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

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  movdqa D15(%rip),%xmm0
  mov    $0x8,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movaps %xmm0,(%rsp)
  movdqa D3(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  call   <build_bottom_layer@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D15
type_infered: byte[16]
name: unpack_data
arguments: label: D3
type_infered: byte[16]

tool: {"D15": [0x01,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: {"D3": [0x07,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x0b,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, const char *argv[]) {
  int test_array[] = {1, 3, 4, 5, 7, 9, 10, 11};
  build_bottom_layer(test_array, 8);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<gauss_jordan>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%r13
  sub    $0x1,%r13
  je     L114
  mov    %rdi,%rbp
  mov    %rdx,%r12
  lea    0x0(,%rdx,8),%r10
  mov    %rdi,%rbx
  mov    $0x0,%r9d
  mov    $0x0,%r11d
  pxor   %xmm3,%xmm3
  lea    -0x1(%rdx),%r8d
  movapd %xmm3,%xmm2
  jmp    L37
L59:
  movsd  (%rsi),%xmm1
  mulsd  (%rdi,%rax,8),%xmm1
  movsd  (%rdx,%rax,8),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  sub    $0x1,%rax
  cmp    %eax,%ecx
  jle    L59
L11:
  add    $0x1,%r14d
  add    %r10,%rdx
  add    %r10,%rsi
  cmp    %r11d,%r14d
  je     L42
L77:
  movslq %r8d,%rax
  cmp    %ecx,%r8d
  jge    L59
  jmp    L11
L42:
  add    $0x1,%r11d
L99:
  add    $0x1,%r9
  add    $0x8,%rbx
  cmp    %r9,%r13
  je     L114
L37:
  movslq %r11d,%rdi
  imul   %r12,%rdi
  lea    (%rdi,%r9,1),%rax
  lea    0x0(%rbp,%rax,8),%rdx
  ucomisd (%rdx),%xmm3
  jp     L124
  ucomisd (%rdx),%xmm2
  je     L99
L124:
  mov    %r9d,%ecx
  cmp    %r9d,%r8d
  jl     L49
  movslq %r8d,%rax
  lea    0x0(%rbp,%rdi,8),%rsi
L66:
  movsd  (%rsi,%rax,8),%xmm0
  divsd  (%rdx),%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  sub    $0x1,%rax
  cmp    %eax,%ecx
  jle    L66
L49:
  test   %r11d,%r11d
  jle    L42
  lea    0x0(%rbp,%rdi,8),%rdi
  mov    %rbx,%rsi
  mov    %rbp,%rdx
  mov    $0x0,%r14d
  jmp    L77
L114:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void gauss_jordan(double *a, const size_t rows, const size_t cols) {
  int row = 0;
  for (int col = 0; col < cols - 1; ++col) {
    if (a[row * cols + col] != 0) {
      for (int i = cols - 1; i > col - 1; --i) {
        a[row * cols + i] /= a[row * cols + col];
      }
      for (int i = 0; i < row; ++i) {
        for (int j = cols - 1; j > col - 1; --j) {
          a[i * cols + j] -= a[i * cols + col] * a[row * cols + j];
        }
      }
      row++;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gauss_jordan>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1,%r13
  je     L86
  mov    %rdi,%rbp
  mov    %rdx,%r12
  mov    %rdi,%rbx
  xor    %r9d,%r9d
  lea    0x0(,%rdx,8),%r10
  pxor   %xmm2,%xmm2
  lea    -0x1(%rdx),%r8d
  xor    %r11d,%r11d
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L51:
  add    $0x1,%r9
  add    $0x8,%rbx
  cmp    %r13,%r9
  je     L86
L14:
  movslq %r11d,%rsi
  imul   %r12,%rsi
  lea    (%rsi,%r9,1),%rax
  lea    0x0(%rbp,%rax,8),%rdi
  movsd  (%rdi),%xmm1
  ucomisd %xmm2,%xmm1
  jp     L30
  je     L51
L30:
  mov    %r9d,%ecx
  cmp    %r9d,%r8d
  jl     L15
  movslq %r8d,%rax
  lea    0x0(%rbp,%rsi,8),%rdx
  jmp    L107
  nop
L79:
  movsd  (%rdi),%xmm1
L107:
  movsd  (%rdx,%rax,8),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  sub    $0x1,%rax
  cmp    %eax,%ecx
  jle    L79
L15:
  test   %r11d,%r11d
  je     L104
  lea    0x0(%rbp,%rsi,8),%rdi
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  xor    %r14d,%r14d
  nopl   (%rax)
L62:
  movslq %r8d,%rax
  cmp    %ecx,%r8d
  jl     L23
  nopl   0x0(%rax,%rax,1)
L92:
  movsd  (%rsi),%xmm1
  mulsd  (%rdi,%rax,8),%xmm1
  movsd  (%rdx,%rax,8),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  sub    $0x1,%rax
  cmp    %eax,%ecx
  jle    L92
L23:
  add    $0x1,%r14d
  add    %r10,%rdx
  add    %r10,%rsi
  cmp    %r11d,%r14d
  jne    L62
L104:
  add    $0x1,%r9
  add    $0x1,%r11d
  add    $0x8,%rbx
  cmp    %r13,%r9
  jne    L14
L86:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void gauss_jordan(double *a, const size_t rows, const size_t cols) {
  int row = 0;
  for (int col = 0; col < cols - 1; ++col) {
    if (a[row * cols + col] != 0) {
      for (int i = cols - 1; i > col - 1; --i) {
        a[row * cols + i] /= a[row * cols + col];
      }
      for (int i = 0; i < row; ++i) {
        for (int j = cols - 1; j > col - 1; --j) {
          a[i * cols + j] -= a[i * cols + col] * a[row * cols + j];
        }
      }
      row++;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rop3_227_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x20(%rbp),%al
  mov    %al,-0x3(%rbp)
  movzbl -0x1c(%rbp),%eax
  or     -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x20(%rbp),%eax
  xor    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  notb   -0x1(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_227_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = *D & T;
  stk2 = S | stk3;
  stk1 = T ^ stk2;
  stk1 = ~stk1;
  *D = stk1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rop3_227_0_0>:
  endbr64
  mov    %edx,%eax
  and    (%rdi),%al
  or     %esi,%eax
  xor    %edx,%eax
  not    %eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_227_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = *D & T;
  stk2 = S | stk3;
  stk1 = T ^ stk2;
  stk1 = ~stk1;
  *D = stk1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rop3_227_0_0>:
  endbr64
  movzbl (%rdi),%eax
  and    %edx,%eax
  or     %esi,%eax
  xor    %edx,%eax
  not    %eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_227_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = *D & T;
  stk2 = S | stk3;
  stk1 = T ^ stk2;
  stk1 = ~stk1;
  *D = stk1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rop3_227_0_0>:
  endbr64
  movzbl (%rdi),%eax
  and    %edx,%eax
  or     %esi,%eax
  xor    %edx,%eax
  not    %eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_227_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = *D & T;
  stk2 = S | stk3;
  stk1 = T ^ stk2;
  stk1 = ~stk1;
  *D = stk1;
}
```
user: What is the c source code of the assembly code below:

