ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D34(%rip),%rdi
  call   <puts@plt>
  lea    D9(%rip),%rdi
  call   <puts@plt>
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%ebx
  lea    D8(%rip),%rbp
L7:
  mov    %ebx,%edx
  imul   %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xb,%ebx
  jne    L7
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D34
datatype: string
name: extract_symbolic_data
arguments: label: D9
datatype: string
name: extract_symbolic_data
arguments: label: D14
datatype: string
name: extract_symbolic_data
arguments: label: D8
datatype: string

tool: D34: Table of Powers of Two


tool: D9:  n  n squared

tool: D14: --- ---------

tool: D8: %3i %9i


assistant: ```c
int main(void) {
  int n;
  printf("Table of Powers of Two\n\n");
  printf(" n  n squared\n");
  printf("--- ---------\n");
  for (n = 1; n <= 10; n++)
    printf("%3i %9i\n", n, n * n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D13(%rip),%rdi
  lea    D2(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <puts@plt>
  lea    D9(%rip),%rdi
  call   <puts@plt>
  lea    D23(%rip),%rdi
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  imul   %ebx,%edx
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xb,%ebx
  jne    L7
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: label: D13
type: string
name: unpack_data
arguments: label: D2
type: string
name: unpack_data
arguments: label: D9
type: string
name: unpack_data
arguments: label: D23
type: string

tool: D13: Table of Powers of Two


tool: D2: %3i %9i


tool: D9:  n  n squared

tool: D23: --- ---------

assistant: ```c
int main(void) {
  int n;
  printf("Table of Powers of Two\n\n");
  printf(" n  n squared\n");
  printf("--- ---------\n");
  for (n = 1; n <= 10; n++)
    printf("%3i %9i\n", n, n * n);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D25(%rip),%rdi
  lea    D2(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <puts@plt>
  lea    D18(%rip),%rdi
  call   <puts@plt>
  lea    D17(%rip),%rdi
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  imul   %ebx,%edx
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xb,%ebx
  jne    L3
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D25
type: string
name: disassemble_data
arguments: label: D2
type: string
name: disassemble_data
arguments: label: D18
type: string
name: disassemble_data
arguments: label: D17
type: string

tool: D25: Table of Powers of Two


tool: D2: %3i %9i


tool: D18:  n  n squared

tool: D17: --- ---------

assistant: ```c
int main(void) {
  int n;
  printf("Table of Powers of Two\n\n");
  printf(" n  n squared\n");
  printf("--- ---------\n");
  for (n = 1; n <= 10; n++)
    printf("%3i %9i\n", n, n * n);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sqmvprod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L17
L34:
  movl   $0x0,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x20(%rbp),%rax
  movsd  (%rax),%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x1,-0x4(%rbp)
  jmp    L21
L5:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L21:
  mov    -0x4(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L5
  addl   $0x1,-0x8(%rbp)
L17:
  mov    -0x8(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L34
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void sqmvprod(const double *x, const double *y, double *z, int m) {
  int i, k;
  for (i = 0; i < m; i++) {
    k = 0;
    z[i] = x[i] * y[0];
    for (k = 1; k < m; k++)
      z[i] += x[k * m + i] * y[k];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sqmvprod>:
  endbr64
  test   %ecx,%ecx
  jle    L37
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r11
  mov    %rsi,%rbx
  mov    %ecx,%r12d
  mov    %rdx,%r8
  movslq %ecx,%rsi
  shl    $0x3,%rsi
  lea    (%rdi,%rsi,1),%r10
  mov    %ecx,%ebp
  lea    -0x2(%rcx),%eax
  lea    0x10(%rbx,%rax,8),%rdi
  mov    $0x0,%r9d
L39:
  mov    %r8,%rcx
  movsd  (%r11,%r9,8),%xmm0
  mulsd  (%rbx),%xmm0
  movsd  %xmm0,(%r8)
  cmp    $0x1,%r12d
  jle    L22
  lea    0x8(%rbx),%rax
  mov    %r10,%rdx
L19:
  movsd  (%rdx),%xmm0
  mulsd  (%rax),%xmm0
  addsd  (%rcx),%xmm0
  movsd  %xmm0,(%rcx)
  add    %rsi,%rdx
  add    $0x8,%rax
  cmp    %rdi,%rax
  jne    L19
L22:
  add    $0x1,%r9
  add    $0x8,%r8
  add    $0x8,%r10
  cmp    %rbp,%r9
  jne    L39
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L37:
  ret
```

assistant: ```c
void sqmvprod(const double *x, const double *y, double *z, int m) {
  int i, k;
  for (i = 0; i < m; i++) {
    k = 0;
    z[i] = x[i] * y[0];
    for (k = 1; k < m; k++)
      z[i] += x[k * m + i] * y[k];
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sqmvprod>:
  endbr64
  test   %ecx,%ecx
  jle    L0
  push   %rbp
  movslq %ecx,%rbp
  lea    -0x2(%rcx),%eax
  mov    %rsi,%r11
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rbp,8),%rdi
  xor    %r9d,%r9d
  lea    (%rbx,%rdi,1),%r10
  lea    0x10(%rsi,%rax,8),%r8
  nopw   0x0(%rax,%rax,1)
L5:
  movsd  (%rbx,%r9,8),%xmm1
  mulsd  (%r11),%xmm1
  lea    0x8(%r11),%rax
  mov    %r10,%rsi
  movsd  %xmm1,(%rdx)
  cmp    $0x1,%ecx
  je     L14
  nopl   0x0(%rax,%rax,1)
L19:
  movsd  (%rsi),%xmm0
  mulsd  (%rax),%xmm0
  add    $0x8,%rax
  add    %rdi,%rsi
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  cmp    %rax,%r8
  jne    L19
L14:
  add    $0x1,%r9
  add    $0x8,%rdx
  add    $0x8,%r10
  cmp    %rbp,%r9
  jne    L5
  pop    %rbx
  pop    %rbp
  ret
L0:
  ret
```

