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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L1
  mov    $0x0,%eax
  jmp    L15
L1:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L15:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D9
data_type: string

tool: D9: %lu


assistant: ```c
int main(int ac, const char *av[]) {
  if (ac < 2)
    return 0;
  printf("%lu\n", strlen(av[1]));
  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
  cmp    $0x1,%edi
  jle    L4
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L4:
  mov    $0x0,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
datatype: string

tool: D0: %lu


assistant: ```c
int main(int ac, const char *av[]) {
  if (ac < 2)
    return 0;
  printf("%lu\n", strlen(av[1]));
  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
  cmp    $0x1,%edi
  jle    L2
  push   %rax
  mov    0x8(%rsi),%rdi
  call   <strlen@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rdx
  ret
L2:
  xor    %eax,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: label: D1
data_type: string

tool: D1: %lu


assistant: ```c
int main(int ac, const char *av[]) {
  if (ac < 2)
    return 0;
  printf("%lu\n", strlen(av[1]));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L7
  push   %rax
  mov    0x8(%rsi),%rdi
  call   <strlen@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rdx
  ret
L7:
  xor    %eax,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
datatype: string

tool: D6: %lu


assistant: ```c
int main(int ac, const char *av[]) {
  if (ac < 2)
    return 0;
  printf("%lu\n", strlen(av[1]));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<gui_nonfatal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L2
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L2:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <produce_message@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_guessed: string

tool: D0: Information

assistant: ```c
void gui_nonfatal(const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  produce_message("Information", fmt, args);
  va_end(args);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<gui_nonfatal>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  lea    D9(%rip),%rdi
  call   <produce_message@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D9
type_infered: string

tool: D9: Information

assistant: ```c
void gui_nonfatal(const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  produce_message("Information", fmt, args);
  va_end(args);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gui_nonfatal>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  lea    D8(%rip),%rdi
  call   <produce_message@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type_guessed: string

tool: D8: Information

assistant: ```c
void gui_nonfatal(const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  produce_message("Information", fmt, args);
  va_end(args);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gui_nonfatal>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  call   <produce_message@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D4
data_type: string

tool: D4: Information

assistant: ```c
void gui_nonfatal(const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  produce_message("Information", fmt, args);
  va_end(args);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    D2(%rip),%eax
  movd   %eax,%xmm0
  call   <bar@plt>
  mov    D8(%rip),%eax
  movd   %eax,%xmm0
  call   <bar@plt>
  push   $0xfffffffffffffff8
  push   $0xfffffffffffffff9
  mov    $0xfffffffa,%r9d
  mov    $0xfffffffb,%r8d
  mov    $0xfffffffc,%ecx
  mov    $0xfffffffd,%edx
  mov    $0xfffffffe,%esi
  mov    $0xffffffff,%edi
  call   <bar8@plt>
  add    $0x10,%rsp
  mov    -0x4(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <bazu@plt>
  mov    $0x80000000,%esi
  mov    $0x8,%edi
  call   <bazi@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D2
data_type: byte[4]
name: retrieve_data
arguments: label: D8
data_type: byte[4]

tool: {"D2": [0x00,0x00,0x80,0x43]}

tool: {"D8": [0xb8,0x1e,0xd5,0x40]}

assistant: ```c
void foo(int i) {
  bar(256);
  bar(6.66f);
  bar8(-1, -2, -3, -4, -5, -6, -7, -8);
  bazu(i, i);
  bazi(0x8, 0x80000000);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  movss  D4(%rip),%xmm0
  call   <bar@plt>
  movss  D12(%rip),%xmm0
  call   <bar@plt>
  push   $0xfffffffffffffff8
  push   $0xfffffffffffffff9
  mov    $0xfffffffa,%r9d
  mov    $0xfffffffb,%r8d
  mov    $0xfffffffc,%ecx
  mov    $0xfffffffd,%edx
  mov    $0xfffffffe,%esi
  mov    $0xffffffff,%edi
  call   <bar8@plt>
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <bazu@plt>
  mov    $0x80000000,%esi
  mov    $0x8,%edi
  call   <bazi@plt>
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type_infered: byte[4]
name: disassemble_data
arguments: tag: D12
type_infered: byte[4]