```asm
<getTimeStamp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffffc22e4507,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x10,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x15180,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x3e8,%rax,%rcx
  mov    -0x18(%rbp),%rax
  shr    $0x3,%rax
  movabs $0x20c49ba5e353f7cf,%rdx
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x4,%rax
  add    %rcx,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
unsigned long getTimeStamp() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (unsigned long)(tv.tv_sec % 86400) * 1000 + (unsigned long)tv.tv_usec / 1000;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getTimeStamp>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    (%rsp),%eax
  movslq %eax,%rdx
  imul   $0xffffffffc22e4507,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x10,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x15180,%edx,%edx
  sub    %edx,%eax
  movslq %eax,%rcx
  imul   $0x3e8,%rcx,%rcx
  mov    0x8(%rsp),%rdx
  shr    $0x3,%rdx
  movabs $0x20c49ba5e353f7cf,%rsi
  mov    %rdx,%rax
  mul    %rsi
  shr    $0x4,%rdx
  lea    (%rcx,%rdx,1),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long getTimeStamp() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (unsigned long)(tv.tv_sec % 86400) * 1000 + (unsigned long)tv.tv_usec / 1000;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getTimeStamp>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  movslq (%rsp),%rdx
  movabs $0x20c49ba5e353f7cf,%rsi
  mov    %rdx,%rax
  imul   $0xffffffffc22e4507,%rdx,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x10,%edx
  sub    %ecx,%edx
  imul   $0x15180,%edx,%edx
  sub    %edx,%eax
  mov    0x8(%rsp),%rdx
  movslq %eax,%rcx
  shr    $0x3,%rdx
  imul   $0x3e8,%rcx,%rcx
  mov    %rdx,%rax
  mul    %rsi
  shr    $0x4,%rdx
  lea    (%rcx,%rdx,1),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long getTimeStamp() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (unsigned long)(tv.tv_sec % 86400) * 1000 + (unsigned long)tv.tv_usec / 1000;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<getTimeStamp>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  movslq (%rsp),%rdx
  movabs $0x20c49ba5e353f7cf,%rsi
  mov    %rdx,%rax
  imul   $0xffffffffc22e4507,%rdx,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x10,%edx
  sub    %ecx,%edx
  imul   $0x15180,%edx,%edx
  sub    %edx,%eax
  mov    0x8(%rsp),%rdx
  movslq %eax,%rcx
  shr    $0x3,%rdx
  imul   $0x3e8,%rcx,%rcx
  mov    %rdx,%rax
  mul    %rsi
  shr    $0x4,%rdx
  lea    (%rcx,%rdx,1),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x28,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long getTimeStamp() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (unsigned long)(tv.tv_sec % 86400) * 1000 + (unsigned long)tv.tv_usec / 1000;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<signal_range>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  je     L6
  cmpl   $0x2,-0x4(%rbp)
  jne    L0
L6:
  mov    $0x0,%eax
  jmp    L24
L0:
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <glb_error@plt>
  mov    $0xffffffff,%eax
L24:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type: string

