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
<b_case>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  jle    L16
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <error@plt>
  jmp    L17
L16:
  mov    $0x0,%eax
L17:
  leave
  ret
```

assistant: ```c
int b_case(int argc, char *argv[]) {
  if (argc > 2)
    return error(1, argv[0]);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<b_case>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x2,%edi
  jg     L5
  ret
L5:
  sub    $0x8,%rsp
  mov    (%rsi),%rsi
  mov    $0x1,%edi
  call   <error@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int b_case(int argc, char *argv[]) {
  if (argc > 2)
    return error(1, argv[0]);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<b_case>:
  endbr64
  cmp    $0x2,%edi
  jg     L5
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L5:
  mov    (%rsi),%rsi
  mov    $0x1,%edi
  jmp    1050 <error@plt>
```

assistant: ```c
int b_case(int argc, char *argv[]) {
  if (argc > 2)
    return error(1, argv[0]);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<b_case>:
  endbr64
  cmp    $0x2,%edi
  jg     L0
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L0:
  mov    (%rsi),%rsi
  mov    $0x1,%edi
  jmp    1050 <error@plt>
```

assistant: ```c
int b_case(int argc, char *argv[]) {
  if (argc > 2)
    return error(1, argv[0]);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<swap_long_integer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl 0x3(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x3(%rax),%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  movzbl 0x1(%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  movzbl 0x2(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x2(%rax),%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap_long_integer(void *ptr) {
  char *tptr = (char *)ptr, tchar;
  {
    tchar = tptr[0];
    tptr[0] = tptr[3];
    tptr[3] = tchar;
  };
  {
    tchar = tptr[1];
    tptr[1] = tptr[2];
    tptr[2] = tchar;
  };
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<swap_long_integer>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x3(%rdi),%edx
  mov    %dl,(%rdi)
  mov    %al,0x3(%rdi)
  movzbl 0x1(%rdi),%eax
  movzbl 0x2(%rdi),%edx
  mov    %dl,0x1(%rdi)
  mov    %al,0x2(%rdi)
  ret
```

assistant: ```c
void swap_long_integer(void *ptr) {
  char *tptr = (char *)ptr, tchar;
  {
    tchar = tptr[0];
    tptr[0] = tptr[3];
    tptr[3] = tchar;
  };
  {
    tchar = tptr[1];
    tptr[1] = tptr[2];
    tptr[2] = tchar;
  };
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<swap_long_integer>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x3(%rdi),%edx
  rolw   $0x8,0x1(%rdi)
  mov    %dl,(%rdi)
  mov    %al,0x3(%rdi)
  ret
```

assistant: ```c
void swap_long_integer(void *ptr) {
  char *tptr = (char *)ptr, tchar;
  {
    tchar = tptr[0];
    tptr[0] = tptr[3];
    tptr[3] = tchar;
  };
  {
    tchar = tptr[1];
    tptr[1] = tptr[2];
    tptr[2] = tchar;
  };
}
```
user: Translate the below assembly code into C programming language:

```asm
<swap_long_integer>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x3(%rdi),%edx
  rolw   $0x8,0x1(%rdi)
  mov    %dl,(%rdi)
  mov    %al,0x3(%rdi)
  ret
```

