ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<Emmc_ReadBlock>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  call   <Emmc_ensureDataMode@plt>
  test   %rax,%rax
  jne    L4
  xor    %esi,%esi
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <Emmc_doDataCommand@plt>
  test   %rax,%rax
  js     L4
  mov    %ebx,%eax
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  mov    $0xffffffff,%eax
  jmp    L15
```

assistant: ```c
int Emmc_ReadBlock(char *buf, unsigned int buflen, unsigned int block_number) {
  if (Emmc_ensureDataMode() != 0)
    return -1;
  if (Emmc_doDataCommand(buf, 0, buflen, block_number) < 0)
    return -1;
  return buflen;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<netd_dbl_sum_op>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L3
L4:
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movsd  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movsd  (%rax),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
L3:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
  jne    L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void netd_dbl_sum_op(void *x, void *y, int nelem) {
  double *a = (double *)x;
  double *b = (double *)y;
  while (nelem--)
    *a++ += *b++;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<netd_dbl_sum_op>:
  endbr64
  test   %edx,%edx
  je     L17
  mov    %edx,%edx
  mov    $0x0,%eax
L9:
  movsd  (%rdi,%rax,8),%xmm0
  addsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rdi,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L9
L17:
  ret
```

assistant: ```c
void netd_dbl_sum_op(void *x, void *y, int nelem) {
  double *a = (double *)x;
  double *b = (double *)y;
  while (nelem--)
    *a++ += *b++;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<netd_dbl_sum_op>:
  endbr64
  test   %edx,%edx
  je     L1
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L17:
  movsd  (%rdi,%rax,8),%xmm0
  addsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rdi,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L17
L1:
  ret
```

assistant: ```c
void netd_dbl_sum_op(void *x, void *y, int nelem) {
  double *a = (double *)x;
  double *b = (double *)y;
  while (nelem--)
    *a++ += *b++;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<netd_dbl_sum_op>:
  endbr64
  mov    %rdi,%rcx
  lea    -0x1(%rdx),%eax
  test   %edx,%edx
  je     L29
  lea    0x8(%rsi),%rdi
  cmp    %rdi,%rcx
  je     L30
  cmp    $0x1,%eax
  jbe    L30
  mov    %edx,%edi
  xor    %eax,%eax
  shr    %edi
  shl    $0x4,%rdi
  nopw   0x0(%rax,%rax,1)
L21:
  movupd (%rsi,%rax,1),%xmm0
  movupd (%rcx,%rax,1),%xmm1
  addpd  %xmm1,%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdi
  jne    L21
  mov    %edx,%edi
  and    $0xfffffffe,%edi
  mov    %edi,%eax
  shl    $0x3,%rax
  add    %rax,%rcx
  add    %rsi,%rax
  cmp    %edi,%edx
  je     L29
  movsd  (%rcx),%xmm0
  addsd  (%rax),%xmm0
  movsd  %xmm0,(%rcx)
  ret
  nopl   (%rax)
L30:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L19:
  movsd  (%rcx,%rax,8),%xmm0
  addsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L19
L29:
  ret
```

assistant: ```c
void netd_dbl_sum_op(void *x, void *y, int nelem) {
  double *a = (double *)x;
  double *b = (double *)y;
  while (nelem--)
    *a++ += *b++;
}
```
user: What is the c source code of the assembly code below:

```asm
<nonzerotest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L10
  mov    $0x0,%edi
  call   <echo@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L12
L10:
  mov    $0x1,%edi
  call   <echo@plt>
  mov    %eax,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int nonzerotest(int a) {
  int retval;
  if (a != 0)
    retval = echo(0);
  else
    retval = echo(1);
  return retval;
}
```
user: Translate the below assembly code into C programming language:

```asm
<nonzerotest>:
  endbr64
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L17
  mov    $0x0,%edi
  call   <echo@plt>
L19:
  add    $0x8,%rsp
  ret
L17:
  mov    $0x1,%edi
  call   <echo@plt>
  jmp    L19
```

assistant: ```c
int nonzerotest(int a) {
  int retval;
  if (a != 0)
    retval = echo(0);
  else
    retval = echo(1);
  return retval;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<nonzerotest>:
  endbr64
  test   %edi,%edi
  je     L4
  xor    %edi,%edi
  jmp    1050 <echo@plt>
  nop
L4:
  mov    $0x1,%edi
  jmp    1050 <echo@plt>
```

assistant: ```c
int nonzerotest(int a) {
  int retval;
  if (a != 0)
    retval = echo(0);
  else
    retval = echo(1);
  return retval;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<nonzerotest>:
  endbr64
  test   %edi,%edi
  je     L0
  xor    %edi,%edi
  jmp    1050 <echo@plt>
  nop
L0:
  mov    $0x1,%edi
  jmp    1050 <echo@plt>
```

assistant: ```c
int nonzerotest(int a) {
  int retval;
  if (a != 0)
    retval = echo(0);
  else
    retval = echo(1);
  return retval;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<md_par_end>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  cmpl   $0x2,-0xc(%rbp)
  je     L65
  cmpl   $0x2,-0xc(%rbp)
  jg     L4
  cmpl   $0x0,-0xc(%rbp)
  je     L21
  cmpl   $0x1,-0xc(%rbp)
  je     L24
  jmp    L4
L21:
  mov    -0x8(%rbp),%rax
  lea    D23(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  jmp    L4
L24:
  mov    -0x8(%rbp),%rax
  lea    D2(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  jmp    L4
L65:
  mov    -0x8(%rbp),%rax
  lea    D23(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  nop
L4:
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D23
datatype: string
name: extract_symbolic_data
arguments: tag: D2
datatype: string

tool: D23: 


tool: D2: </p>


assistant: ```c
void md_par_end(FILE *fw, int term) {
  switch (term) {
  case 0:
    fprintf(fw, "\n");
    break;
  case 1:
    fprintf(fw, "</p>\n");
    break;
  case 2:
    fprintf(fw, "\n");
    break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<md_par_end>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%esi
  je     L31
  cmp    $0x2,%esi
  je     L3
  test   %esi,%esi
  je     L14
L42:
  add    $0x8,%rsp
  ret
L14:
  lea    D28(%rip),%rsi
  call   <fprintf@plt>
  jmp    L42
L31:
  lea    D15(%rip),%rsi
  call   <fprintf@plt>
  jmp    L42
L3:
  lea    D28(%rip),%rsi
  call   <fprintf@plt>
  jmp    L42
```

tool_calls: name: retrieve_data
arguments: label: D28
type_infered: string
name: retrieve_data
arguments: label: D15
type_infered: string

tool: D28: 