tool: {"D4": [0x00,0x00,0x80,0x43]}

tool: {"D12": [0xb8,0x1e,0xd5,0x40]}

assistant: ```c
void foo(int i) {
  bar(256);
  bar(6.66f);
  bar8(-1, -2, -3, -4, -5, -6, -7, -8);
  bazu(i, i);
  bazi(0x8, 0x80000000);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<foo>:
  endbr64
  push   %rbp
  movss  D19(%rip),%xmm0
  mov    %edi,%ebp
  call   <bar@plt>
  movss  D7(%rip),%xmm0
  call   <bar@plt>
  push   $0xfffffffffffffff8
  mov    $0xfffffffd,%edx
  mov    $0xfffffffa,%r9d
  push   $0xfffffffffffffff9
  mov    $0xfffffffb,%r8d
  mov    $0xfffffffc,%ecx
  mov    $0xfffffffe,%esi
  mov    $0xffffffff,%edi
  call   <bar8@plt>
  mov    %ebp,%esi
  mov    %ebp,%edi
  call   <bazu@plt>
  pop    %rax
  mov    $0x80000000,%esi
  pop    %rdx
  mov    $0x8,%edi
  pop    %rbp
  jmp    10a0 <bazi@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D19
type_guessed: byte[4]
name: disassemble_data
arguments: data_tag: D7
type_guessed: byte[4]

tool: {"D19": [0x00,0x00,0x80,0x43]}

tool: {"D7": [0xb8,0x1e,0xd5,0x40]}

assistant: ```c
void foo(int i) {
  bar(256);
  bar(6.66f);
  bar8(-1, -2, -3, -4, -5, -6, -7, -8);
  bazu(i, i);
  bazi(0x8, 0x80000000);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<foo>:
  endbr64
  push   %rbp
  movss  D7(%rip),%xmm0
  mov    %edi,%ebp
  call   <bar@plt>
  movss  D13(%rip),%xmm0
  call   <bar@plt>
  push   $0xfffffffffffffff8
  mov    $0xfffffffd,%edx
  mov    $0xfffffffa,%r9d
  push   $0xfffffffffffffff9
  mov    $0xfffffffb,%r8d
  mov    $0xfffffffc,%ecx
  mov    $0xfffffffe,%esi
  mov    $0xffffffff,%edi
  call   <bar8@plt>
  mov    %ebp,%esi
  mov    %ebp,%edi
  call   <bazu@plt>
  pop    %rax
  mov    $0x80000000,%esi
  pop    %rdx
  mov    $0x8,%edi
  pop    %rbp
  jmp    10a0 <bazi@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D13
data_type: byte[4]

tool: {"D7": [0x00,0x00,0x80,0x43]}

tool: {"D13": [0xb8,0x1e,0xd5,0x40]}

assistant: ```c
void foo(int i) {
  bar(256);
  bar(6.66f);
  bar8(-1, -2, -3, -4, -5, -6, -7, -8);
  bazu(i, i);
  bazi(0x8, 0x80000000);
}
```
user: Translate the below assembly code into C programming language:

```asm
<soma_matriz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x18(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%rcx
  mov    $0x0,%ebx
  movl   $0x0,-0x24(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L10
L17:
  movl   $0x0,-0x1c(%rbp)
  jmp    L37
L7:
  mov    -0x20(%rbp),%edx
  cmp    -0x1c(%rbp),%edx
  jge    L31
  mov    -0x20(%rbp),%edx
  movslq %edx,%rcx
  movslq %eax,%rdx
  imul   %rcx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rdx,%rcx
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rcx,%rdx,4),%edx
  add    %edx,-0x24(%rbp)
L31:
  addl   $0x1,-0x1c(%rbp)
L37:
  mov    -0x1c(%rbp),%edx
  cmp    -0x34(%rbp),%edx
  jl     L7
  addl   $0x1,-0x20(%rbp)
L10:
  mov    -0x20(%rbp),%edx
  cmp    -0x34(%rbp),%edx
  jl     L17
  mov    -0x24(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int soma_matriz(int n, int mat[n][n]) {
  int soma = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i < j)
        soma += mat[i][j];
  return soma;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<soma_matriz>:
  endbr64
  movslq %edi,%r10
  test   %edi,%edi
  jle    L60
  shl    $0x2,%r10
  mov    $0x0,%edx
  mov    $0x0,%ecx
  mov    %edi,%r8d
  jmp    L65
L32:
  add    $0x1,%rax
  cmp    %r8,%rax
  je     L69
L53:
  cmp    %eax,%edx
  jge    L32
  add    (%r9,%rax,4),%ecx
  jmp    L32
L69:
  add    $0x1,%edx
  add    %r10,%rsi
  cmp    %edx,%edi
  je     L22
L65:
  mov    %rsi,%r9
  mov    $0x0,%eax
  jmp    L53
L60:
  mov    $0x0,%ecx
L22:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int soma_matriz(int n, int mat[n][n]) {
  int soma = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i < j)
        soma += mat[i][j];
  return soma;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<soma_matriz>:
  endbr64
  movslq %edi,%rcx
  test   %edi,%edi
  jle    L18
  lea    0x0(,%rcx,4),%r9
  xor    %edx,%edx
  xor    %r8d,%r8d
  nopl   0x0(%rax,%rax,1)
L26:
  mov    $0x1,%eax
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L5:
  cmp    %eax,%edx
  jge    L8
  add    (%rsi,%rax,4),%r8d
L8:
  add    $0x1,%rax
L15:
  cmp    %rax,%rcx
  jne    L5
  add    $0x1,%edx
  add    %r9,%rsi
  cmp    %edx,%edi
  jne    L26
  mov    %r8d,%eax
  ret
L18:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int soma_matriz(int n, int mat[n][n]) {
  int soma = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i < j)
        soma += mat[i][j];
  return soma;
}
```
user: What is the c source code of the assembly code below?

