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

```asm
<load>:
  endbr64
  push   %r15
  push   %r14
  lea    D15(%rip),%r14
  push   %r13
  mov    %r14,%rsi
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x24(%rsp),%rdx
  call   <fscanf@plt>
  mov    0x24(%rsp),%ecx
  test   %ecx,%ecx
  jle    L63
  lea    0x28(%rsp),%rax
  movl   $0x0,0xc(%rsp)
  lea    D2(%rip),%r13
  mov    %rax,0x10(%rsp)
  lea    0x2c(%rsp),%rax
  mov    %rax,0x18(%rsp)
  nopl   0x0(%rax,%rax,1)
L16:
  xor    %eax,%eax
  call   <init_DM@plt>
  mov    0x10(%rsp),%rdx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <fscanf@plt>
  mov    0x28(%rsp),%edx
  test   %edx,%edx
  jle    L5
  xor    %r15d,%r15d
  lea    0x34(%rsp),%r12
  lea    0x30(%rsp),%rbp
  xchg   %ax,%ax
L7:
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <fscanf@plt>
  mov    0x34(%rsp),%eax
  lea    -0x1(%rax),%esi
  mov    0x30(%rsp),%eax
  lea    -0x1(%rax),%edi
  call   <portal_D@plt>
  cmp    %r15d,0x28(%rsp)
  jg     L7
L5:
  mov    0x18(%rsp),%rdx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <fscanf@plt>
  mov    0x2c(%rsp),%eax
  test   %eax,%eax
  jle    L53
  xor    %r15d,%r15d
  lea    0x34(%rsp),%r12
  lea    0x30(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <fscanf@plt>
  mov    0x34(%rsp),%eax
  lea    -0x1(%rax),%esi
  mov    0x30(%rsp),%eax
  lea    -0x1(%rax),%edi
  call   <portal_D@plt>
  cmp    %r15d,0x2c(%rsp)
  jg     L2
L53:
  xor    %eax,%eax
  call   <connect_D@plt>
  xor    %eax,%eax
  call   <solve@plt>
  mov    %r14,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%eax
  cmp    %eax,0x24(%rsp)
  jg     L16
L63:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D15
datatype: string
name: retrieve_data
arguments: label: D2
datatype: string

tool: D15: %d


tool: D2: %d %d


assistant: ```c
void load(FILE *in) {
  int T, NL, NS;
  fscanf(in, "%d\n", &T);
  for (int i = 0; i < T; i++) {
    init_DM();
    fscanf(in, "%d\n", &NL);
    for (int l = 0; l < NL; l++) {
      int a, b;
      fscanf(in, "%d %d\n", &a, &b);
      portal_D(a - 1, b - 1);
    }
    fscanf(in, "%d\n", &NS);
    for (int s = 0; s < NS; s++) {
      int a, b;
      fscanf(in, "%d %d\n", &a, &b);
      portal_D(a - 1, b - 1);
    }
    connect_D();
    printf("%d\n", solve());
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<fg_detach_client>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <pr_info@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <i2c_detach_client@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
data_type: string
name: disassemble_data
arguments: data_tag: D13
data_type: string

tool: D2: fg_detach_client

tool: D13: %s


assistant: ```c
int fg_detach_client(struct i2c_client *client) {
  pr_info("%s\n", __func__);
  i2c_detach_client(client);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<fg_detach_client>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D12(%rip),%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <pr_info@plt>
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <i2c_detach_client@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D12
type_infered: string
name: extract_symbolic_data
arguments: tag: D1
type_infered: string

tool: D12: fg_detach_client

tool: D1: %s


assistant: ```c
int fg_detach_client(struct i2c_client *client) {
  pr_info("%s\n", __func__);
  i2c_detach_client(client);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fg_detach_client>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rsi
  mov    %rdi,%rbp
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <pr_info@plt>
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <i2c_detach_client@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type_guessed: string
name: retrieve_data
arguments: data_tag: D11
type_guessed: string

tool: D4: fg_detach_client

tool: D11: %s


assistant: ```c
int fg_detach_client(struct i2c_client *client) {
  pr_info("%s\n", __func__);
  i2c_detach_client(client);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fg_detach_client>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rsi
  mov    %rdi,%rbp
  xor    %eax,%eax
  lea    D12(%rip),%rdi
  call   <pr_info@plt>
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <i2c_detach_client@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D18
type: string
name: unpack_data
arguments: tag: D12
type: string

tool: D18: fg_detach_client

tool: D12: %s


assistant: ```c
int fg_detach_client(struct i2c_client *client) {
  pr_info("%s\n", __func__);
  i2c_detach_client(client);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cbrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  fld1
  fstpt  -0x20(%rbp)
  fldz
  fstpt  -0x10(%rbp)
  fldz
  fldt   0x10(%rbp)
  fcomip %st(1),%st
  fstp   %st(0)
  jb     L7
  jmp    L2
L27:
  fstp   %st(0)
L51:
  fldt   -0x10(%rbp)
  fld    %st(0)
  fmulp  %st,%st(1)
  fldt   -0x10(%rbp)
  fmulp  %st,%st(1)
  fldt   0x10(%rbp)
  fxch   %st(1)
  fcomip %st(1),%st
  fstp   %st(0)
  jbe    L39
  fldt   -0x10(%rbp)
  fldt   -0x20(%rbp)
  fsubrp %st,%st(1)
  fstpt  -0x10(%rbp)
  fldt   -0x20(%rbp)
  fldt   0xebb(%rip)        # 2000 <_fini+0xe88>
  fdivrp %st,%st(1)
  fstpt  -0x20(%rbp)
  jmp    L2
L39:
  fldt   -0x10(%rbp)
  fldt   -0x20(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x10(%rbp)
L2:
  fldt   -0x10(%rbp)
  fld    %st(0)
  fmulp  %st,%st(1)
  fldt   -0x10(%rbp)
  fmulp  %st,%st(1)
  fldt   0x10(%rbp)
  fucomip %st(1),%st
  jp     L27
  fldt   0x10(%rbp)
  fucomip %st(1),%st
  fstp   %st(0)
  jne    L51
L7:
  fldt   -0x10(%rbp)
  pop    %rbp
  ret
```