tool: D15: </p>


assistant: ```c
void md_par_end(FILE *fw, int term) {
  switch (term) {
  case 0:
    fprintf(fw, "\n");
    break;
  case 1:
    fprintf(fw, "</p>\n");
    break;
  case 2:
    fprintf(fw, "\n");
    break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<md_par_end>:
  endbr64
  cmp    $0x1,%esi
  je     L2
  and    $0xfffffffd,%esi
  je     L11
  ret
  nop
L11:
  lea    D15(%rip),%rsi
  jmp    1050 <fprintf@plt>
  nopl   0x0(%rax)
L2:
  lea    D18(%rip),%rsi
  jmp    1050 <fprintf@plt>
```

tool_calls: name: parse_data
arguments: label: D15
datatype: string
name: parse_data
arguments: label: D18
datatype: string

tool: D15: 


tool: D18: </p>


assistant: ```c
void md_par_end(FILE *fw, int term) {
  switch (term) {
  case 0:
    fprintf(fw, "\n");
    break;
  case 1:
    fprintf(fw, "</p>\n");
    break;
  case 2:
    fprintf(fw, "\n");
    break;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<md_par_end>:
  endbr64
  cmp    $0x1,%esi
  je     L12
  and    $0xfffffffd,%esi
  je     L1
  ret
  nop
L1:
  lea    D18(%rip),%rsi
  jmp    1050 <fprintf@plt>
  nopl   0x0(%rax)
L12:
  lea    D9(%rip),%rsi
  jmp    1050 <fprintf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D18
type_guessed: string
name: retrieve_data
arguments: data_label: D9
type_guessed: string

tool: D18: 


tool: D9: </p>


assistant: ```c
void md_par_end(FILE *fw, int term) {
  switch (term) {
  case 0:
    fprintf(fw, "\n");
    break;
  case 1:
    fprintf(fw, "</p>\n");
    break;
  case 2:
    fprintf(fw, "\n");
    break;
  }
}
```
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    $0xfd0,%rsp
  mov    %edi,-0xfc4(%rbp)
  mov    %rsi,-0xfd0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0xfb4(%rbp)
  movl   $0x0,-0xfb4(%rbp)
  jmp    L23
L69:
  mov    -0xfb4(%rbp),%eax
  cltq
  mov    -0xfb4(%rbp),%edx
  mov    %edx,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0xfb4(%rbp)
L23:
  cmpl   $0x3e7,-0xfb4(%rbp)
  jle    L69
  cmpl   $0xffffffff,-0xfb4(%rbp)
  je     L52
  mov    $0x0,%eax
  call   <abort@plt>
L52:
  movl   $0x0,-0xfb4(%rbp)
  jmp    L4
L20:
  mov    -0xfb4(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  cmp    %eax,-0xfb4(%rbp)
  je     L22
  mov    $0x0,%eax
  call   <abort@plt>
L22:
  addl   $0x1,-0xfb4(%rbp)
L4:
  cmpl   $0x3e7,-0xfb4(%rbp)
  jle    L20
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i = -1;
  int c[1000];
#pragma omp parallel for
  for (i = 0; i < 1000; i++) {
    c[i] = i;
  }
  if (i != -1)
    abort();
  for (i = 0; i < 1000; i++) {
    if (c[i] != i)
      abort();
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
L40:
  mov    %eax,(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x3e8,%rax
  jne    L40
  mov    $0x0,%eax
  call   <abort@plt>
  mov    $0x0,%ebx
  mov    %rsp,%rbp
  jmp    L18
L3:
  add    $0x1,%rbx
  cmp    $0x3e8,%rbx
  je     L15
L18:
  cmp    %ebx,0x0(%rbp,%rbx,4)
  je     L3
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L3
L15:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  mov    $0x0,%eax
  add    $0xfb8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i = -1;
  int c[1000];
#pragma omp parallel for
  for (i = 0; i < 1000; i++) {
    c[i] = i;
  }
  if (i != -1)
    abort();
  for (i = 0; i < 1000; i++) {
    if (c[i] != i)
      abort();
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nopl   0x0(%rax,%rax,1)
L27:
  mov    %eax,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    $0x3e8,%rax
  jne    L27
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <abort@plt>
  nopl   0x0(%rax)
L32:
  cmp    %ebx,0x0(%rbp,%rbx,4)
  je     L16
  xor    %eax,%eax
  call   <abort@plt>
L16:
  add    $0x1,%rbx
  cmp    $0x3e8,%rbx
  jne    L32
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0xfb8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i = -1;
  int c[1000];
#pragma omp parallel for
  for (i = 0; i < 1000; i++) {
    c[i] = i;
  }
  if (i != -1)
    abort();
  for (i = 0; i < 1000; i++) {
    if (c[i] != i)
      abort();
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xfb8,%rsp
  movdqa D15(%rip),%xmm0
  movdqa D9(%rip),%xmm2
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0xfa0(%rsp),%rdx
  mov    %rbp,%rax
  xchg   %ax,%ax
L33:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  movaps %xmm1,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L33
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <abort@plt>
  xchg   %ax,%ax
L30:
  cmp    %ebx,0x0(%rbp,%rbx,4)
  je     L34
  xor    %eax,%eax
  call   <abort@plt>
L34:
  add    $0x1,%rbx
  cmp    $0x3e8,%rbx
  jne    L30
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0xfb8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D15
data_type: byte[16]
name: parse_data
arguments: tag: D9
data_type: byte[16]