assistant: ```c
void swap_long_integer(void *ptr) {
  char *tptr = (char *)ptr, tchar;
  {
    tchar = tptr[0];
    tptr[0] = tptr[3];
    tptr[3] = tchar;
  };
  {
    tchar = tptr[1];
    tptr[1] = tptr[2];
    tptr[2] = tchar;
  };
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<col_stride_mmul_2x2_2x2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,-0x10(%rbp)
  mov    %r8,-0x20(%rbp)
  mov    %r9d,-0x24(%rbp)
  mov    -0x24(%rbp),%r8d
  mov    -0x20(%rbp),%rdi
  mov    -0x10(%rbp),%ecx
  mov    -0x18(%rbp),%rdx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %r8d,%r9d
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <col_stride_mmul_2x2_2x1@plt>
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%r8
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rdi
  mov    -0x24(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %ecx,%r9d
  mov    %edx,%ecx
  mov    %rdi,%rdx
  mov    %rax,%rdi
  call   <col_stride_mmul_2x2_2x1@plt>
  nop
  leave
  ret
```

assistant: ```c
void col_stride_mmul_2x2_2x2(const double *a, int a_col_stride, const double *b, int b_col_stride, double *c, int c_col_stride) {
  col_stride_mmul_2x2_2x1(&a[(0) + (0) * a_col_stride], a_col_stride, &b[(0) + (0) * b_col_stride], b_col_stride, &c[(0) + (0) * c_col_stride], c_col_stride);
  col_stride_mmul_2x2_2x1(&a[(0) + (0) * a_col_stride], a_col_stride, &b[(0) + (1) * b_col_stride], b_col_stride, &c[(0) + (1) * c_col_stride], c_col_stride);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<col_stride_mmul_2x2_2x2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %rdx,%r15
  mov    %ecx,%ebx
  mov    %r8,%r14
  mov    %r9d,%ebp
  call   <col_stride_mmul_2x2_2x1@plt>
  movslq %ebp,%rax
  movslq %ebx,%rdx
  lea    (%r15,%rdx,8),%rdx
  mov    %ebp,%r9d
  lea    (%r14,%rax,8),%r8
  mov    %ebx,%ecx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <col_stride_mmul_2x2_2x1@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void col_stride_mmul_2x2_2x2(const double *a, int a_col_stride, const double *b, int b_col_stride, double *c, int c_col_stride) {
  col_stride_mmul_2x2_2x1(&a[(0) + (0) * a_col_stride], a_col_stride, &b[(0) + (0) * b_col_stride], b_col_stride, &c[(0) + (0) * c_col_stride], c_col_stride);
  col_stride_mmul_2x2_2x1(&a[(0) + (0) * a_col_stride], a_col_stride, &b[(0) + (1) * b_col_stride], b_col_stride, &c[(0) + (1) * c_col_stride], c_col_stride);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<col_stride_mmul_2x2_2x2>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %r9d,%ebx
  sub    $0x8,%rsp
  call   <col_stride_mmul_2x2_2x1@plt>
  add    $0x8,%rsp
  movslq %ebx,%rax
  movslq %r12d,%rdx
  lea    (%r15,%rdx,8),%rdx
  mov    %ebx,%r9d
  lea    (%r14,%rax,8),%r8
  pop    %rbx
  mov    %r12d,%ecx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <col_stride_mmul_2x2_2x1@plt>
```