assistant: ```c
long double cbrt(long double x) {
  long double mod = 1, n = 0;
  if (x >= 0) {
    while (n * n * n != x) {
      if (n * n * n > x) {
        n -= mod;
        mod /= 10;
      } else {
        n += mod;
      }
    }
  }
  return n;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cbrt>:
  endbr64
  fldz
  fldt   0x8(%rsp)
  fcomi  %st(1),%st
  jb     L79
  fucomi %st(1),%st
  fstp   %st(1)
  jp     L86
  je     L45
L86:
  fldz
  fld    %st(0)
  fld1
  flds   0xee5(%rip)        # 2000 <_fini+0xeb0>
  fxch   %st(3)
  jmp    L38
L93:
  fadd   %st,%st(1)
L30:
  fld    %st(1)
  fmul   %st(2),%st
  fmul   %st(2),%st
  fucomi %st(4),%st
  jp     L38
  je     L39
L38:
  fcomip %st(4),%st
  jbe    L93
  fsubr  %st,%st(1)
  fdiv   %st(2),%st
  jmp    L30
L79:
  fstp   %st(0)
  fstp   %st(0)
  fldz
  jmp    L52
L39:
  fstp   %st(0)
  fstp   %st(0)
  fstp   %st(1)
  fstp   %st(1)
L52:
  ret
L45:
  fstp   %st(0)
  fldz
  jmp    L52
```

assistant: ```c
long double cbrt(long double x) {
  long double mod = 1, n = 0;
  if (x >= 0) {
    while (n * n * n != x) {
      if (n * n * n > x) {
        n -= mod;
        mod /= 10;
      } else {
        n += mod;
      }
    }
  }
  return n;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cbrt>:
  endbr64
  fldz
  fldt   0x8(%rsp)
  fcomi  %st(1),%st
  jb     L36
  fucomi %st(1),%st
  fstp   %st(1)
  jp     L99
  je     L32
L99:
  fldz
  fld    %st(0)
  fld1
  fxch   %st(2)
  jmp    L28
L98:
  fsub   %st(1),%st
  fxch   %st(1)
  fdivs  0xed6(%rip)        # 2000 <_fini+0xe88>
L90:
  fld    %st(1)
  fmul   %st(2),%st
  fmul   %st(2),%st
  fucomi %st(3),%st
  jp     L66
  je     L15
  fxch   %st(1)
  fxch   %st(2)
  fxch   %st(1)
  jmp    L28
L66:
  fxch   %st(1)
  fxch   %st(2)
  fxch   %st(1)
  nopl   0x0(%rax)
L28:
  fcomip %st(3),%st
  ja     L98
  fadd   %st(1),%st
  fxch   %st(1)
  jmp    L90
  nopw   0x0(%rax,%rax,1)
L15:
  fstp   %st(0)
  fstp   %st(0)
  fstp   %st(1)
  ret
  nop
L36:
  fstp   %st(0)
  fstp   %st(0)
  jmp    L94
  cs nopw 0x0(%rax,%rax,1)
L32:
  fstp   %st(0)
L94:
  fldz
  ret
```