assistant: ```c
void sqmvprod(const double *x, const double *y, double *z, int m) {
  int i, k;
  for (i = 0; i < m; i++) {
    k = 0;
    z[i] = x[i] * y[0];
    for (k = 1; k < m; k++)
      z[i] += x[k * m + i] * y[k];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<sqmvprod>:
  endbr64
  test   %ecx,%ecx
  jle    L17
  push   %rbp
  movslq %ecx,%rbp
  lea    -0x2(%rcx),%eax
  mov    %rsi,%r11
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rbp,8),%rdi
  xor    %r9d,%r9d
  lea    (%rbx,%rdi,1),%r10
  lea    0x10(%rsi,%rax,8),%r8
  nopw   0x0(%rax,%rax,1)
L25:
  movsd  (%rbx,%r9,8),%xmm1
  mulsd  (%r11),%xmm1
  lea    0x8(%r11),%rax
  mov    %r10,%rsi
  movsd  %xmm1,(%rdx)
  cmp    $0x1,%ecx
  je     L8
  nopl   0x0(%rax,%rax,1)
L38:
  movsd  (%rsi),%xmm0
  mulsd  (%rax),%xmm0
  add    $0x8,%rax
  add    %rdi,%rsi
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  cmp    %rax,%r8
  jne    L38
L8:
  add    $0x1,%r9
  add    $0x8,%rdx
  add    $0x8,%r10
  cmp    %rbp,%r9
  jne    L25
  pop    %rbx
  pop    %rbp
  ret
L17:
  ret
```

assistant: ```c
void sqmvprod(const double *x, const double *y, double *z, int m) {
  int i, k;
  for (i = 0; i < m; i++) {
    k = 0;
    z[i] = x[i] * y[0];
    for (k = 1; k < m; k++)
      z[i] += x[k * m + i] * y[k];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<btp_parse_char_span>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strspn@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L0
  mov    $0x0,%eax
  jmp    L5
L0:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <btp_strndup@plt>
  mov    -0x28(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
L5:
  leave
  ret
```