assistant: ```c
void col_stride_mmul_2x2_2x2(const double *a, int a_col_stride, const double *b, int b_col_stride, double *c, int c_col_stride) {
  col_stride_mmul_2x2_2x1(&a[(0) + (0) * a_col_stride], a_col_stride, &b[(0) + (0) * b_col_stride], b_col_stride, &c[(0) + (0) * c_col_stride], c_col_stride);
  col_stride_mmul_2x2_2x1(&a[(0) + (0) * a_col_stride], a_col_stride, &b[(0) + (1) * b_col_stride], b_col_stride, &c[(0) + (1) * c_col_stride], c_col_stride);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<col_stride_mmul_2x2_2x2>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %r9d,%ebx
  sub    $0x8,%rsp
  call   <col_stride_mmul_2x2_2x1@plt>
  add    $0x8,%rsp
  movslq %ebx,%rax
  movslq %r12d,%rdx
  lea    (%r15,%rdx,8),%rdx
  mov    %ebx,%r9d
  lea    (%r14,%rax,8),%r8
  pop    %rbx
  mov    %r12d,%ecx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <col_stride_mmul_2x2_2x1@plt>
```

assistant: ```c
void col_stride_mmul_2x2_2x2(const double *a, int a_col_stride, const double *b, int b_col_stride, double *c, int c_col_stride) {
  col_stride_mmul_2x2_2x1(&a[(0) + (0) * a_col_stride], a_col_stride, &b[(0) + (0) * b_col_stride], b_col_stride, &c[(0) + (0) * c_col_stride], c_col_stride);
  col_stride_mmul_2x2_2x1(&a[(0) + (0) * a_col_stride], a_col_stride, &b[(0) + (1) * b_col_stride], b_col_stride, &c[(0) + (1) * c_col_stride], c_col_stride);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<utime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x48(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <_DIAGASSERT@plt>
  cmpq   $0x0,-0x50(%rbp)
  jne    L14
  movq   $0x0,-0x38(%rbp)
  jmp    L17
L14:
  mov    -0x50(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x50(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x18(%rbp)
  movq   $0x0,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x38(%rbp)
L17:
  mov    -0x38(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <utimes@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

assistant: ```c
int utime(const char *path, const struct utimbuf *times) {
  struct timeval tv[2], *tvp;
  _DIAGASSERT(path != NULL);
  if (times == (struct utimbuf *)NULL)
    tvp = NULL;
  else {
    tv[0].tv_sec = times->actime;
    tv[1].tv_sec = times->modtime;
    tv[0].tv_usec = tv[1].tv_usec = 0;
    tvp = tv;
  }
  return (utimes(path, tvp));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<utime>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <_DIAGASSERT@plt>
  test   %rbx,%rbx
  je     L9
  mov    0x4(%rbx),%eax
  mov    %eax,0x8(%rsp)
  mov    (%rbx),%eax
  mov    %eax,0x18(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,(%rsp)
  mov    %rsp,%rbx
L9:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <utimes@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utime(const char *path, const struct utimbuf *times) {
  struct timeval tv[2], *tvp;
  _DIAGASSERT(path != NULL);
  if (times == (struct utimbuf *)NULL)
    tvp = NULL;
  else {
    tv[0].tv_sec = times->actime;
    tv[1].tv_sec = times->modtime;
    tv[0].tv_usec = tv[1].tv_usec = 0;
    tvp = tv;
  }
  return (utimes(path, tvp));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<utime>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <_DIAGASSERT@plt>
  test   %rbp,%rbp
  je     L10
  mov    0x4(%rbp),%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    %eax,0x8(%rsp)
  mov    0x0(%rbp),%eax
  mov    %rsp,%rbp
  mov    %eax,0x18(%rsp)
L10:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <utimes@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utime(const char *path, const struct utimbuf *times) {
  struct timeval tv[2], *tvp;
  _DIAGASSERT(path != NULL);
  if (times == (struct utimbuf *)NULL)
    tvp = NULL;
  else {
    tv[0].tv_sec = times->actime;
    tv[1].tv_sec = times->modtime;
    tv[0].tv_usec = tv[1].tv_usec = 0;
    tvp = tv;
  }
  return (utimes(path, tvp));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<utime>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <_DIAGASSERT@plt>
  test   %rbp,%rbp
  je     L10
  mov    0x4(%rbp),%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    %eax,0x8(%rsp)
  mov    0x0(%rbp),%eax
  mov    %rsp,%rbp
  mov    %eax,0x18(%rsp)
L10:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <utimes@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utime(const char *path, const struct utimbuf *times) {
  struct timeval tv[2], *tvp;
  _DIAGASSERT(path != NULL);
  if (times == (struct utimbuf *)NULL)
    tvp = NULL;
  else {
    tv[0].tv_sec = times->actime;
    tv[1].tv_sec = times->modtime;
    tv[0].tv_usec = tv[1].tv_usec = 0;
    tvp = tv;
  }
  return (utimes(path, tvp));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L3
L15:
  mov    -0x8(%rbp),%eax
  movd   %eax,%xmm0
  call   <cube@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x2,%edi
  movd   %eax,%xmm0
  call   <pow@plt>
  movq   %xmm0,%rdx
  movss  -0x4(%rbp),%xmm0
  mov    -0x8(%rbp),%eax
  movaps %xmm0,%xmm2
  movq   %rdx,%xmm1
  movd   %eax,%xmm0
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L3:
  movss  -0x8(%rbp),%xmm0
  movss  D25(%rip),%xmm1
  addss  %xmm0,%xmm1
  movss  %xmm1,-0x8(%rbp)
  movss  D10(%rip),%xmm1
  comiss %xmm0,%xmm1
  ja     L15
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D29
data_type: string
name: unpack_data
arguments: data_label: D25
data_type: byte[4]
name: unpack_data
arguments: data_label: D10
data_type: f32

tool: D29: %f %f %f 


tool: {"D25": [0x00,0x00,0x80,0x3f]}

tool: D10: 3

assistant: ```c
int main() {
  float y;
  float x;
  while (x++ < 3.0) {
    y = cube(x);
    printf("%f %f %f \n", x, pow(x, 2), y);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  pxor   %xmm0,%xmm0
  movaps %xmm0,%xmm5
  addss  D43(%rip),%xmm5
  movss  %xmm5,0x8(%rsp)
  lea    D37(%rip),%rbx
  movss  D30(%rip),%xmm1
  comiss %xmm0,%xmm1
  jbe    L19
L14:
  movss  0x8(%rsp),%xmm0
  call   <cube@plt>
  movss  %xmm0,0xc(%rsp)
  mov    $0x2,%edi
  movss  0x8(%rsp),%xmm0
  call   <pow@plt>
  movapd %xmm0,%xmm1
  movss  0xc(%rsp),%xmm2
  movss  0x8(%rsp),%xmm0
  mov    %rbx,%rdi
  call   <printf@plt>
  movss  0x8(%rsp),%xmm3
  movaps %xmm3,%xmm0
  addss  D43(%rip),%xmm3
  movss  %xmm3,0x8(%rsp)
  movss  D30(%rip),%xmm4
  comiss %xmm0,%xmm4
  ja     L14
L19:
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D43
datatype: byte[4]
name: extract_symbolic_data
arguments: data_label: D37
datatype: string
name: extract_symbolic_data
arguments: data_label: D30
datatype: f32

tool: {"D43": [0x00,0x00,0x80,0x3f]}

tool: D37: %f %f %f 


tool: D30: 3

assistant: ```c
int main() {
  float y;
  float x;
  while (x++ < 3.0) {
    y = cube(x);
    printf("%f %f %f \n", x, pow(x, 2), y);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  movss  D38(%rip),%xmm5
  pxor   %xmm0,%xmm0
  movss  D43(%rip),%xmm3
  comiss %xmm0,%xmm5
  addss  %xmm0,%xmm3
  jbe    L0
  push   %rbx
  lea    D31(%rip),%rbx
  sub    $0x10,%rsp
  nopl   (%rax)
L17:
  movaps %xmm3,%xmm0
  movss  %xmm3,0x8(%rsp)
  call   <cube@plt>
  movss  0x8(%rsp),%xmm3
  mov    $0x2,%edi
  movss  %xmm0,0xc(%rsp)
  movaps %xmm3,%xmm0
  call   <pow@plt>
  movss  0x8(%rsp),%xmm3
  movss  0xc(%rsp),%xmm2
  mov    %rbx,%rdi
  movapd %xmm0,%xmm1
  movaps %xmm3,%xmm0
  call   <printf@plt>
  movss  0x8(%rsp),%xmm3
  movss  D38(%rip),%xmm4
  movaps %xmm3,%xmm0
  addss  D43(%rip),%xmm3
  comiss %xmm0,%xmm4
  ja     L17
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L0:
  xor    %eax,%eax
  ret
```

