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

```asm
<stack_push4>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x8,%esi
  sub    $0x8,%rsp
  call   <tcb_regs_stack_push@plt>
  mov    %rbp,-0x8(%rbx)
  add    $0x8,%rsp
  lea    -0x8(%rbx),%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *stack_push4(struct tcb_regs *regs, void *stack, unsigned long value) {
  tcb_regs_stack_push(regs, sizeof(value));
  unsigned long *top = ((unsigned long *)stack) - 1;
  top[0] = value;
  return top;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stack_push4>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x8,%esi
  sub    $0x8,%rsp
  call   <tcb_regs_stack_push@plt>
  mov    %rbp,-0x8(%rbx)
  add    $0x8,%rsp
  lea    -0x8(%rbx),%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *stack_push4(struct tcb_regs *regs, void *stack, unsigned long value) {
  tcb_regs_stack_push(regs, sizeof(value));
  unsigned long *top = ((unsigned long *)stack) - 1;
  top[0] = value;
  return top;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movss  D0(%rip),%xmm0
  movss  %xmm0,-0x1c(%rbp)
  lea    -0x1c(%rbp),%rax
  movss  D5(%rip),%xmm0
  movss  %xmm0,(%rax)
  movss  D2(%rip),%xmm0
  movss  %xmm0,-0x18(%rbp)
  movss  D8(%rip),%xmm0
  movss  %xmm0,-0x14(%rbp)
  movl   $0xc,-0x10(%rbp)
  movb   $0x41,-0x1d(%rbp)
  movl   $0x34,-0xc(%rbp)
  movzbl -0x1d(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1d(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
data_type: f32
name: extract_symbolic_data
arguments: label: D5
data_type: byte[4]
name: extract_symbolic_data
arguments: label: D2
data_type: f32
name: extract_symbolic_data
arguments: label: D8
data_type: byte[4]

tool: D0: 3.141592

tool: {"D5": [0x00,0x00,0x00,0x40]}

tool: D2: 3.14

tool: {"D8": [0xeb,0x9f,0xa5,0xb5]}

assistant: ```c
int main(void) {
  const float PI = 3.141592;
  *((float *)&PI) = 2;
  float gleitZahl;
  float floatVariable = 3.14;
  float floatVariable2 = -1.234E-6;
  int zahl1, zahl2 = 12;
  char zeichen1 = 'A';
  zahl1 = 52;
  zeichen1 = zeichen1 + 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  const float PI = 3.141592;
  *((float *)&PI) = 2;
  float gleitZahl;
  float floatVariable = 3.14;
  float floatVariable2 = -1.234E-6;
  int zahl1, zahl2 = 12;
  char zeichen1 = 'A';
  zahl1 = 52;
  zeichen1 = zeichen1 + 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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  const float PI = 3.141592;
  *((float *)&PI) = 2;
  float gleitZahl;
  float floatVariable = 3.14;
  float floatVariable2 = -1.234E-6;
  int zahl1, zahl2 = 12;
  char zeichen1 = 'A';
  zahl1 = 52;
  zeichen1 = zeichen1 + 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  const float PI = 3.141592;
  *((float *)&PI) = 2;
  float gleitZahl;
  float floatVariable = 3.14;
  float floatVariable2 = -1.234E-6;
  int zahl1, zahl2 = 12;
  char zeichen1 = 'A';
  zahl1 = 52;
  zeichen1 = zeichen1 + 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<LexIsBinDigit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x30,-0x4(%rbp)
  je     L22
  cmpb   $0x31,-0x4(%rbp)
  jne    L15
L22:
  mov    $0x1,%eax
  jmp    L12
L15:
  mov    $0x0,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int LexIsBinDigit(char c) {
  if (c == '0' || c == '1') {
    return 1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<LexIsBinDigit>:
  endbr64
  sub    $0x30,%edi
  cmp    $0x1,%dil
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int LexIsBinDigit(char c) {
  if (c == '0' || c == '1') {
    return 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<LexIsBinDigit>:
  endbr64
  sub    $0x30,%edi
  xor    %eax,%eax
  cmp    $0x1,%dil
  setbe  %al
  ret
```

assistant: ```c
int LexIsBinDigit(char c) {
  if (c == '0' || c == '1') {
    return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<LexIsBinDigit>:
  endbr64
  sub    $0x30,%edi
  xor    %eax,%eax
  cmp    $0x1,%dil
  setbe  %al
  ret
```

assistant: ```c
int LexIsBinDigit(char c) {
  if (c == '0' || c == '1') {
    return 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hToUTF8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jle    L23
  cmpq   $0x0,-0x18(%rbp)
  jne    L7
  mov    -0xc(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L7
  mov    $0x0,%eax
  jmp    L3
L7:
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fb_hCharToUTF8@plt>
  jmp    L6
L23:
  mov    -0x20(%rbp),%rax
  movl   $0x0,(%rax)
L6:
  mov    -0x18(%rbp),%rax
L3:
  leave
  ret
```

