ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<getdelim>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    (%rdi),%r15
  mov    %rsi,0x8(%rsp)
  test   %r15,%r15
  je     L25
  mov    (%rsi),%r13
L25:
  xor    %r14d,%r14d
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L72:
  lea    0x1(%r14),%rsi
  mov    %al,(%r15,%r14,1)
  mov    %rsi,%r14
  cmp    %r12d,%eax
  je     L20
L11:
  mov    %r13,%rax
  sub    %r14,%rax
  cmp    $0x2,%rax
  ja     L40
  lea    0x0(%r13,%r13,1),%rax
  test   %r13,%r13
  mov    $0x100,%r13d
  mov    (%rbx),%rdi
  cmovne %rax,%r13
  mov    %r13,%rsi
  call   <realloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L0
  mov    %rax,(%rbx)
  mov    0x8(%rsp),%rax
  mov    %r13,(%rax)
L40:
  mov    %rbp,%rdi
  call   <fgetc@plt>
  cmp    $0xffffffff,%eax
  jne    L72
  test   %r14,%r14
  je     L0
  mov    %rbp,%rdi
  call   <ferror@plt>
  test   %rax,%rax
  jne    L0
L20:
  movb   $0x0,(%r15,%r14,1)
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  add    $0x18,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
ssize_t getdelim(char **restrict lineptr, size_t *restrict n, int delimiter, FILE *restrict stream) {
  char *ptr = *lineptr;
  size_t size = (ptr != NULL) ? *n : 0;
  size_t len = 0;
  for (;;) {
    if ((size - len) <= 2) {
      size = size ? (size * 2) : 256;
      ptr = realloc(*lineptr, size);
      if (ptr == NULL)
        return -1;
      *lineptr = ptr;
      *n = size;
    }
    int c = fgetc(stream);
    if (c == -1) {
      if (len == 0 || ferror(stream))
        return -1;
      break;
    }
    ptr[len++] = c;
    if (c == delimiter)
      break;
  }
  ptr[len] = '\0';
  return len;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<padcon_get_dobldobl_numerator_coefficient>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %edx,-0x4c(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x30(%rbp)
  movl   $0x1,-0x2c(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x48(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  lea    -0x20(%rbp),%rcx
  lea    -0x4c(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x366,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x34(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x58(%rbp),%rax
  movsd  %xmm0,(%rax)
  movsd  -0x18(%rbp),%xmm0
  mov    -0x60(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x34(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int padcon_get_dobldobl_numerator_coefficient(int leadidx, int idx, int verbose, double *cre, double *cim) {
  int fail;
  int inpars[4];
  double nbr[2];
  inpars[0] = 1;
  inpars[1] = 1;
  inpars[2] = leadidx;
  inpars[3] = idx;
  fail = _ada_use_c2phc4c(870, inpars, &verbose, nbr, 0);
  *cre = nbr[0];
  *cim = nbr[1];
  return fail;
}
```
user: Translate the below assembly code into C programming language:

```asm
<padcon_get_dobldobl_numerator_coefficient>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edx,0xc(%rsp)
  mov    %rcx,%rbp
  mov    %r8,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  movl   $0x1,0x14(%rsp)
  mov    %edi,0x18(%rsp)
  mov    %esi,0x1c(%rsp)
  lea    0x20(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    $0x0,%r8d
  mov    $0x366,%edi
  call   <_ada_use_c2phc4c@plt>
  movsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int padcon_get_dobldobl_numerator_coefficient(int leadidx, int idx, int verbose, double *cre, double *cim) {
  int fail;
  int inpars[4];
  double nbr[2];
  inpars[0] = 1;
  inpars[1] = 1;
  inpars[2] = leadidx;
  inpars[3] = idx;
  fail = _ada_use_c2phc4c(870, inpars, &verbose, nbr, 0);
  *cre = nbr[0];
  *cim = nbr[1];
  return fail;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<padcon_get_dobldobl_numerator_coefficient>:
  endbr64
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %r8,%rbx
  xor    %r8d,%r8d
  sub    $0x48,%rsp
  mov    %edx,0xc(%rsp)