assistant: ```c
long double cbrt(long double x) {
  long double mod = 1, n = 0;
  if (x >= 0) {
    while (n * n * n != x) {
      if (n * n * n > x) {
        n -= mod;
        mod /= 10;
      } else {
        n += mod;
      }
    }
  }
  return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cbrt>:
  endbr64
  fldz
  fldt   0x8(%rsp)
  fcomi  %st(1),%st
  jb     L103
  fucomi %st(1),%st
  fstp   %st(1)
  jp     L47
  je     L62
L47:
  fldz
  fld    %st(0)
  fld1
  fxch   %st(2)
  jmp    L40
L53:
  fxch   %st(3)
  fxch   %st(1)
  fxch   %st(2)
  fxch   %st(1)
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L34:
  fxch   %st(3)
  fxch   %st(1)
  fxch   %st(2)
  fxch   %st(1)
L40:
  fcomip %st(3),%st
  jbe    L28
  fsub   %st(1),%st
  fxch   %st(1)
  fdivs  0xeba(%rip)        # 2000 <_fini+0xe78>
  fld    %st(1)
  fmul   %st(2),%st
  fmul   %st(2),%st
  fxch   %st(3)
  fucomi %st(3),%st
  jp     L53
  jne    L34
  fstp   %st(0)
  fstp   %st(0)
  fstp   %st(1)
  ret
  nopl   0x0(%rax,%rax,1)
L28:
  fadd   %st(1),%st
  fld    %st(0)
  fmul   %st(1),%st
  fmul   %st(1),%st
  fucomi %st(3),%st
  jp     L40
  jne    L40
  fstp   %st(0)
  fstp   %st(1)
  fstp   %st(1)
  ret
  nopl   (%rax)
L103:
  fstp   %st(0)
  fstp   %st(0)
  jmp    L33
  xchg   %ax,%ax
L62:
  fstp   %st(0)
L33:
  fldz
  ret
```

assistant: ```c
long double cbrt(long double x) {
  long double mod = 1, n = 0;
  if (x >= 0) {
    while (n * n * n != x) {
      if (n * n * n > x) {
        n -= mod;
        mod /= 10;
      } else {
        n += mod;
      }
    }
  }
  return n;
}
```
user: What is the c source code of the assembly code below?

```asm
<hdlc_get_len>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  shl    $0xb,%eax
  and    $0x1800,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint16_t hdlc_get_len(const uint8_t *buf) {
  uint16_t len = (buf[0] & (0x03)) << 11;
  return (len + buf[1]);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<hdlc_get_len>:
  endbr64
  mov    (%rdi),%eax
  shl    $0xb,%eax
  and    $0x1800,%eax
  add    0x4(%rdi),%eax
  ret
```

assistant: ```c
uint16_t hdlc_get_len(const uint8_t *buf) {
  uint16_t len = (buf[0] & (0x03)) << 11;
  return (len + buf[1]);
}
```
user: Translate the below assembly code into C programming language:

```asm
<hdlc_get_len>:
  endbr64
  mov    (%rdi),%eax
  shl    $0xb,%eax
  and    $0x1800,%eax
  add    0x4(%rdi),%eax
  ret
```

assistant: ```c
uint16_t hdlc_get_len(const uint8_t *buf) {
  uint16_t len = (buf[0] & (0x03)) << 11;
  return (len + buf[1]);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<hdlc_get_len>:
  endbr64
  mov    (%rdi),%eax
  shl    $0xb,%eax
  and    $0x1800,%eax
  add    0x4(%rdi),%eax
  ret
```

assistant: ```c
uint16_t hdlc_get_len(const uint8_t *buf) {
  uint16_t len = (buf[0] & (0x03)) << 11;
  return (len + buf[1]);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xffffffff,-0x8(%rbp)
  movl   $0x80000000,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
data_type: string
name: retrieve_data
arguments: data_tag: D3
data_type: string

tool: D6: x= %u = %d


tool: D3: u = %u = %d 


assistant: ```c
int main() {
  int x = -1;
  unsigned u = 2147483648;
  printf("x= %u = %d\n", x, x);
  printf("u = %u = %d \n", u, u);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%edx
  mov    $0xffffffff,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    $0x80000000,%edx
  mov    $0x80000000,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D10
type_infered: string
name: retrieve_data
arguments: label: D18
type_infered: string

tool: D10: x= %u = %d


tool: D18: u = %u = %d 


assistant: ```c
int main() {
  int x = -1;
  unsigned u = 2147483648;
  printf("x= %u = %d\n", x, x);
  printf("u = %u = %d \n", u, u);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%edx
  mov    $0xffffffff,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x80000000,%edx
  mov    $0x80000000,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
data_type: string
name: disassemble_data
arguments: label: D3
data_type: string

tool: D5: x= %u = %d


tool: D3: u = %u = %d 


assistant: ```c
int main() {
  int x = -1;
  unsigned u = 2147483648;
  printf("x= %u = %d\n", x, x);
  printf("u = %u = %d \n", u, u);
  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
  sub    $0x8,%rsp
  mov    $0xffffffff,%edx
  mov    $0xffffffff,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x80000000,%edx
  mov    $0x80000000,%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D16
type_infered: string
name: parse_data
arguments: label: D15
type_infered: string

tool: D16: x= %u = %d


tool: D15: u = %u = %d 


assistant: ```c
int main() {
  int x = -1;
  unsigned u = 2147483648;
  printf("x= %u = %d\n", x, x);
  printf("u = %u = %d \n", u, u);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sgn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x4(%rbp)
  pxor   %xmm0,%xmm0
  comiss -0x4(%rbp),%xmm0
  jbe    L28
  mov    $0xffffffff,%eax
  jmp    L2
L28:
  pxor   %xmm0,%xmm0
  ucomiss -0x4(%rbp),%xmm0
  jp     L17
  pxor   %xmm0,%xmm0
  ucomiss -0x4(%rbp),%xmm0
  jne    L17
  mov    $0x0,%eax
  jmp    L2
L17:
  mov    $0x1,%eax
L2:
  pop    %rbp
  ret
```

assistant: ```c
int sgn(float n) {
  if (n < 0)
    return -1;
  if (n == 0)
    return 0;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sgn>:
  endbr64
  mov    $0xffffffff,%eax
  pxor   %xmm1,%xmm1
  comiss %xmm0,%xmm1
  ja     L6
  ucomiss %xmm1,%xmm0
  setp   %al
  movzbl %al,%eax
  mov    $0x1,%edx
  cmovne %edx,%eax
L6:
  ret
```