```asm
<soma_matriz>:
  endbr64
  movslq %edi,%rcx
  test   %edi,%edi
  jle    L36
  lea    0x0(,%rcx,4),%r9
  xor    %edx,%edx
  xor    %r8d,%r8d
  nopl   0x0(%rax,%rax,1)
L40:
  mov    $0x1,%eax
  cmp    $0x1,%edi
  je     L6
  nopw   0x0(%rax,%rax,1)
L5:
  cmp    %eax,%edx
  jge    L32
  add    (%rsi,%rax,4),%r8d
L32:
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L5
L6:
  add    $0x1,%edx
  add    %r9,%rsi
  cmp    %edx,%edi
  jne    L40
  mov    %r8d,%eax
  ret
L36:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int soma_matriz(int n, int mat[n][n]) {
  int soma = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i < j)
        soma += mat[i][j];
  return soma;
}
```
user: What is the c source code of the assembly code below?

```asm
<ehPalindroma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L30
L10:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L24
  mov    $0x0,%eax
  jmp    L20
L24:
  addl   $0x1,-0x8(%rbp)
L30:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x8(%rbp)
  jl     L10
  mov    $0x1,%eax
L20:
  leave
  ret
```

assistant: ```c
int ehPalindroma(const char *palavra) {
  int length, i;
  length = strlen(palavra);
  for (i = 0; i < length / 2; i++) {
    if (palavra[i] != palavra[length - i - 1]) {
      return 0;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ehPalindroma>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  cmp    $0x1,%eax
  jle    L18
  cltq
  lea    -0x1(%rbx,%rax,1),%rdx
  mov    $0x0,%eax
L46:
  movzbl (%rdx),%esi
  cmp    %sil,(%rbx,%rax,1)
  jne    L20
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%ecx
  jg     L46
  mov    $0x1,%eax
  jmp    L26
L18:
  mov    $0x1,%eax
  jmp    L26
L20:
  mov    $0x0,%eax
L26:
  pop    %rbx
  ret
```

assistant: ```c
int ehPalindroma(const char *palavra) {
  int length, i;
  length = strlen(palavra);
  for (i = 0; i < length / 2; i++) {
    if (palavra[i] != palavra[length - i - 1]) {
      return 0;
    }
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<ehPalindroma>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  cmp    $0x1,%eax
  jle    L32
  cltq
  lea    -0x1(%rbx,%rax,1),%rdx
  xor    %eax,%eax
  jmp    L6
  cs nopw 0x0(%rax,%rax,1)
L15:
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%ecx
  jle    L32
L6:
  movzbl (%rdx),%esi
  cmp    %sil,(%rbx,%rax,1)
  je     L15
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L32:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ehPalindroma(const char *palavra) {
  int length, i;
  length = strlen(palavra);
  for (i = 0; i < length / 2; i++) {
    if (palavra[i] != palavra[length - i - 1]) {
      return 0;
    }
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ehPalindroma>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  cmp    $0x1,%eax
  jle    L5
  cltq
  lea    -0x1(%rbx,%rax,1),%rdx
  xor    %eax,%eax
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
L0:
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%ecx
  jle    L5
L14:
  movzbl (%rdx),%esi
  cmp    %sil,(%rbx,%rax,1)
  je     L0
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L5:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ehPalindroma(const char *palavra) {
  int length, i;
  length = strlen(palavra);
  for (i = 0; i < length / 2; i++) {
    if (palavra[i] != palavra[length - i - 1]) {
      return 0;
    }
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<convert_string_to_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int convert_string_to_int(char buf[]) {
  int num;
  num = atoi(buf);
  return num;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<convert_string_to_int>:
  endbr64
  sub    $0x8,%rsp
  call   <atoi@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int convert_string_to_int(char buf[]) {
  int num;
  num = atoi(buf);
  return num;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<convert_string_to_int>:
  endbr64
  jmp    1050 <atoi@plt>
```

assistant: ```c
int convert_string_to_int(char buf[]) {
  int num;
  num = atoi(buf);
  return num;
}
```
user: Translate the below assembly code into C programming language:

```asm
<convert_string_to_int>:
  endbr64
  jmp    1050 <atoi@plt>
```