  lea    0x20(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x100000001,%rax
  mov    %edi,0x18(%rsp)
  mov    $0x366,%edi
  mov    %esi,0x1c(%rsp)
  lea    0x10(%rsp),%rsi
  mov    %rax,0x10(%rsp)
  call   <_ada_use_c2phc4c@plt>
  movsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int padcon_get_dobldobl_numerator_coefficient(int leadidx, int idx, int verbose, double *cre, double *cim) {
  int fail;
  int inpars[4];
  double nbr[2];
  inpars[0] = 1;
  inpars[1] = 1;
  inpars[2] = leadidx;
  inpars[3] = idx;
  fail = _ada_use_c2phc4c(870, inpars, &verbose, nbr, 0);
  *cre = nbr[0];
  *cim = nbr[1];
  return fail;
}
```
user: Translate the below assembly code into C programming language:

```asm
<padcon_get_dobldobl_numerator_coefficient>:
  endbr64
  push   %rbp
  movd   %esi,%xmm1
  movd   %edi,%xmm0
  mov    %rcx,%rbp
  push   %rbx
  punpckldq %xmm1,%xmm0
  mov    %r8,%rbx
  mov    $0x366,%edi
  xor    %r8d,%r8d
  sub    $0x48,%rsp
  mov    %edx,0xc(%rsp)
  lea    0x20(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    D1(%rip),%rax
  movq   %xmm0,0x18(%rsp)
  mov    %rax,0x10(%rsp)
  call   <_ada_use_c2phc4c@plt>
  movsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D1
type_infered: byte[8]

tool: {"D1": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int padcon_get_dobldobl_numerator_coefficient(int leadidx, int idx, int verbose, double *cre, double *cim) {
  int fail;
  int inpars[4];
  double nbr[2];
  inpars[0] = 1;
  inpars[1] = 1;
  inpars[2] = leadidx;
  inpars[3] = idx;
  fail = _ada_use_c2phc4c(870, inpars, &verbose, nbr, 0);
  *cre = nbr[0];
  *cim = nbr[1];
  return fail;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L16
L1:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x5,%eax
  sub    %edx,%eax
  add    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  addq   $0x1,-0x18(%rbp)
L16:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L1
  mov    -0x4(%rbp),%edx
  mov    %edx,%ecx
  mov    $0x9a8feff7,%eax
  imul   %rcx,%rax
  shr    $0x20,%rax
  mov    %rax,%rcx
  mov    %edx,%eax
  sub    %ecx,%eax
  shr    %eax
  add    %ecx,%eax
  shr    $0xa,%eax
  imul   $0x4fd,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned Hash(char *name) {
  unsigned hashval;
  for (hashval = 0; *name != '\0'; name++)
    hashval = *name + 31 * hashval;
  return hashval % 1277;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Hash>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L15
  mov    $0x0,%ecx
L27:
  mov    %ecx,%edx
  shl    $0x5,%edx
  sub    %ecx,%edx
  movsbl %al,%eax
  lea    (%rax,%rdx,1),%ecx
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L27
L2:
  mov    %ecx,%edx
  mov    $0x9a8feff7,%eax
  imul   %rax,%rdx
  shr    $0x20,%rdx
  mov    %ecx,%eax
  sub    %edx,%eax
  shr    %eax
  add    %edx,%eax
  shr    $0xa,%eax
  imul   $0x4fd,%eax,%edx
  mov    %ecx,%eax
  sub    %edx,%eax
  ret
L15:
  mov    $0x0,%ecx
  jmp    L2
```

assistant: ```c
unsigned Hash(char *name) {
  unsigned hashval;
  for (hashval = 0; *name != '\0'; name++)
    hashval = *name + 31 * hashval;
  return hashval % 1277;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Hash>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L6
  xor    %ecx,%ecx
  nopl   (%rax)
L14:
  mov    %ecx,%edx
  add    $0x1,%rdi
  shl    $0x5,%edx
  sub    %ecx,%edx
  lea    (%rax,%rdx,1),%ecx
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L14
  mov    %ecx,%edx
  mov    $0x9a8feff7,%eax
  imul   %rax,%rdx
  mov    %ecx,%eax
  shr    $0x20,%rdx
  sub    %edx,%eax
  shr    %eax
  add    %edx,%eax
  shr    $0xa,%eax
  imul   $0x4fd,%eax,%edx
  mov    %ecx,%eax
  sub    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned Hash(char *name) {
  unsigned hashval;
  for (hashval = 0; *name != '\0'; name++)
    hashval = *name + 31 * hashval;
  return hashval % 1277;
}
```
user: What is the c source code of the assembly code below?

