edx,%edx
  nop
L7:
  sub    $0x1,%eax
  lea    (%rdx,%rdi,1),%r8d
  add    %ecx,%edx
  cmp    $0xffffffff,%eax
  jne    L7
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int wkd_cksum(const unsigned char *b, int len) {
  unsigned int ret = 0;
  while (len--)
    ret += b;
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wkd_cksum>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %esi,%esi
  je     L7
  mov    %edi,%ecx
  xor    %edx,%edx
  nop
L6:
  sub    $0x1,%eax
  lea    (%rdx,%rdi,1),%r8d
  add    %ecx,%edx
  cmp    $0xffffffff,%eax
  jne    L6
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int wkd_cksum(const unsigned char *b, int len) {
  unsigned int ret = 0;
  while (len--)
    ret += b;
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L1
  mov    $0x0,%eax
  jmp    L9
L1:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
L9:
  leave
  ret
```

assistant: ```c
void *memdup(void *m, int len) {
  void *ret;
  if (len == 0)
    return NULL;
  ret = malloc(len);
  memcpy(ret, m, len);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<memdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%ebp
  test   %esi,%esi
  je     L5
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %esi,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L5:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void *memdup(void *m, int len) {
  void *ret;
  if (len == 0)
    return NULL;
  ret = malloc(len);
  memcpy(ret, m, len);
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<memdup>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  test   %esi,%esi
  je     L0
  mov    %rdi,%rbp
  mov    %esi,%edi
  mov    %esi,%r12d
  call   <malloc@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memcpy@plt>
L0:
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *memdup(void *m, int len) {
  void *ret;
  if (len == 0)
    return NULL;
  ret = malloc(len);
  memcpy(ret, m, len);
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memdup>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  test   %esi,%esi
  je     L1
  mov    %rdi,%rbp
  mov    %esi,%edi
  mov    %esi,%r12d
  call   <malloc@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memcpy@plt>
L1:
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *memdup(void *m, int len) {
  void *ret;
  if (len == 0)
    return NULL;
  ret = malloc(len);
  memcpy(ret, m, len);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x1c(%rbp)
  lea    -0x18(%rbp),%rdx
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L3
L22:
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm0
  mov    -0x10(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <newtonRaphson@plt>
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L3:
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  test   %eax,%eax
  jne    L22
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D8
type_infered: string
name: parse_data
arguments: label: D13
type_infered: string
name: parse_data
arguments: label: D22
type_infered: string

tool: D8: %i %lf

tool: D13: 
%lf

tool: D22: %lf


assistant: ```c
int main(int argc, char const *argv[]) {
  double r = 0, num = 0;
  int n = 0;
  scanf("%i %lf", &n, &r);
  while (n--) {
    scanf("\n%lf", &num);
    printf("%lf\n", newtonRaphson(num, r));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movl   $0x0,0x4(%rsp)
  lea    0x8(%rsp),%rdx
  lea    0x4(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L11
  lea    0x10(%rsp),%r12
  lea    D12(%rip),%rbp
  lea    D27(%rip),%rbx
L6:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm0
  call   <newtonRaphson@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L6
L11:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D29
type: string
name: parse_data
arguments: data_label: D12
type: string
name: parse_data
arguments: data_label: D27
type: string

tool: D29: %i %lf

tool: D12: 
%lf

tool: D27: %lf


assistant: ```c
int main(int argc, char const *argv[]) {
  double r = 0, num = 0;
  int n = 0;
  scanf("%i %lf", &n, &r);
  while (n--) {
    scanf("\n%lf", &num);
    printf("%lf\n", newtonRaphson(num, r));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D4(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdx
  lea    0x4(%rsp),%rsi
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movl   $0x0,0x4(%rsp)
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L21
  lea    0x10(%rsp),%r12
  lea    D28(%rip),%rbp
  lea    D6(%rip),%rbx
  xchg   %ax,%ax
L8:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm0
  call   <newtonRaphson@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L8
L21:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
data_type: string
name: extract_symbolic_data
arguments: label: D28
data_type: string
name: extract_symbolic_data
arguments: label: D6
data_type: string

tool: D4: %i %lf

tool: D28: 
%lf

tool: D6: %lf


assistant: ```c
int main(int argc, char const *argv[]) {
  double r = 0, num = 0;
  int n = 0;
  scanf("%i %lf", &n, &r);
  while (n--) {
    scanf("\n%lf", &num);
    printf("%lf\n", newtonRaphson(num, r));
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  lea    D11(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdx
  lea    0x4(%rsp),%rsi
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movl   $0x0,0x4(%rsp)
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L20
  lea    0x10(%rsp),%r12
  lea    D6(%rip),%rbp
  lea    D16(%rip),%rbx
  xchg   %ax,%ax
L13:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm0
  call   <newtonRaphson@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L13
L20:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
data_type: string
name: retrieve_data
arguments: data_tag: D6
data_type: string
name: retrieve_data
arguments: data_tag: D16
data_type: string

tool: D11: %i %lf

tool: D6: 
%lf

tool: D16: %lf


assistant: ```c
int main(int argc, char const *argv[]) {
  double r = 0, num = 0;
  int n = 0;
  scanf("%i %lf", &n, &r);
  while (n--) {
    scanf("\n%lf", &num);
    printf("%lf\n", newtonRaphson(num, r));
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<unit_name_to_prefix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x40,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L5
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strndup@plt>
  jmp    L0
L5:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <unit_name_to_prefix_and_instance@plt>
L0:
  leave
  ret
```