tool: {"D15": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D9": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, char *argv[]) {
  int i = -1;
  int c[1000];
#pragma omp parallel for
  for (i = 0; i < 1000; i++) {
    c[i] = i;
  }
  if (i != -1)
    abort();
  for (i = 0; i < 1000; i++) {
    if (c[i] != i)
      abort();
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<splitspec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %r9d,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L11
L12:
  mov    -0x8(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cltd
  idivl  0x18(%rbp)
  mov    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cltd
  idivl  0x18(%rbp)
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%eax
  cmp    0x10(%rbp),%eax
  jl     L12
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void splitspec(int *binsplit, int bin_offset, int table_offset, float *inmag, float *dest_mag, int start, int end, int n) {
  int i;
  int bindex;
  for (i = start; i < end; i++) {
    bindex = binsplit[(i + table_offset) % n];
    bindex = (bindex + bin_offset) % n;
    dest_mag[bindex] = inmag[bindex];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<splitspec>:
  endbr64
  mov    %esi,%r10d
  mov    %rcx,%r11
  mov    0x10(%rsp),%esi
  cmp    0x8(%rsp),%r9d
  jge    L4
  lea    (%r9,%rdx,1),%ecx
  add    0x8(%rsp),%edx
  mov    %edx,%r9d
L3:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  movslq %edx,%rdx
  mov    %r10d,%eax
  add    (%rdi,%rdx,4),%eax
  cltd
  idiv   %esi
  movslq %edx,%rdx
  movss  (%r11,%rdx,4),%xmm0
  movss  %xmm0,(%r8,%rdx,4)
  add    $0x1,%ecx
  cmp    %r9d,%ecx
  jne    L3
L4:
  ret
```

assistant: ```c
void splitspec(int *binsplit, int bin_offset, int table_offset, float *inmag, float *dest_mag, int start, int end, int n) {
  int i;
  int bindex;
  for (i = start; i < end; i++) {
    bindex = binsplit[(i + table_offset) % n];
    bindex = (bindex + bin_offset) % n;
    dest_mag[bindex] = inmag[bindex];
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<splitspec>:
  endbr64
  mov    %esi,%r10d
  mov    %rcx,%r11
  mov    0x10(%rsp),%esi
  cmp    0x8(%rsp),%r9d
  jge    L18
  lea    (%r9,%rdx,1),%ecx
  add    0x8(%rsp),%edx
  mov    %edx,%r9d
L9:
  mov    %ecx,%eax
  add    $0x1,%ecx
  cltd
  idiv   %esi
  movslq %edx,%rdx
  mov    (%rdi,%rdx,4),%eax
  add    %r10d,%eax
  cltd
  idiv   %esi
  movslq %edx,%rdx
  movss  (%r11,%rdx,4),%xmm0
  movss  %xmm0,(%r8,%rdx,4)
  cmp    %ecx,%r9d
  jne    L9
L18:
  ret
```