tool_calls: name: parse_data
arguments: tag: D38
datatype: f32
name: parse_data
arguments: tag: D43
datatype: byte[4]
name: parse_data
arguments: tag: D31
datatype: string

tool: D38: 3

tool: {"D43": [0x00,0x00,0x80,0x3f]}

tool: D31: %f %f %f 


assistant: ```c
int main() {
  float y;
  float x;
  while (x++ < 3.0) {
    y = cube(x);
    printf("%f %f %f \n", x, pow(x, 2), y);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  movss  D47(%rip),%xmm5
  pxor   %xmm0,%xmm0
  movss  D37(%rip),%xmm3
  comiss %xmm0,%xmm5
  addss  %xmm0,%xmm3
  jbe    L10
  push   %rbx
  lea    D5(%rip),%rbx
  sub    $0x10,%rsp
  nopl   (%rax)
L17:
  movaps %xmm3,%xmm0
  movss  %xmm3,0x8(%rsp)
  call   <cube@plt>
  movss  0x8(%rsp),%xmm3
  mov    $0x2,%edi
  movss  %xmm0,0xc(%rsp)
  movaps %xmm3,%xmm0
  call   <pow@plt>
  movss  0x8(%rsp),%xmm3
  movss  0xc(%rsp),%xmm2
  mov    %rbx,%rdi
  movapd %xmm0,%xmm1
  movaps %xmm3,%xmm0
  call   <printf@plt>
  movss  0x8(%rsp),%xmm3
  movss  D47(%rip),%xmm4
  movaps %xmm3,%xmm0
  addss  D37(%rip),%xmm3
  comiss %xmm0,%xmm4
  ja     L17
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L10:
  xor    %eax,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D47
data_type: f32
name: disassemble_data
arguments: data_tag: D37
data_type: byte[4]
name: disassemble_data
arguments: data_tag: D5
data_type: string

