ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<addps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  cmpl   $0x0,-0x14(%rbp)
  je     L19
  mov    -0x10(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  jmp    L0
L19:
  mov    -0x10(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
L0:
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D8
type: string
name: retrieve_data
arguments: label: D7
type: string

tool: D8: .eps

tool: D7: .ps

assistant: ```c
void addps(char *instring, char *outstring, int eps) {
  strcpy(outstring, instring);
  if (eps) {
    strcat(outstring, ".eps");
  } else {
    strcat(outstring, ".ps");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<addps>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  mov    %edx,%ebp
  mov    %rdi,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  test   %ebp,%ebp
  je     L2
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
L15:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  lea    D13(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  jmp    L15
```

tool_calls: name: retrieve_data
arguments: tag: D0
type: string
name: retrieve_data
arguments: tag: D13
type: string

tool: D0: .eps

tool: D13: .ps

assistant: ```c
void addps(char *instring, char *outstring, int eps) {
  strcpy(outstring, instring);
  if (eps) {
    strcat(outstring, ".eps");
  } else {
    strcat(outstring, ".ps");
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<addps>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdi,%rsi
  push   %rbx
  mov    %rbp,%rdi
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <strcpy@plt>
  lea    D6(%rip),%rsi
  test   %ebx,%ebx
  jne    L0
  lea    D4(%rip),%rsi
L0:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1070 <strcat@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D6
data_type: string
name: parse_data
arguments: data_tag: D4
data_type: string

tool: D6: .eps

tool: D4: .ps

assistant: ```c
void addps(char *instring, char *outstring, int eps) {
  strcpy(outstring, instring);
  if (eps) {
    strcat(outstring, ".eps");
  } else {
    strcat(outstring, ".ps");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<addps>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdi,%rsi
  push   %rbx
  mov    %rbp,%rdi
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <strcpy@plt>
  lea    D2(%rip),%rsi
  test   %ebx,%ebx
  jne    L3
  lea    D18(%rip),%rsi
L3:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1070 <strcat@plt>
```

tool_calls: name: unpack_data
arguments: label: D2
type_infered: string
name: unpack_data
arguments: label: D18
type_infered: string

tool: D2: .eps

tool: D18: .ps

assistant: ```c
void addps(char *instring, char *outstring, int eps) {
  strcpy(outstring, instring);
  if (eps) {
    strcat(outstring, ".eps");
  } else {
    strcat(outstring, ".ps");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Clock_Clicks>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <clock@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double Clock_Clicks(void) {
  double ans;
  ans = (double)clock();
  return ans;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Clock_Clicks>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <clock@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  add    $0x8,%rsp
  ret
```

assistant: ```c
double Clock_Clicks(void) {
  double ans;
  ans = (double)clock();
  return ans;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Clock_Clicks>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clock@plt>
  pxor   %xmm0,%xmm0
  add    $0x8,%rsp
  cvtsi2sd %rax,%xmm0
  ret
```