assistant: ```c
int convert_string_to_int(char buf[]) {
  int num;
  num = atoi(buf);
  return num;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<lastoccurence>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L11
L7:
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %eax,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L11:
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int lastoccurence(char *s1, char *s2) {
  char *ptr;
  int i;
  ptr = s1;
  while (ptr = strstr(ptr, s2)) {
    i = (ptr - s1);
    ptr++;
  }
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<lastoccurence>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  jmp    L15
L17:
  mov    %eax,%r12d
  sub    %ebx,%r12d
  lea    0x1(%rax),%rdi
L15:
  mov    %rbp,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L17
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int lastoccurence(char *s1, char *s2) {
  char *ptr;
  int i;
  ptr = s1;
  while (ptr = strstr(ptr, s2)) {
    i = (ptr - s1);
    ptr++;
  }
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<lastoccurence>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L0
L2:
  mov    %eax,%r12d
  lea    0x1(%rax),%rdi
  sub    %ebx,%r12d
L0:
  mov    %rbp,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L2
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int lastoccurence(char *s1, char *s2) {
  char *ptr;
  int i;
  ptr = s1;
  while (ptr = strstr(ptr, s2)) {
    i = (ptr - s1);
    ptr++;
  }
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<lastoccurence>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L6
L15:
  mov    %eax,%r12d
  lea    0x1(%rax),%rdi
  sub    %ebx,%r12d
L6:
  mov    %rbp,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L15
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int lastoccurence(char *s1, char *s2) {
  char *ptr;
  int i;
  ptr = s1;
  while (ptr = strstr(ptr, s2)) {
    i = (ptr - s1);
    ptr++;
  }
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<CANDisable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <_CANBaseValid@plt>
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  cltq
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  or     $0x1,%eax
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void CANDisable(uint32_t ui32Base) {
  ASSERT(_CANBaseValid(ui32Base));
  (*((volatile uint32_t *)(ui32Base + 0x00000000))) |= 0x00000001;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CANDisable>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <_CANBaseValid@plt>
  mov    %eax,%edi
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  or     $0x1,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void CANDisable(uint32_t ui32Base) {
  ASSERT(_CANBaseValid(ui32Base));
  (*((volatile uint32_t *)(ui32Base + 0x00000000))) |= 0x00000001;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<CANDisable>:
  endbr64
  push   %rbx
  movslq %edi,%rbx
  mov    %ebx,%edi
  call   <_CANBaseValid@plt>
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    (%rbx),%eax
  or     $0x1,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void CANDisable(uint32_t ui32Base) {
  ASSERT(_CANBaseValid(ui32Base));
  (*((volatile uint32_t *)(ui32Base + 0x00000000))) |= 0x00000001;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<CANDisable>:
  endbr64
  push   %rbx
  movslq %edi,%rbx
  mov    %ebx,%edi
  call   <_CANBaseValid@plt>
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    (%rbx),%eax
  or     $0x1,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void CANDisable(uint32_t ui32Base) {
  ASSERT(_CANBaseValid(ui32Base));
  (*((volatile uint32_t *)(ui32Base + 0x00000000))) |= 0x00000001;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<xdr_int16_t>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x82,%eax
  je     L13
  cmp    $0x82,%eax
  jg     L88
  cmp    $0x80,%eax
  je     L23
  cmp    $0x81,%eax
  jne    L88
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <XDR_PUTINT32@plt>
  jmp    L95
L13:
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <XDR_GETINT32@plt>
  test   %eax,%eax
  jne    L32
  mov    $0x0,%eax
  jmp    L95
L32:
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  jmp    L95
L23:
  mov    $0x1,%eax
  jmp    L95
L88:
  mov    $0x0,%eax
L95:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int xdr_int16_t(XDR *xdrs, int16_t *ip) {
  int32_t t;
  switch (xdrs->x_op) {
  case XDR_ENCODE:
    t = (int32_t)*ip;
    return XDR_PUTINT32(xdrs, &t);
  case XDR_DECODE:
    if (!XDR_GETINT32(xdrs, &t))
      return 0;
    *ip = (int16_t)t;
    return 1;
  case XDR_FREE:
    return 1;
  default:
    return 0;
  }
}
```