tool: D47: 3

tool: {"D37": [0x00,0x00,0x80,0x3f]}

tool: D5: %f %f %f 


assistant: ```c
int main() {
  float y;
  float x;
  while (x++ < 3.0) {
    y = cube(x);
    printf("%f %f %f \n", x, pow(x, 2), y);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rcx
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jle    L23
  mov    -0x14(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jge    L23
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L23:
  mov    -0x14(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jle    L77
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jge    L77
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L77:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jle    L12
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jge    L12
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L12:
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jle    L75
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jge    L75
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L75:
  mov    -0xc(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jle    L65
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jge    L65
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L65:
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jle    L27
  mov    -0xc(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jge    L27
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L27:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D21
data_type: string
name: parse_data
arguments: data_label: D74
data_type: string
name: parse_data
arguments: data_label: D41
data_type: string

tool: D21: Pleaes enter 3 numbers separated by spaces > 

tool: D74: %d%d%d

tool: D41: %d is the median


assistant: ```c
int main(void) {
  int A, B, C;
  printf("Pleaes enter 3 numbers separated by spaces > ");
  scanf("%d%d%d", &A, &B, &C);
  if (A > B && A < C)
    printf("%d is the median\n", A);
  if (A > C && A < B)
    printf("%d is the median\n", A);
  if (B > C && B < A)
    printf("%d is the median\n", B);
  if (B > A && B < C)
    printf("%d is the median\n", B);
  if (C > A && C < B)
    printf("%d is the median\n", C);
  if (C > B && C < A)
    printf("%d is the median\n", C);
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D67(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D57(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    0x10(%rsp),%esi
  jle    L93
  cmp    0x14(%rsp),%esi
  jl     L162
  mov    0x14(%rsp),%eax
  cmp    %eax,0xc(%rsp)
  jg     L30
L152:
  mov    0x10(%rsp),%esi
  cmp    0x14(%rsp),%esi
  jle    L171
  cmp    0xc(%rsp),%esi
  jl     L2
L171:
  mov    0x10(%rsp),%eax
  mov    0xc(%rsp),%edx
  cmp    %edx,%eax
  jle    L104
  mov    0x14(%rsp),%esi
  cmp    %esi,%eax
  jl     L128
  cmp    %esi,%edx
  jl     L7
  jmp    L68
L162:
  lea    D72(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L93:
  mov    0xc(%rsp),%edx
  mov    0x14(%rsp),%eax
  cmp    %eax,%edx
  jle    L152
  mov    0x10(%rsp),%esi
  cmp    %esi,%edx
  jl     L144
L163:
  cmp    %esi,%eax
  jge    L123
  cmp    %esi,0xc(%rsp)
  jle    L104
L2:
  lea    D72(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L171
L144:
  mov    %edx,%esi
  lea    D72(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L152
L128:
  mov    %eax,%esi
  lea    D72(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L104:
  mov    0x14(%rsp),%esi
  cmp    0xc(%rsp),%esi
  jle    L123
L7:
  cmp    %esi,0x10(%rsp)
  jg     L166
L68:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L166:
  lea    D72(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L123:
  mov    0x14(%rsp),%esi
  cmp    0x10(%rsp),%esi
  jle    L68
  cmp    0xc(%rsp),%esi
  jge    L68
  lea    D72(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L68
L30:
  mov    0x10(%rsp),%esi
  jmp    L163
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D67
type_infered: string
name: extract_symbolic_data
arguments: tag: D57
type_infered: string
name: extract_symbolic_data
arguments: tag: D72
type_infered: string

tool: D67: Pleaes enter 3 numbers separated by spaces > 

tool: D57: %d%d%d

tool: D72: %d is the median


assistant: ```c
int main(void) {
  int A, B, C;
  printf("Pleaes enter 3 numbers separated by spaces > ");
  scanf("%d%d%d", &A, &B, &C);
  if (A > B && A < C)
    printf("%d is the median\n", A);
  if (A > C && A < B)
    printf("%d is the median\n", A);
  if (B > C && B < A)
    printf("%d is the median\n", B);
  if (B > A && B < C)
    printf("%d is the median\n", B);
  if (C > A && C < B)
    printf("%d is the median\n", C);
  if (C > B && C < A)
    printf("%d is the median\n", C);
  return (0);
}
```