assistant: ```c
double Clock_Clicks(void) {
  double ans;
  ans = (double)clock();
  return ans;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Clock_Clicks>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clock@plt>
  pxor   %xmm0,%xmm0
  add    $0x8,%rsp
  cvtsi2sd %rax,%xmm0
  ret
```

assistant: ```c
double Clock_Clicks(void) {
  double ans;
  ans = (double)clock();
  return ans;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  D13(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D13
type: f64
name: disassemble_data
arguments: data_tag: D11
type: string

tool: D13: 4.4

tool: D11: %f


assistant: ```c
int main(void) {
  double d;
  d = 4.4;
  printf("%f\n", d);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movsd  D0(%rip),%xmm0
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D0
type: f64
name: retrieve_data
arguments: label: D5
type: string

tool: D0: 4.4

tool: D5: %f


assistant: ```c
int main(void) {
  double d;
  d = 4.4;
  printf("%f\n", d);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movsd  D14(%rip),%xmm0
  lea    D16(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D14
type_infered: f64
name: extract_symbolic_data
arguments: data_label: D16
type_infered: string

tool: D14: 4.4

tool: D16: %f


assistant: ```c
int main(void) {
  double d;
  d = 4.4;
  printf("%f\n", d);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movsd  D1(%rip),%xmm0
  lea    D18(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D1
type: f64
name: disassemble_data
arguments: label: D18
type: string

tool: D1: 4.4

tool: D18: %f


assistant: ```c
int main(void) {
  double d;
  d = 4.4;
  printf("%f\n", d);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void Swap(uint64_t *x, uint64_t *y) {
  uint64_t t = *x;
  *x = *y;
  *y = t;
}
```
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>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void Swap(uint64_t *x, uint64_t *y) {
  uint64_t t = *x;
  *x = *y;
  *y = t;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void Swap(uint64_t *x, uint64_t *y) {
  uint64_t t = *x;
  *x = *y;
  *y = t;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void Swap(uint64_t *x, uint64_t *y) {
  uint64_t t = *x;
  *x = *y;
  *y = t;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L44
  movl   $0x1,-0x8(%rbp)
  jmp    L28
L24:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L64
  mov    $0x1,%edi
  call   <hash@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
L64:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  cmp    %eax,-0x8(%rbp)
  jg     L26
  mov    $0x1,%edi
  call   <hash@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
L26:
  movl   $0x1,-0x4(%rbp)
  jmp    L86
L25:
  mov    $0x2,%edi
  call   <hash@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  addl   $0x1,-0x4(%rbp)
L86:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L25
  addl   $0x1,-0x8(%rbp)
L28:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L24
L44:
  cmpl   $0x0,-0x14(%rbp)
  jg     L74
  movl   $0x1,-0x8(%rbp)
  jmp    L51
L34:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  addl   $0x1,-0x8(%rbp)
L51:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L34
L74:
  nop
  leave
  ret
```

assistant: ```c
void test(int m, int n) {
  int i, j;
  if (m >= 1) {
    for (i = 1; i <= n; i++) {
      if (i >= m) {
        {
          hash(1);
          hash(i);
        };
      }
      if (i <= ((2 * m) < (m - 1) ? (2 * m) : (m - 1))) {
        {
          hash(1);
          hash(i);
        };
      }
      for (j = 1; j <= m; j++) {
        {
          hash(2);
          hash(i);
          hash(j);
        };
      }
    }
  }
  if (m <= 0) {
    for (i = 1; i <= n; i++) {
      {
        hash(1);
        hash(i);
      };
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r13d
  test   %edi,%edi
  jle    L71
  mov    %edi,%r12d
  test   %esi,%esi
  jle    L28
  lea    -0x1(%rdi),%r14d
  lea    (%rdi,%rdi,1),%eax
  cmp    %eax,%r14d
  cmovg  %eax,%r14d
  mov    $0x1,%ebp
  jmp    L42
L39:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  jmp    L74
L21:
  mov    $0x1,%ebx
L70:
  mov    $0x2,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  call   <hash@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jge    L70
  add    $0x1,%ebp
  cmp    %ebp,%r13d
  jl     L28
L42:
  cmp    %ebp,%r12d
  jle    L39
L74:
  cmp    %ebp,%r14d
  jl     L21
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  jmp    L21
L71:
  mov    $0x1,%ebx
  test   %esi,%esi
  jle    L28
L78:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  call   <hash@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r13d
  jge    L78
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void test(int m, int n) {
  int i, j;
  if (m >= 1) {
    for (i = 1; i <= n; i++) {
      if (i >= m) {
        {
          hash(1);
          hash(i);
        };
      }
      if (i <= ((2 * m) < (m - 1) ? (2 * m) : (m - 1))) {
        {
          hash(1);
          hash(i);
        };
      }
      for (j = 1; j <= m; j++) {