tool: D3: Signal flag is neither GLB_SIG nor GLB_BG

assistant: ```c
int signal_range(int signal) {
  if ((signal == 3) || (signal == 2))
    return 0;
  glb_error("Signal flag is neither GLB_SIG nor GLB_BG");
  return -1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<signal_range>:
  endbr64
  sub    $0x2,%edi
  mov    $0x0,%eax
  cmp    $0x1,%edi
  ja     L7
  ret
L7:
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <glb_error@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_guessed: string

tool: D3: Signal flag is neither GLB_SIG nor GLB_BG

assistant: ```c
int signal_range(int signal) {
  if ((signal == 3) || (signal == 2))
    return 0;
  glb_error("Signal flag is neither GLB_SIG nor GLB_BG");
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<signal_range>:
  endbr64
  sub    $0x2,%edi
  cmp    $0x1,%edi
  ja     L4
  xor    %eax,%eax
  ret
  nop
L4:
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <glb_error@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type_infered: string

tool: D9: Signal flag is neither GLB_SIG nor GLB_BG

assistant: ```c
int signal_range(int signal) {
  if ((signal == 3) || (signal == 2))
    return 0;
  glb_error("Signal flag is neither GLB_SIG nor GLB_BG");
  return -1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<signal_range>:
  endbr64
  sub    $0x2,%edi
  cmp    $0x1,%edi
  ja     L0
  xor    %eax,%eax
  ret
  nop
L0:
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <glb_error@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D9
type: string