```asm
<Hash>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L10
  xor    %ecx,%ecx
  nopl   (%rax)
L15:
  mov    %ecx,%edx
  add    $0x1,%rdi
  shl    $0x5,%edx
  sub    %ecx,%edx
  lea    (%rax,%rdx,1),%ecx
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L15
  mov    %ecx,%edx
  mov    $0x9a8feff7,%eax
  imul   %rax,%rdx
  mov    %ecx,%eax
  shr    $0x20,%rdx
  sub    %edx,%eax
  shr    %eax
  add    %edx,%eax
  shr    $0xa,%eax
  imul   $0x4fd,%eax,%edx
  mov    %ecx,%eax
  sub    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned Hash(char *name) {
  unsigned hashval;
  for (hashval = 0; *name != '\0'; name++)
    hashval = *name + 31 * hashval;
  return hashval % 1277;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<myvsnprintf_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L23
L0:
  mov    -0x28(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L23:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L19
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
L19:
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void myvsnprintf_string(char **pdest, char *de, const char *str) {
  char *dest = *pdest;
  while (dest < de && *str)
    *dest++ = *str++;
  *pdest = dest;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<myvsnprintf_string>:
  endbr64
  mov    (%rdi),%rax
  cmp    %rsi,%rax
  jae    L2
  sub    %rax,%rsi
  add    %rdx,%rsi
L26:
  movzbl (%rdx),%ecx
  test   %cl,%cl
  je     L2
  add    $0x1,%rdx
  add    $0x1,%rax
  mov    %cl,-0x1(%rax)
  cmp    %rsi,%rdx
  jne    L26
L2:
  mov    %rax,(%rdi)
  ret
```

assistant: ```c
void myvsnprintf_string(char **pdest, char *de, const char *str) {
  char *dest = *pdest;
  while (dest < de && *str)
    *dest++ = *str++;
  *pdest = dest;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<myvsnprintf_string>:
  endbr64
  mov    (%rdi),%rax
  cmp    %rsi,%rax
  jb     L30
  jmp    L15
  xchg   %ax,%ax
L27:
  add    $0x1,%rax
  add    $0x1,%rdx
  mov    %cl,-0x1(%rax)
  cmp    %rax,%rsi
  je     L15
L30:
  movzbl (%rdx),%ecx
  test   %cl,%cl
  jne    L27
L15:
  mov    %rax,(%rdi)
  ret
```

assistant: ```c
void myvsnprintf_string(char **pdest, char *de, const char *str) {
  char *dest = *pdest;
  while (dest < de && *str)
    *dest++ = *str++;
  *pdest = dest;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<myvsnprintf_string>:
  endbr64
  mov    (%rdi),%rax
  cmp    %rsi,%rax
  jb     L34
  jmp    L3
  xchg   %ax,%ax
L10:
  add    $0x1,%rax
  add    $0x1,%rdx
  mov    %cl,-0x1(%rax)
  cmp    %rax,%rsi
  je     L3
L34:
  movzbl (%rdx),%ecx
  test   %cl,%cl
  jne    L10
L3:
  mov    %rax,(%rdi)
  ret
```

assistant: ```c
void myvsnprintf_string(char **pdest, char *de, const char *str) {
  char *dest = *pdest;
  while (dest < de && *str)
    *dest++ = *str++;
  *pdest = dest;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<basPositionSound>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <popNumber@plt>
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%eax
  call   <popNumber@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <popNumber@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <positionsound@plt>
  mov    %eax,%edi
  call   <pushNumber@plt>
  nop
  leave
  ret
```

assistant: ```c
void basPositionSound() {
  int c, angle, dist;
  dist = popNumber();
  angle = popNumber();
  c = popNumber();
  pushNumber(positionsound(c, angle, dist));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<basPositionSound>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <popNumber@plt>
  mov    %eax,%ebp
  mov    $0x0,%eax
  call   <popNumber@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <popNumber@plt>
  mov    %eax,%edi
  mov    %ebp,%edx
  mov    %ebx,%esi
  call   <positionsound@plt>
  mov    %eax,%edi
  call   <pushNumber@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void basPositionSound() {