        {
          hash(2);
          hash(i);
          hash(j);
        };
      }
    }
  }
  if (m <= 0) {
    for (i = 1; i <= n; i++) {
      {
        hash(1);
        hash(i);
      };
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test>:
  endbr64
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L8
  test   %esi,%esi
  jle    L99
  lea    -0x1(%rdi),%r14d
  lea    (%rdi,%rdi,1),%eax
  mov    %edi,%r12d
  mov    $0x1,%ebp
  cmp    %eax,%r14d
  cmovg  %eax,%r14d
  nopl   0x0(%rax)
L52:
  cmp    %ebp,%r12d
  jle    L15
  cmp    %ebp,%r14d
  jge    L26
L0:
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L47:
  mov    $0x2,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <hash@plt>
  cmp    %ebx,%r12d
  jge    L47
  add    $0x1,%ebp
  cmp    %ebp,%r13d
  jge    L52
L99:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L15:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  cmp    %ebp,%r14d
  jl     L0
L26:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  jmp    L0
L8:
  mov    $0x1,%ebx
  test   %esi,%esi
  jle    L99
  nopw   0x0(%rax,%rax,1)
L60:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <hash@plt>
  cmp    %ebx,%r13d
  jge    L60
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void test(int m, int n) {
  int i, j;
  if (m >= 1) {
    for (i = 1; i <= n; i++) {
      if (i >= m) {
        {
          hash(1);
          hash(i);
        };
      }
      if (i <= ((2 * m) < (m - 1) ? (2 * m) : (m - 1))) {
        {
          hash(1);
          hash(i);
        };
      }
      for (j = 1; j <= m; j++) {
        {
          hash(2);
          hash(i);
          hash(j);
        };
      }
    }
  }
  if (m <= 0) {
    for (i = 1; i <= n; i++) {
      {
        hash(1);
        hash(i);
      };
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test>:
  endbr64
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L74
  test   %esi,%esi
  jle    L86
  lea    -0x1(%rdi),%r14d
  lea    (%rdi,%rdi,1),%eax
  mov    %edi,%r12d
  mov    $0x1,%ebp
  cmp    %eax,%r14d
  cmovg  %eax,%r14d
  nopl   0x0(%rax)
L83:
  cmp    %ebp,%r12d
  jle    L5
  cmp    %ebp,%r14d
  jge    L80
L22:
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L44:
  mov    $0x2,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <hash@plt>
  cmp    %ebx,%r12d
  jge    L44
  add    $0x1,%ebp
  cmp    %ebp,%r13d
  jge    L83
L86:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L5:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  cmp    %ebp,%r14d
  jl     L22
L80:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebp,%edi
  call   <hash@plt>
  jmp    L22
L74:
  mov    $0x1,%ebx
  test   %esi,%esi
  jle    L86
  nopw   0x0(%rax,%rax,1)
L55:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <hash@plt>
  cmp    %ebx,%r13d
  jge    L55
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void test(int m, int n) {
  int i, j;
  if (m >= 1) {
    for (i = 1; i <= n; i++) {
      if (i >= m) {
        {
          hash(1);
          hash(i);
        };
      }
      if (i <= ((2 * m) < (m - 1) ? (2 * m) : (m - 1))) {
        {
          hash(1);
          hash(i);
        };
      }
      for (j = 1; j <= m; j++) {
        {
          hash(2);
          hash(i);
          hash(j);
        };
      }
    }
  }
  if (m <= 0) {
    for (i = 1; i <= n; i++) {
      {
        hash(1);
        hash(i);
      };
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<writeRam>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x130,%rsp
  mov    %edi,-0x124(%rbp)
  mov    %esi,-0x128(%rbp)
  mov    %rdx,-0x130(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x16,-0x110(%rbp)
  movb   $0xfd,-0x10f(%rbp)
  mov    -0x124(%rbp),%eax
  mov    %al,-0x10e(%rbp)
  mov    -0x124(%rbp),%eax
  sar    $0x8,%eax
  mov    %al,-0x10d(%rbp)
  mov    -0x124(%rbp),%eax
  sar    $0x10,%eax
  mov    %al,-0x10c(%rbp)
  mov    -0x128(%rbp),%eax
  mov    %al,-0x10b(%rbp)
  mov    -0x128(%rbp),%eax
  shr    $0x8,%eax
  mov    %al,-0x10a(%rbp)
  lea    -0x110(%rbp),%rax
  mov    $0x7,%r8d
  mov    %rax,%rcx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <SMBWrite@plt>
  mov    %eax,-0x114(%rbp)
  cmpl   $0x0,-0x114(%rbp)
  jns    L8
  mov    -0x114(%rbp),%eax
  jmp    L0
L8:
  mov    -0x128(%rbp),%edx
  mov    -0x130(%rbp),%rax
  mov    %edx,%r8d
  mov    %rax,%rcx
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <SMBWrite@plt>
  mov    %eax,-0x114(%rbp)
  mov    -0x114(%rbp),%eax
L0:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
int writeRam(int address, unsigned int size, unsigned char *buf) {
  int status, i;
  unsigned char block[0xFF];
  block[0] = 0x16;
  block[1] = 0xFD;
  block[2] = address & 0xFF;
  block[3] = (address >> 8) & 0xFF;
  block[4] = (address >> 16) & 0xFF;
  block[5] = size & 0xFF;
  block[6] = (size >> 8) & 0xFF;
  status = SMBWrite(1, 0, 0, block, 7);
  if (status < 0)
    return status;
  status = SMBWrite(0, 0, 1, buf, size);
  return status;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<writeRam>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %edi,%eax