assistant: ```c
int sgn(float n) {
  if (n < 0)
    return -1;
  if (n == 0)
    return 0;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<sgn>:
  endbr64
  pxor   %xmm1,%xmm1
  mov    $0xffffffff,%eax
  comiss %xmm0,%xmm1
  ja     L5
  xor    %eax,%eax
  ucomiss %xmm1,%xmm0
  mov    $0x1,%edx
  setp   %al
  cmovne %edx,%eax
L5:
  ret
```

assistant: ```c
int sgn(float n) {
  if (n < 0)
    return -1;
  if (n == 0)
    return 0;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sgn>:
  endbr64
  pxor   %xmm1,%xmm1
  mov    $0xffffffff,%eax
  comiss %xmm0,%xmm1
  ja     L5
  xor    %eax,%eax
  ucomiss %xmm1,%xmm0
  mov    $0x1,%edx
  setp   %al
  cmovne %edx,%eax
L5:
  ret
```

assistant: ```c
int sgn(float n) {
  if (n < 0)
    return -1;
  if (n == 0)
    return 0;
  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
<toUpperChar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x60,-0x4(%rbp)
  jbe    L22
  cmpb   $0x7a,-0x4(%rbp)
  ja     L22
  movzbl -0x4(%rbp),%eax
  sub    $0x20,%eax
  jmp    L0
L22:
  movzbl -0x4(%rbp),%eax
L0:
  pop    %rbp
  ret
```

assistant: ```c
unsigned char toUpperChar(unsigned char c) {
  if (c > 96 && c < 123) {
    return (c - 32);
  }
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<toUpperChar>:
  endbr64
  lea    -0x61(%rdi),%edx
  lea    -0x20(%rdi),%eax
  cmp    $0x1a,%dl
  cmovae %edi,%eax
  ret
```

assistant: ```c
unsigned char toUpperChar(unsigned char c) {
  if (c > 96 && c < 123) {
    return (c - 32);
  }
  return c;
}
```
user: Translate the below assembly code into C programming language:

```asm
<toUpperChar>:
  endbr64
  lea    -0x61(%rdi),%edx
  lea    -0x20(%rdi),%eax
  cmp    $0x1a,%dl
  cmovae %edi,%eax
  ret
```

assistant: ```c
unsigned char toUpperChar(unsigned char c) {
  if (c > 96 && c < 123) {
    return (c - 32);
  }
  return c;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<toUpperChar>:
  endbr64
  lea    -0x61(%rdi),%edx
  lea    -0x20(%rdi),%eax
  cmp    $0x1a,%dl
  cmovae %edi,%eax
  ret
```