assistant: ```c
char *unit_name_to_prefix(const char *n) {
  const char *p;
  if ((p = strchr(n, '@')))
    return strndup(n, p - n);
  return unit_name_to_prefix_and_instance(n);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<unit_name_to_prefix>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x40,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L14
  mov    %eax,%esi
  sub    %ebx,%esi
  mov    %rbx,%rdi
  call   <strndup@plt>
L1:
  pop    %rbx
  ret
L14:
  mov    %rbx,%rdi
  call   <unit_name_to_prefix_and_instance@plt>
  jmp    L1
```

assistant: ```c
char *unit_name_to_prefix(const char *n) {
  const char *p;
  if ((p = strchr(n, '@')))
    return strndup(n, p - n);
  return unit_name_to_prefix_and_instance(n);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<unit_name_to_prefix>:
  endbr64
  push   %rbp
  mov    $0x40,%esi
  mov    %rdi,%rbp
  call   <strchr@plt>
  test   %rax,%rax
  je     L9
  mov    %eax,%esi
  mov    %rbp,%rdi
  sub    %ebp,%esi
  pop    %rbp
  jmp    1080 <strndup@plt>
  nopl   0x0(%rax)
L9:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1090 <unit_name_to_prefix_and_instance@plt>
```

assistant: ```c
char *unit_name_to_prefix(const char *n) {
  const char *p;
  if ((p = strchr(n, '@')))
    return strndup(n, p - n);
  return unit_name_to_prefix_and_instance(n);
}
```
user: Translate the below assembly code into C programming language:

```asm
<unit_name_to_prefix>:
  endbr64
  push   %rbp
  mov    $0x40,%esi
  mov    %rdi,%rbp
  call   <strchr@plt>
  test   %rax,%rax
  je     L6
  mov    %eax,%esi
  mov    %rbp,%rdi
  sub    %ebp,%esi
  pop    %rbp
  jmp    1080 <strndup@plt>
  nopl   0x0(%rax)
L6:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1090 <unit_name_to_prefix_and_instance@plt>
```

assistant: ```c
char *unit_name_to_prefix(const char *n) {
  const char *p;
  if ((p = strchr(n, '@')))
    return strndup(n, p - n);
  return unit_name_to_prefix_and_instance(n);
}
```