  mov    %esi,%ebx
  mov    %rdx,%rbp
  mov    %fs:0x28,%rdx
  mov    %rdx,0x108(%rsp)
  xor    %edx,%edx
  movb   $0x16,(%rsp)
  movb   $0xfd,0x1(%rsp)
  mov    %dil,0x2(%rsp)
  movzbl %ah,%ecx
  mov    %cl,0x3(%rsp)
  sar    $0x10,%eax
  mov    %al,0x4(%rsp)
  mov    %sil,0x5(%rsp)
  movzbl %bh,%eax
  mov    %al,0x6(%rsp)
  mov    %rsp,%rcx
  mov    $0x7,%r8d
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <SMBWrite@plt>
  test   %eax,%eax
  js     L9
  mov    %ebx,%r8d
  mov    %rbp,%rcx
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <SMBWrite@plt>
L9:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int writeRam(int address, unsigned int size, unsigned char *buf) {
  int status, i;
  unsigned char block[0xFF];
  block[0] = 0x16;
  block[1] = 0xFD;
  block[2] = address & 0xFF;
  block[3] = (address >> 8) & 0xFF;
  block[4] = (address >> 16) & 0xFF;
  block[5] = size & 0xFF;
  block[6] = (size >> 8) & 0xFF;
  status = SMBWrite(1, 0, 0, block, 7);
  if (status < 0)
    return status;
  status = SMBWrite(0, 0, 1, buf, size);
  return status;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<writeRam>:
  endbr64
  push   %r12
  mov    $0x7,%r8d
  mov    %rdx,%r12
  xor    %edx,%edx
  push   %rbx
  mov    %esi,%ebx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %di,0x2(%rsp)
  sar    $0x10,%edi
  mov    $0xfffffd16,%eax
  mov    %dil,0x4(%rsp)
  mov    %rsp,%rcx
  mov    $0x1,%edi
  mov    %si,0x5(%rsp)
  xor    %esi,%esi
  mov    %ax,(%rsp)
  call   <SMBWrite@plt>
  test   %eax,%eax
  js     L1
  mov    %ebx,%r8d
  mov    %r12,%rcx
  mov    $0x1,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <SMBWrite@plt>
L1:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x118,%rsp
  pop    %rbx
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int writeRam(int address, unsigned int size, unsigned char *buf) {
  int status, i;
  unsigned char block[0xFF];
  block[0] = 0x16;
  block[1] = 0xFD;
  block[2] = address & 0xFF;
  block[3] = (address >> 8) & 0xFF;
  block[4] = (address >> 16) & 0xFF;
  block[5] = size & 0xFF;
  block[6] = (size >> 8) & 0xFF;
  status = SMBWrite(1, 0, 0, block, 7);
  if (status < 0)
    return status;
  status = SMBWrite(0, 0, 1, buf, size);
  return status;
}
```
user: Translate the below assembly code into C programming language:

```asm
<writeRam>:
  endbr64
  push   %r12
  mov    $0x7,%r8d
  mov    %rdx,%r12
  xor    %edx,%edx
  push   %rbx
  mov    %esi,%ebx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %di,0x2(%rsp)
  sar    $0x10,%edi
  mov    $0xfffffd16,%eax
  mov    %dil,0x4(%rsp)
  mov    %rsp,%rcx
  mov    $0x1,%edi
  mov    %si,0x5(%rsp)
  xor    %esi,%esi
  mov    %ax,(%rsp)
  call   <SMBWrite@plt>
  test   %eax,%eax
  js     L7
  mov    %ebx,%r8d
  mov    %r12,%rcx
  mov    $0x1,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <SMBWrite@plt>
L7:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x118,%rsp
  pop    %rbx
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int writeRam(int address, unsigned int size, unsigned char *buf) {
  int status, i;
  unsigned char block[0xFF];
  block[0] = 0x16;
  block[1] = 0xFD;
  block[2] = address & 0xFF;
  block[3] = (address >> 8) & 0xFF;
  block[4] = (address >> 16) & 0xFF;
  block[5] = size & 0xFF;
  block[6] = (size >> 8) & 0xFF;
  status = SMBWrite(1, 0, 0, block, 7);
  if (status < 0)
    return status;
  status = SMBWrite(0, 0, 1, buf, size);
  return status;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getrange_min_max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L19
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x2,%eax
  jg     L39
L19:
  mov    $0xffffffff,%eax
  jmp    L116
L39:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L54
  mov    $0xffffffff,%eax
  jmp    L116
L54:
  mov    -0x10(%rbp),%rax
  mov    $0x3a,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L85
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L60
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L36
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L129
  mov    -0x30(%rbp),%rax
  mov    -0x18(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x38(%rbp),%rax
  mov    -0x14(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L116
L85:
  nop
  jmp    L121
L60:
  nop
  jmp    L121
L36:
  nop
  jmp    L121
L129:
  nop
L121:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0xffffffff,%eax
L116:
  leave
  ret
```

assistant: ```c
int getrange_min_max(char *buf, u_int32_t *range_min, u_int32_t *range_max) {
  int i;
  char *dup, *ptr;
  u_int32_t min, max;
  if (!buf || strlen(buf) < 3)
    return -1;
  dup = strdup(buf);
  if (!dup)
    return -1;
  ptr = strchr(dup, ':');
  if (!ptr)
    goto error;
  *ptr++ = 0;
  i = atoi(dup);
  if (i < 1)
    goto error;
  min = i;
  i = atoi(ptr);
  if (i < 1)
    goto error;
  max = i;
  if (min >= max)
    goto error;
  *range_min = min;
  *range_max = max;
  free(dup);
  return 0;
error:
  free(dup);
  return -1;
}
```