assistant: ```c
unsigned char toUpperChar(unsigned char c) {
  if (c > 96 && c < 123) {
    return (c - 32);
  }
  return c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<req_handler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%eax
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %al,-0x14(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  je     L27
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  cmp    $0x1,%rax
  jbe    L27
  cmpq   $0x0,-0x28(%rbp)
  jne    L105
L27:
  mov    $0xffffffff,%eax
  jmp    L114
L105:
  cmpl   $0x0,-0x18(%rbp)
  je     L8
  cmpq   $0x0,-0x20(%rbp)
  je     L29
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rbx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x3,%eax
  cltq
  cmp    %rax,%rbx
  jb     L29
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0xff,%eax
  jle    L8
L29:
  mov    $0xffffffff,%eax
  jmp    L114
L8:
  cmpl   $0x0,-0x18(%rbp)
  jne    L73
  mov    -0x28(%rbp),%rax
  movzbl -0x14(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  mov    -0x30(%rbp),%rax
  movq   $0x2,(%rax)
  mov    $0x0,%eax
  jmp    L114
L73:
  mov    -0x28(%rbp),%rax
  movzbl -0x14(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  movb   $0x2,(%rax)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  add    $0x2,%rax
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  lea    0x3(%rax),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x3,%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  cmpb   $0x51,-0x14(%rbp)
  jne    L96
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x30,%al
  jne    L49
  mov    -0x28(%rbp),%rax
  add    $0x3,%rax
  movb   $0x0,(%rax)
  jmp    L96
L49:
  mov    -0x28(%rbp),%rax
  add    $0x3,%rax
  movb   $0x1,(%rax)
L96:
  mov    $0x0,%eax
L114:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int req_handler(char cmd, int arg_cnt, const char *arg, char *req, size_t *req_len) {
  if (NULL == req_len || *req_len < 2 || NULL == req)
    return -1;
  if (arg_cnt != 0 && (NULL == arg || *req_len < 3 + strlen(arg) || strlen(arg) > 255))
    return -1;
  if (arg_cnt == 0) {
    req[0] = cmd;
    req[1] = 0;
    *req_len = 2;
    return 0;
  } else {
    req[0] = cmd;
    req[1] = '\x2';
    req[2] = strlen(arg);
    memcpy(req + 3, arg, strlen(arg));
    *req_len = 3 + strlen(arg);
    if (cmd == 'Q') {
      if (arg[0] == '0') {
        req[3] = 0;
      } else {
        req[3] = 1;
      }
    }
    return 0;
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<req_handler>:
  endbr64
  test   %r8,%r8
  je     L5
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r13d
  mov    %rdx,%r12
  mov    %rcx,%rbx
  mov    %r8,%rbp
  mov    (%r8),%r14
  test   %rcx,%rcx
  je     L136
  cmp    $0x1,%r14
  jbe    L136
  test   %esi,%esi
  je     L65
  test   %rdx,%rdx
  je     L29
  mov    %rdx,%rdi
  call   <strlen@plt>
  add    $0x3,%eax
  cltq
  cmp    %rax,%r14
  jb     L16
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0xff,%eax
  jg     L108
  mov    %r13b,(%rbx)
  movb   $0x2,0x1(%rbx)
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %al,0x2(%rbx)
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    0x3(%rbx),%rdi
  mov    %r12,%rsi
  call   <memcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  add    $0x3,%eax
  cltq
  mov    %rax,0x0(%rbp)
  cmp    $0x51,%r13b
  je     L21
  mov    $0x0,%eax
  jmp    L126
L21:
  cmpb   $0x30,(%r12)
  je     L70
  movb   $0x1,0x3(%rbx)
  mov    $0x0,%eax
  jmp    L126
L70:
  movb   $0x0,0x3(%rbx)
  mov    $0x0,%eax
  jmp    L126
L5:
  mov    $0xffffffff,%eax
  ret
L136:
  mov    $0xffffffff,%eax
  jmp    L126
L29:
  mov    $0xffffffff,%eax
  jmp    L126
L16:
  mov    $0xffffffff,%eax
  jmp    L126
L108:
  mov    $0xffffffff,%eax
  jmp    L126
L65:
  mov    %dil,(%rbx)
  movb   $0x0,0x1(%rcx)
  movq   $0x2,(%r8)
  mov    %esi,%eax
L126:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int req_handler(char cmd, int arg_cnt, const char *arg, char *req, size_t *req_len) {
  if (NULL == req_len || *req_len < 2 || NULL == req)
    return -1;
  if (arg_cnt != 0 && (NULL == arg || *req_len < 3 + strlen(arg) || strlen(arg) > 255))
    return -1;
  if (arg_cnt == 0) {
    req[0] = cmd;
    req[1] = 0;
    *req_len = 2;
    return 0;
  } else {
    req[0] = cmd;
    req[1] = '\x2';
    req[2] = strlen(arg);
    memcpy(req + 3, arg, strlen(arg));
    *req_len = 3 + strlen(arg);
    if (cmd == 'Q') {
      if (arg[0] == '0') {
        req[3] = 0;
      } else {
        req[3] = 1;
      }
    }
    return 0;
  }
  return -1;
}
```