  int c, angle, dist;
  dist = popNumber();
  angle = popNumber();
  c = popNumber();
  pushNumber(positionsound(c, angle, dist));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<basPositionSound>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  sub    $0x8,%rsp
  call   <popNumber@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <popNumber@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <popNumber@plt>
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <positionsound@plt>
  add    $0x8,%rsp
  pop    %rbp
  mov    %eax,%edi
  pop    %r12
  jmp    1070 <pushNumber@plt>
```

assistant: ```c
void basPositionSound() {
  int c, angle, dist;
  dist = popNumber();
  angle = popNumber();
  c = popNumber();
  pushNumber(positionsound(c, angle, dist));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<basPositionSound>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  sub    $0x8,%rsp
  call   <popNumber@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <popNumber@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <popNumber@plt>
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <positionsound@plt>
  add    $0x8,%rsp
  pop    %rbp
  mov    %eax,%edi
  pop    %r12
  jmp    1070 <pushNumber@plt>
```

assistant: ```c
void basPositionSound() {
  int c, angle, dist;
  dist = popNumber();
  angle = popNumber();
  c = popNumber();
  pushNumber(positionsound(c, angle, dist));
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_itoa_base>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x1,-0x18(%rbp)
  jle    L111
  cmpl   $0x10,-0x18(%rbp)
  jle    L117
L111:
  mov    $0x0,%eax
  jmp    L80
L117:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <get_length@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpl   $0xa,-0x18(%rbp)
  jne    L92
  cmpl   $0x0,-0x14(%rbp)
  jns    L92
  mov    -0x8(%rbp),%rax
  movb   $0x2d,(%rax)
L92:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L86
L41:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jns    L31
  negl   -0xc(%rbp)
L31:
  cmpl   $0x9,-0xc(%rbp)
  jle    L87
  mov    -0xc(%rbp),%eax
  lea    0x37(%rax),%ecx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L39
L87:
  mov    -0xc(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L39:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L95
L86:
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  test   %eax,%eax
  jg     L41
  jmp    L63
L95:
  nop
L63:
  mov    -0x8(%rbp),%rax
L80:
  leave
  ret
```

assistant: ```c
char *ft_itoa_base(int value, int base) {
  int len;
  int index;
  char *str;
  if (base < 2 || base > 16)
    return (0);
  len = get_length(value, base);
  str = (char *)malloc(sizeof(char) * (len + 1));
  if (base == 10 && value < 0)
    str[0] = '-';
  str[len] = '\0';
  while (len-- > 0) {
    index = value % base;
    if (index < 0)
      index = -index;
    if (index > 9)
      str[len] = index + 'A' - 10;
    else
      str[len] = index + '0';
    value /= base;
    if (value == 0)
      break;
  }
  return (str);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ft_itoa_base>:
  endbr64
  lea    -0x2(%rsi),%eax
  mov    $0x0,%r8d
  cmp    $0xe,%eax
  ja     L73
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%ebp
  call   <get_length@plt>
  mov    %eax,%r12d
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  cmp    $0xa,%ebp
  jne    L7
  test   %ebx,%ebx
  jns    L7
  movb   $0x2d,(%rax)
L7:
  movslq %r12d,%rdx
  movb   $0x0,(%rax,%rdx,1)
  lea    -0x1(%rax,%rdx,1),%rax
  mov    %rax,%rsi
  mov    %r12d,%edi
  sub    %eax,%edi
  jmp    L2
L17:
  add    $0x30,%edx
  mov    %dl,(%rsi)
L70:
  mov    %ebx,%eax
  cltd
  idiv   %ebp
  mov    %eax,%ebx
  sub    $0x1,%rsi
  test   %eax,%eax
  je     L64
L2:
  lea    (%rdi,%rsi,1),%eax
  test   %eax,%eax
  jle    L64
  mov    %ebx,%eax