assistant: ```c
int btp_parse_char_span(const char **input, const char *accept, char **result) {
  size_t count = strspn(*input, accept);
  if (count == 0)
    return 0;
  *result = btp_strndup(*input, count);
  *input += count;
  return count;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<btp_parse_char_span>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rdx,%r12
  mov    (%rdi),%r13
  mov    %r13,%rdi
  call   <strspn@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  test   %rbx,%rbx
  jne    L0
L7:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <btp_strndup@plt>
  mov    %rax,(%r12)
  add    %rbx,0x0(%rbp)
  mov    %ebx,%eax
  jmp    L7
```

assistant: ```c
int btp_parse_char_span(const char **input, const char *accept, char **result) {
  size_t count = strspn(*input, accept);
  if (count == 0)
    return 0;
  *result = btp_strndup(*input, count);
  *input += count;
  return count;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<btp_parse_char_span>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r13
  mov    %r13,%rdi
  call   <strspn@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  test   %rbx,%rbx
  je     L1
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <btp_strndup@plt>
  mov    %rax,(%r12)
  mov    %ebx,%eax
  add    %rbx,0x0(%rbp)
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int btp_parse_char_span(const char **input, const char *accept, char **result) {
  size_t count = strspn(*input, accept);
  if (count == 0)
    return 0;
  *result = btp_strndup(*input, count);
  *input += count;
  return count;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<btp_parse_char_span>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r13
  mov    %r13,%rdi
  call   <strspn@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  test   %rbx,%rbx
  je     L5
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <btp_strndup@plt>
  mov    %rax,(%r12)
  mov    %ebx,%eax
  add    %rbx,0x0(%rbp)
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int btp_parse_char_span(const char **input, const char *accept, char **result) {
  size_t count = strspn(*input, accept);
  if (count == 0)
    return 0;
  *result = btp_strndup(*input, count);
  *input += count;
  return count;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<onintr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x0,%eax
  call   <ignoresig@plt>
  mov    $0x0,%eax
  call   <tty_reset@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void onintr(int sig) {
  ignoresig();
  tty_reset();
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<onintr>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ignoresig@plt>
  mov    $0x0,%eax
  call   <tty_reset@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void onintr(int sig) {
  ignoresig();
  tty_reset();
  exit(1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<onintr>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ignoresig@plt>
  xor    %eax,%eax
  call   <tty_reset@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1080 <exit@plt>
```

assistant: ```c
void onintr(int sig) {
  ignoresig();
  tty_reset();
  exit(1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<onintr>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ignoresig@plt>
  xor    %eax,%eax
  call   <tty_reset@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1080 <exit@plt>
```

assistant: ```c
void onintr(int sig) {
  ignoresig();
  tty_reset();
  exit(1);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ft_striter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L9
  jmp    L5
L11:
  mov    -0x8(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rax,%rdi
  call   *%rdx
  addq   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L11
  jmp    L0
L9:
  nop
L0:
  leave
  ret
```

assistant: ```c
void ft_striter(char *s, void (*f)(char *)) {
  if (!s)
    return;
  while (*s) {
    f(&(*s));
    s++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_striter>:
  endbr64
  test   %rdi,%rdi
  je     L2
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  cmpb   $0x0,(%rdi)
  je     L5
L27:
  mov    %rbx,%rdi
  call   *%rbp
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L27
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  ret
```

assistant: ```c
void ft_striter(char *s, void (*f)(char *)) {
  if (!s)
    return;
  while (*s) {
    f(&(*s));
    s++;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ft_striter>:
  endbr64
  test   %rdi,%rdi
  je     L29
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmpb   $0x0,(%rdi)
  je     L4
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %rbx,%rdi
  add    $0x1,%rbx
  call   *%rbp
  cmpb   $0x0,(%rbx)
  jne    L26
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L29:
  ret
```