assistant: ```c
void splitspec(int *binsplit, int bin_offset, int table_offset, float *inmag, float *dest_mag, int start, int end, int n) {
  int i;
  int bindex;
  for (i = start; i < end; i++) {
    bindex = binsplit[(i + table_offset) % n];
    bindex = (bindex + bin_offset) % n;
    dest_mag[bindex] = inmag[bindex];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<splitspec>:
  endbr64
  mov    %esi,%r10d
  mov    %rcx,%r11
  mov    0x10(%rsp),%esi
  cmp    0x8(%rsp),%r9d
  jge    L4
  lea    (%r9,%rdx,1),%ecx
  add    0x8(%rsp),%edx
  mov    %edx,%r9d
L9:
  mov    %ecx,%eax
  add    $0x1,%ecx
  cltd
  idiv   %esi
  movslq %edx,%rdx
  mov    (%rdi,%rdx,4),%eax
  add    %r10d,%eax
  cltd
  idiv   %esi
  movslq %edx,%rdx
  movss  (%r11,%rdx,4),%xmm0
  movss  %xmm0,(%r8,%rdx,4)
  cmp    %ecx,%r9d
  jne    L9
L4:
  ret
```

assistant: ```c
void splitspec(int *binsplit, int bin_offset, int table_offset, float *inmag, float *dest_mag, int start, int end, int n) {
  int i;
  int bindex;
  for (i = start; i < end; i++) {
    bindex = binsplit[(i + table_offset) % n];
    bindex = (bindex + bin_offset) % n;
    dest_mag[bindex] = inmag[bindex];
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<firstUser>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  sub    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L5
L8:
  cmpl   $0x0,-0xc(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <getChoosing@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  sub    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L5:
  cmpl   $0x1,-0x14(%rbp)
  jg     L8
  cmpl   $0x0,-0xc(%rbp)
  jne    L14
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L0
L14:
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L0:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D41
data_type: string
name: parse_data
arguments: data_label: D16
data_type: string
name: parse_data
arguments: data_label: D1
data_type: string
name: parse_data
arguments: data_label: D57
data_type: string
name: parse_data
arguments: data_label: D37
data_type: string

tool: D41: 
users turn :

tool: D16: %d

tool: D1: 
sticks left :%d


tool: D57: 
 computer lost

tool: D37: 
user lost

assistant: ```c
int firstUser(int sticks) {
  int choosing = 0, turn = 0;
  printf("\nusers turn :\n");
  scanf("%d", &choosing);
  sticks = sticks - choosing;
  printf("\nsticks left :%d\n", sticks);
  while (sticks > 1) {
    turn = turn == 0 ? 1 : 0;
    choosing = getChoosing(turn, sticks);
    sticks -= choosing;
    printf("\nsticks left :%d\n", sticks);
  }
  if (turn == 0) {
    printf("\n computer lost\n");
  } else
    printf("\nuser lost\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<firstUser>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    D12(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D56(%rip),%rdi
  call   <scanf@plt>
  sub    0x4(%rsp),%ebx
  mov    %ebx,%esi
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x1,%ebx
  jle    L25
  mov    $0x0,%ebp
  lea    D32(%rip),%r13
L8:
  mov    %ebp,%r12d
  xor    $0x1,%ebp
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <getChoosing@plt>
  mov    %eax,0x4(%rsp)
  sub    %eax,%ebx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x1,%ebx
  jg     L8
  test   %r12d,%r12d
  jne    L25
  lea    D59(%rip),%rdi
  call   <puts@plt>
  jmp    L46
L25:
  lea    D46(%rip),%rdi
  call   <puts@plt>
L46:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D12
datatype: string
name: extract_symbolic_data
arguments: data_tag: D56
datatype: string
name: extract_symbolic_data
arguments: data_tag: D32
datatype: string
name: extract_symbolic_data
arguments: data_tag: D59
datatype: string
name: extract_symbolic_data
arguments: data_tag: D46
datatype: string

tool: D12: 
users turn :

tool: D56: %d

tool: D32: 
sticks left :%d


tool: D59: 
user lost

tool: D46: 
 computer lost

assistant: ```c
int firstUser(int sticks) {
  int choosing = 0, turn = 0;
  printf("\nusers turn :\n");
  scanf("%d", &choosing);
  sticks = sticks - choosing;
  printf("\nsticks left :%d\n", sticks);
  while (sticks > 1) {
    turn = turn == 0 ? 1 : 0;
    choosing = getChoosing(turn, sticks);
    sticks -= choosing;
    printf("\nsticks left :%d\n", sticks);
  }
  if (turn == 0) {
    printf("\n computer lost\n");
  } else
    printf("\nuser lost\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<firstUser>:
  endbr64
  push   %r13
  push   %r12
  lea    D48(%rip),%r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  lea    D41(%rip),%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  sub    0x4(%rsp),%ebx
  xor    %eax,%eax
  mov    %r12,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  cmp    $0x1,%ebx
  jle    L19
  xor    %ebp,%ebp
  nopl   0x0(%rax,%rax,1)
L39:
  mov    %ebp,%r13d
  xor    $0x1,%ebp
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <getChoosing@plt>
  mov    %r12,%rdi
  sub    %eax,%ebx
  mov    %eax,0x4(%rsp)
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  cmp    $0x1,%ebx
  jg     L39
  cmp    $0x1,%r13d
  je     L19
  lea    D1(%rip),%rdi
  call   <puts@plt>
L11:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L19:
  lea    D36(%rip),%rdi
  call   <puts@plt>
  jmp    L11
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D48
type_infered: string
name: disassemble_data
arguments: label: D41
type_infered: string
name: disassemble_data
arguments: label: D31
type_infered: string
name: disassemble_data
arguments: label: D1
type_infered: string
name: disassemble_data
arguments: label: D36
type_infered: string

tool: D48: 
sticks left :%d


tool: D41: 
users turn :

tool: D31: %d

tool: D1: 
user lost

tool: D36: 
 computer lost

assistant: ```c
int firstUser(int sticks) {
  int choosing = 0, turn = 0;
  printf("\nusers turn :\n");
  scanf("%d", &choosing);
  sticks = sticks - choosing;
  printf("\nsticks left :%d\n", sticks);
  while (sticks > 1) {
    turn = turn == 0 ? 1 : 0;
    choosing = getChoosing(turn, sticks);
    sticks -= choosing;
    printf("\nsticks left :%d\n", sticks);
  }
  if (turn == 0) {
    printf("\n computer lost\n");
  } else
    printf("\nuser lost\n");
  return 0;
}
```