tool: D9: Signal flag is neither GLB_SIG nor GLB_BG

assistant: ```c
int signal_range(int signal) {
  if ((signal == 3) || (signal == 2))
    return 0;
  glb_error("Signal flag is neither GLB_SIG nor GLB_BG");
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strtoupper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L13:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L13
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strtoupper(char *str) {
  int i = 0;
  while (str[i] != '\0') {
    str[i] = (char)toupper((int)str[i]);
    i++;
  }
  return str;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strtoupper>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L2
  mov    %rbp,%rbx
L16:
  movsbl %dil,%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L16
L2:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strtoupper(char *str) {
  int i = 0;
  while (str[i] != '\0') {
    str[i] = (char)toupper((int)str[i]);
    i++;
  }
  return str;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strtoupper>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L0
  mov    %r12,%rbx
  nopl   0x0(%rax)
L5:
  call   <toupper@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L5
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *strtoupper(char *str) {
  int i = 0;
  while (str[i] != '\0') {
    str[i] = (char)toupper((int)str[i]);
    i++;
  }
  return str;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strtoupper>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L14
  mov    %r12,%rbx
  nopl   0x0(%rax)
L4:
  call   <toupper@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L4
L14:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *strtoupper(char *str) {
  int i = 0;
  while (str[i] != '\0') {
    str[i] = (char)toupper((int)str[i]);
    i++;
  }
  return str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hashEncodeHex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %esi,-0x6c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D7(%rip),%rax
  mov    %rax,-0x58(%rbp)
  mov    -0x6c(%rbp),%eax
  cmp    $0x28,%eax
  ja     L15
  mov    -0x6c(%rbp),%eax
  mov    $0x29,%edx
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <ERROR@plt>
  mov    $0xffffffff,%eax
  jmp    L22
L15:
  movl   $0x0,-0x5c(%rbp)
  jmp    L3
L10:
  mov    -0x5c(%rbp),%edx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  shr    $0x4,%al
  movzbl %al,%edx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  mov    -0x5c(%rbp),%edx
  add    %edx,%edx
  movzbl (%rax),%eax
  mov    %edx,%edx
  mov    %al,-0x50(%rbp,%rdx,1)
  mov    -0x5c(%rbp),%edx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0xf,%eax
  mov    %rax,%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  mov    -0x5c(%rbp),%edx
  add    %edx,%edx
  add    $0x1,%edx
  movzbl (%rax),%eax
  mov    %edx,%edx
  mov    %al,-0x50(%rbp,%rdx,1)
  addl   $0x1,-0x5c(%rbp)
L3:
  cmpl   $0x13,-0x5c(%rbp)
  jbe    L10
  mov    -0x5c(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%eax
  movb   $0x0,-0x50(%rbp,%rax,1)
  mov    -0x68(%rbp),%rax
  mov    -0x50(%rbp),%rcx
  mov    -0x48(%rbp),%rbx
  mov    %rcx,(%rax)
  mov    %rbx,0x8(%rax)
  mov    -0x40(%rbp),%rcx
  mov    -0x38(%rbp),%rbx
  mov    %rcx,0x10(%rax)
  mov    %rbx,0x18(%rax)
  mov    -0x30(%rbp),%rdx
  mov    %rdx,0x20(%rax)
  movzbl -0x28(%rbp),%edx
  mov    %dl,0x28(%rax)
  mov    $0x0,%eax
L22:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
type: string
name: unpack_data
arguments: data_label: D15
type: string

tool: D7: 0123456789abcdef

tool: D15: Could not hexencode hash: Insufficient buffer size (Supplied: %d, Internal: %d)


assistant: ```c
int hashEncodeHex(unsigned char *hash, int buffer_length) {
  unsigned i;
  unsigned char buffer[(20 * 2) + 1];
  unsigned char *hexdict = (unsigned char *)"0123456789abcdef";
  if (buffer_length < (sizeof(buffer) / sizeof(unsigned char))) {
    ERROR(printf("Could not hexencode hash: Insufficient buffer size (Supplied: %d, Internal: %d)\n", buffer_length, (sizeof(buffer) / sizeof(unsigned char))));
    return -1;
  }
  for (i = 0; i < 20; i++) {
    buffer[i * 2] = hexdict[(hash[i]) >> 4];
    buffer[(i * 2) + 1] = hexdict[(hash[i]) & 0xF];
  }
  buffer[i * 2] = 0;
  memcpy(hash, buffer, (20 * 2) + 1);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hashEncodeHex>:
  endbr64
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  cmp    $0x28,%esi
  jbe    L13
  mov    %rdi,%r9
  mov    %rdi,%rsi
  lea    0x1(%rsp),%rax
  lea    0x29(%rsp),%r8
  lea    D14(%rip),%rdi
L39:
  movzbl (%rsi),%edx
  mov    %edx,%ecx
  shr    $0x4,%cl
  movzbl %cl,%ecx
  movzbl (%rdi,%rcx,1),%ecx
  mov    %cl,-0x1(%rax)
  and    $0xf,%edx
  movzbl (%rdi,%rdx,1),%edx
  mov    %dl,(%rax)
  add    $0x1,%rsi
  add    $0x2,%rax
  cmp    %r8,%rax
  jne    L39
  movb   $0x0,0x28(%rsp)
  movdqa (%rsp),%xmm0
  movups %xmm0,(%r9)
  movdqa 0x10(%rsp),%xmm1
  movups %xmm1,0x10(%r9)
  mov    0x20(%rsp),%rax
  mov    %rax,0x20(%r9)
  movb   $0x0,0x28(%r9)
  mov    $0x0,%eax
L37:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L35
  add    $0x48,%rsp
  ret
L13:
  mov    $0x29,%edx
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <ERROR@plt>
  mov    $0xffffffff,%eax
  jmp    L37
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D19
type_guessed: string