assistant: ```c
void ft_striter(char *s, void (*f)(char *)) {
  if (!s)
    return;
  while (*s) {
    f(&(*s));
    s++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ft_striter>:
  endbr64
  test   %rdi,%rdi
  je     L28
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmpb   $0x0,(%rdi)
  je     L21
  nopw   0x0(%rax,%rax,1)
L29:
  mov    %rbx,%rdi
  add    $0x1,%rbx
  call   *%rbp
  cmpb   $0x0,(%rbx)
  jne    L29
L21:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L28:
  ret
```

assistant: ```c
void ft_striter(char *s, void (*f)(char *)) {
  if (!s)
    return;
  while (*s) {
    f(&(*s));
    s++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<solcon_write_quaddobl_solution_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x48(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L27:
  cmp    %rdx,%rsp
  je     L46
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L27
L46:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L36
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L36:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x48(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x1a5,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x3c(%rbp)
  movl   $0x0,-0x40(%rbp)
  jmp    L56
L41:
  mov    -0x30(%rbp),%rax
  mov    -0x40(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%edx
  mov    -0x40(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x50(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x40(%rbp)
L56:
  mov    -0x40(%rbp),%eax
  cmp    -0x48(%rbp),%eax
  jl     L41
  mov    -0x48(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x3c(%rbp),%eax
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int solcon_write_quaddobl_solution_string(int k, int n, char *s) {
  int a[2], b[n], fail, i;
  double *c;
  a[0] = k;
  a[1] = n;
  fail = _ada_use_c2phc4c(421, a, b, c, 0);
  for (i = 0; i < n; i++)
    s[i] = b[i];
  s[n] = '\0';
  return fail;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<solcon_write_quaddobl_solution_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%r14d
  mov    %rdx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq %esi,%r13
  lea    0xf(,%r13,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L3:
  cmp    %rdx,%rsp
  je     L0
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L3
L0:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L2
  orq    $0x0,-0x8(%rsp,%rax,1)
L2:
  mov    %rsp,%r12
  mov    %edi,-0x30(%rbp)
  mov    %r14d,-0x2c(%rbp)
  lea    -0x30(%rbp),%rsi
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    %r12,%rdx
  mov    $0x1a5,%edi
  call   <_ada_use_c2phc4c@plt>
  test   %r14d,%r14d
  jle    L35
  mov    %r14d,%esi
  mov    $0x0,%edx
L11:
  mov    (%r12,%rdx,4),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L11
L35:
  movb   $0x0,(%rbx,%r13,1)
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int solcon_write_quaddobl_solution_string(int k, int n, char *s) {
  int a[2], b[n], fail, i;
  double *c;
  a[0] = k;
  a[1] = n;
  fail = _ada_use_c2phc4c(421, a, b, c, 0);
  for (i = 0; i < n; i++)
    s[i] = b[i];
  s[n] = '\0';
  return fail;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<solcon_write_quaddobl_solution_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  movslq %esi,%r12
  push   %rbx
  mov    %rdx,%rbx
  mov    %r12,%r14
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    0xf(,%r12,4),%rax
  mov    %rsp,%rcx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L46
L25:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L25
L46:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L0
L59:
  mov    %rsp,%r13
  mov    %edi,-0x30(%rbp)
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  lea    -0x30(%rbp),%rsi
  mov    %r13,%rdx
  mov    $0x1a5,%edi
  mov    %r14d,-0x2c(%rbp)
  call   <_ada_use_c2phc4c@plt>
  test   %r14d,%r14d
  jle    L8
  xor    %edx,%edx
  nopl   (%rax)
L39:
  mov    0x0(%r13,%rdx,4),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %r12,%rdx
  jne    L39
L8:
  movb   $0x0,(%rbx,%r12,1)
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L40
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L59
L40:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int solcon_write_quaddobl_solution_string(int k, int n, char *s) {
  int a[2], b[n], fail, i;
  double *c;
  a[0] = k;
  a[1] = n;
  fail = _ada_use_c2phc4c(421, a, b, c, 0);
  for (i = 0; i < n; i++)
    s[i] = b[i];
  s[n] = '\0';
  return fail;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movb   $0x61,-0x1(%rbp)
  jmp    L8
L21:
  movsbl -0x1(%rbp),%edx
  movsbl -0x1(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movzbl -0x1(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1(%rbp)
L8:
  cmpb   $0x7a,-0x1(%rbp)
  jle    L21
  movb   $0x41,-0x1(%rbp)
  jmp    L57
L40:
  movsbl -0x1(%rbp),%edx
  movsbl -0x1(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movzbl -0x1(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1(%rbp)
L57:
  cmpb   $0x5a,-0x1(%rbp)
  jle    L40
  movb   $0x30,-0x1(%rbp)
  jmp    L29
L7:
  movsbl -0x1(%rbp),%edx
  movsbl -0x1(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movzbl -0x1(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1(%rbp)
L29:
  cmpb   $0x39,-0x1(%rbp)
  jle    L7
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type_guessed: string

tool: D7: %d - %c


assistant: ```c
int main() {
  char c;
  c = 'a';
  while (c <= 'z') {
    printf("%d - %c\n", c, c);
    c++;
  }
  c = 'A';
  while (c <= 'Z') {
    printf("%d - %c\n", c, c);
    c++;
  }
  c = '0';
  while (c <= '9') {
    printf("%d - %c\n", c, c);
    c++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x61,%ebx
  lea    D5(%rip),%rbp
L29:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x7b,%ebx
  jne    L29
  mov    $0x41,%ebx
  lea    D5(%rip),%rbp
L27:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x5b,%ebx
  jne    L27
  mov    $0x30,%ebx
  lea    D5(%rip),%rbp
L15:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x3a,%ebx
  jne    L15
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
datatype: string

tool: D5: %d - %c


assistant: ```c
int main() {
  char c;
  c = 'a';
  while (c <= 'z') {
    printf("%d - %c\n", c, c);
    c++;
  }
  c = 'A';
  while (c <= 'Z') {
    printf("%d - %c\n", c, c);
    c++;
  }
  c = '0';
  while (c <= '9') {
    printf("%d - %c\n", c, c);
    c++;
  }
  return 0;
}
```