  cltd
  idiv   %ebp
  mov    %edx,%ecx
  neg    %ecx
  cmovns %ecx,%edx
  cmp    $0x9,%edx
  jle    L17
  add    $0x37,%edx
  mov    %dl,(%rsi)
  jmp    L70
L64:
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L73:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *ft_itoa_base(int value, int base) {
  int len;
  int index;
  char *str;
  if (base < 2 || base > 16)
    return (0);
  len = get_length(value, base);
  str = (char *)malloc(sizeof(char) * (len + 1));
  if (base == 10 && value < 0)
    str[0] = '-';
  str[len] = '\0';
  while (len-- > 0) {
    index = value % base;
    if (index < 0)
      index = -index;
    if (index > 9)
      str[len] = index + 'A' - 10;
    else
      str[len] = index + '0';
    value /= base;
    if (value == 0)
      break;
  }
  return (str);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ft_itoa_base>:
  endbr64
  lea    -0x2(%rsi),%eax
  xor    %r8d,%r8d
  cmp    $0xe,%eax
  ja     L1
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  call   <get_length@plt>
  lea    0x1(%rax),%edi
  mov    %eax,%ebx
  call   <malloc@plt>
  mov    %rax,%r8
  cmp    $0xa,%ebp
  jne    L18
  test   %r12d,%r12d
  jns    L18
  movb   $0x2d,(%rax)
L18:
  movslq %ebx,%rcx
  lea    -0x1(%rax),%rdi
  movb   $0x0,(%rax,%rcx,1)
  jmp    L2
  nopl   0x0(%rax)
L50:
  mov    %r12d,%eax
  cltd
  idiv   %ebp
  mov    %eax,%r12d
  mov    %edx,%eax
  neg    %eax
  cmovns %eax,%edx
  cmp    $0xa,%edx
  lea    0x37(%rdx),%esi
  lea    0x30(%rdx),%eax
  cmovge %esi,%eax
  mov    %al,(%rdi,%rcx,1)
  sub    $0x1,%rcx
  test   %r12d,%r12d
  je     L6
L2:
  test   %ecx,%ecx
  jg     L50
L6:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L1:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *ft_itoa_base(int value, int base) {
  int len;
  int index;
  char *str;
  if (base < 2 || base > 16)
    return (0);
  len = get_length(value, base);
  str = (char *)malloc(sizeof(char) * (len + 1));
  if (base == 10 && value < 0)
    str[0] = '-';
  str[len] = '\0';
  while (len-- > 0) {
    index = value % base;
    if (index < 0)
      index = -index;
    if (index > 9)
      str[len] = index + 'A' - 10;
    else
      str[len] = index + '0';
    value /= base;
    if (value == 0)
      break;
  }
  return (str);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_itoa_base>:
  endbr64
  lea    -0x2(%rsi),%eax
  xor    %r8d,%r8d
  cmp    $0xe,%eax
  ja     L42
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  call   <get_length@plt>
  lea    0x1(%rax),%edi
  mov    %eax,%ebx
  call   <malloc@plt>
  mov    %rax,%r8
  cmp    $0xa,%ebp
  jne    L52
  test   %r12d,%r12d
  jns    L52
  movb   $0x2d,(%rax)
L52:
  movslq %ebx,%rsi
  lea    -0x1(%rax),%rdi
  movb   $0x0,(%rax,%rsi,1)
  cs nopw 0x0(%rax,%rax,1)
L14:
  test   %esi,%esi
  jle    L58
  mov    %r12d,%eax
  cltd
  idiv   %ebp
  mov    %edx,%ecx
  mov    %eax,%r12d
  neg    %ecx
  cmovns %ecx,%edx
  lea    (%rdi,%rsi,1),%rcx
  cmp    $0x9,%edx
  jle    L12
  add    $0x37,%edx
  sub    $0x1,%rsi
  mov    %dl,(%rcx)
  test   %eax,%eax
  jne    L14
L58:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  add    $0x30,%edx
  sub    $0x1,%rsi
  mov    %dl,(%rcx)
  test   %eax,%eax
  jne    L14
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L42:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *ft_itoa_base(int value, int base) {
  int len;
  int index;
  char *str;
  if (base < 2 || base > 16)
    return (0);
  len = get_length(value, base);
  str = (char *)malloc(sizeof(char) * (len + 1));
  if (base == 10 && value < 0)
    str[0] = '-';
  str[len] = '\0';
  while (len-- > 0) {
    index = value % base;
    if (index < 0)
      index = -index;
    if (index > 9)
      str[len] = index + 'A' - 10;
    else
      str[len] = index + '0';
    value /= base;
    if (value == 0)
      break;
  }
  return (str);
}
```