tool: D14: 0123456789abcdef

tool: D19: Could not hexencode hash: Insufficient buffer size (Supplied: %d, Internal: %d)


assistant: ```c
int hashEncodeHex(unsigned char *hash, int buffer_length) {
  unsigned i;
  unsigned char buffer[(20 * 2) + 1];
  unsigned char *hexdict = (unsigned char *)"0123456789abcdef";
  if (buffer_length < (sizeof(buffer) / sizeof(unsigned char))) {
    ERROR(printf("Could not hexencode hash: Insufficient buffer size (Supplied: %d, Internal: %d)\n", buffer_length, (sizeof(buffer) / sizeof(unsigned char))));
    return -1;
  }
  for (i = 0; i < 20; i++) {
    buffer[i * 2] = hexdict[(hash[i]) >> 4];
    buffer[(i * 2) + 1] = hexdict[(hash[i]) & 0xF];
  }
  buffer[i * 2] = 0;
  memcpy(hash, buffer, (20 * 2) + 1);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hashEncodeHex>:
  endbr64
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  cmp    $0x28,%esi
  jbe    L21
  mov    %rdi,%r9
  lea    0x1(%rsp),%rax
  lea    0x29(%rsp),%r8
  mov    %rdi,%rsi
  lea    D18(%rip),%rdi
  nopl   0x0(%rax,%rax,1)
L30:
  movzbl (%rsi),%edx
  add    $0x2,%rax
  add    $0x1,%rsi
  mov    %edx,%ecx
  and    $0xf,%edx
  shr    $0x4,%cl
  movzbl (%rdi,%rdx,1),%edx
  movzbl %cl,%ecx
  movzbl (%rdi,%rcx,1),%ecx
  mov    %dl,-0x2(%rax)
  mov    %cl,-0x3(%rax)
  cmp    %rax,%r8
  jne    L30
  mov    0x20(%rsp),%rax
  movdqa (%rsp),%xmm0
  movb   $0x0,0x28(%rsp)
  movdqa 0x10(%rsp),%xmm1
  movb   $0x0,0x28(%r9)
  mov    %rax,0x20(%r9)
  xor    %eax,%eax
  movups %xmm0,(%r9)
  movups %xmm1,0x10(%r9)
L29:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x48,%rsp
  ret
L21:
  mov    $0x29,%edx
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    %eax,%edi
  xor    %eax,%eax
  call   <ERROR@plt>
  mov    $0xffffffff,%eax
  jmp    L29
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
data_type: string
name: extract_symbolic_data
arguments: tag: D2
data_type: string

tool: D18: 0123456789abcdef

tool: D2: Could not hexencode hash: Insufficient buffer size (Supplied: %d, Internal: %d)


assistant: ```c
int hashEncodeHex(unsigned char *hash, int buffer_length) {
  unsigned i;
  unsigned char buffer[(20 * 2) + 1];
  unsigned char *hexdict = (unsigned char *)"0123456789abcdef";
  if (buffer_length < (sizeof(buffer) / sizeof(unsigned char))) {
    ERROR(printf("Could not hexencode hash: Insufficient buffer size (Supplied: %d, Internal: %d)\n", buffer_length, (sizeof(buffer) / sizeof(unsigned char))));
    return -1;
  }
  for (i = 0; i < 20; i++) {
    buffer[i * 2] = hexdict[(hash[i]) >> 4];
    buffer[(i * 2) + 1] = hexdict[(hash[i]) & 0xF];
  }
  buffer[i * 2] = 0;
  memcpy(hash, buffer, (20 * 2) + 1);
  return 0;
}
```