assistant: ```c
char *hToUTF8(const char *src, int chars, char *dst, int *bytes) {
  if (chars > 0) {
    if (dst == 0) {
      dst = malloc(chars * 2);
      if (dst == 0)
        return 0;
    }
    fb_hCharToUTF8(src, chars, dst, bytes);
  } else
    *bytes = 0;
  return dst;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hToUTF8>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%rbx
  mov    %rcx,%r13
  test   %esi,%esi
  jle    L25
  mov    %rdi,%r12
  mov    %esi,%ebp
  test   %rdx,%rdx
  je     L3
L40:
  mov    %r13,%rcx
  mov    %rbx,%rdx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <fb_hCharToUTF8@plt>
  mov    %rbx,%rax
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  lea    (%rsi,%rsi,1),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L40
  jmp    L11
L25:
  movl   $0x0,(%rcx)
  mov    %rdx,%rax
  jmp    L11
```

assistant: ```c
char *hToUTF8(const char *src, int chars, char *dst, int *bytes) {
  if (chars > 0) {
    if (dst == 0) {
      dst = malloc(chars * 2);
      if (dst == 0)
        return 0;
    }
    fb_hCharToUTF8(src, chars, dst, bytes);
  } else
    *bytes = 0;
  return dst;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<hToUTF8>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  test   %esi,%esi
  jle    L32
  mov    %rdi,%rbp
  test   %rdx,%rdx
  je     L7
L3:
  mov    %rbx,%rdx
  mov    %rbp,%rdi
  call   <fb_hCharToUTF8@plt>
  mov    %rbx,%rax
L8:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L32:
  movl   $0x0,(%rcx)
  add    $0x18,%rsp
  mov    %rdx,%rax
  pop    %rbx
  pop    %rbp
  ret
L7:
  lea    (%rsi,%rsi,1),%edi
  mov    %rcx,0x8(%rsp)
  mov    %esi,0x4(%rsp)
  call   <malloc@plt>
  mov    0x4(%rsp),%esi
  mov    0x8(%rsp),%rcx
  test   %rax,%rax
  mov    %rax,%rbx
  jne    L3
  xor    %eax,%eax
  jmp    L8
```

assistant: ```c
char *hToUTF8(const char *src, int chars, char *dst, int *bytes) {
  if (chars > 0) {
    if (dst == 0) {
      dst = malloc(chars * 2);
      if (dst == 0)
        return 0;
    }
    fb_hCharToUTF8(src, chars, dst, bytes);
  } else
    *bytes = 0;
  return dst;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<hToUTF8>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  test   %esi,%esi
  jle    L32
  mov    %rdi,%rbp
  test   %rdx,%rdx
  je     L22
L5:
  mov    %rbx,%rdx
  mov    %rbp,%rdi
  call   <fb_hCharToUTF8@plt>
  mov    %rbx,%rax
L34:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L32:
  movl   $0x0,(%rcx)
  add    $0x18,%rsp
  mov    %rdx,%rax
  pop    %rbx
  pop    %rbp
  ret
L22:
  lea    (%rsi,%rsi,1),%edi
  mov    %rcx,0x8(%rsp)
  mov    %esi,0x4(%rsp)
  call   <malloc@plt>
  mov    0x4(%rsp),%esi
  mov    0x8(%rsp),%rcx
  test   %rax,%rax
  mov    %rax,%rbx
  jne    L5
  xor    %eax,%eax
  jmp    L34
```

assistant: ```c
char *hToUTF8(const char *src, int chars, char *dst, int *bytes) {
  if (chars > 0) {
    if (dst == 0) {
      dst = malloc(chars * 2);
      if (dst == 0)
        return 0;
    }
    fb_hCharToUTF8(src, chars, dst, bytes);
  } else
    *bytes = 0;
  return dst;
}
```
user: What is the c source code of the assembly code below?

```asm
<parse_field>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x3d,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L11
  mov    -0x18(%rbp),%rax
  jmp    L13
L11:
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <pl_trim_whitespaces@plt>
L13:
  leave
  ret
```

assistant: ```c
char *parse_field(char *line) {
  char *its;
  its = strchr(line, '=');
  if (!its)
    return line;
  its++;
  return pl_trim_whitespaces(its);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_field>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x3d,%esi
  call   <strchr@plt>
  mov    %rax,%rdi
  mov    %rbx,%rax
  test   %rdi,%rdi
  je     L0
  add    $0x1,%rdi
  call   <pl_trim_whitespaces@plt>
L0:
  pop    %rbx
  ret
```

assistant: ```c
char *parse_field(char *line) {
  char *its;
  its = strchr(line, '=');
  if (!its)
    return line;
  its++;
  return pl_trim_whitespaces(its);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<parse_field>:
  endbr64
  push   %r12
  mov    $0x3d,%esi
  mov    %rdi,%r12
  call   <strchr@plt>
  test   %rax,%rax
  je     L9
  lea    0x1(%rax),%rdi
  pop    %r12
  jmp    1070 <pl_trim_whitespaces@plt>
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *parse_field(char *line) {
  char *its;
  its = strchr(line, '=');
  if (!its)
    return line;
  its++;
  return pl_trim_whitespaces(its);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<parse_field>:
  endbr64
  push   %r12
  mov    $0x3d,%esi
  mov    %rdi,%r12
  call   <strchr@plt>
  test   %rax,%rax
  je     L8
  lea    0x1(%rax),%rdi
  pop    %r12
  jmp    1070 <pl_trim_whitespaces@plt>
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *parse_field(char *line) {
  char *its;
  its = strchr(line, '=');
  if (!its)
    return line;
  its++;
  return pl_trim_whitespaces(its);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<encrypt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %rsi,%rbp
  call   <strlen@plt>
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    $0x0,%ebx
  mov    $0x1a,%r15d
  jmp    L10
L28:
  movsbl (%r14,%rbx,1),%edi
  call   <toupper@plt>
  movsbl %al,%edx
  sub    $0x41,%edx
  cmp    $0x20,%al
  mov    %edx,%eax
  cmove  %r15d,%eax
  mov    %eax,(%r12,%rbx,4)
  add    $0x1,%rbx
L10:
  mov    %r14,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L28
  mov    $0x0,%ebx
  mov    $0x1a,%r14d
  jmp    L23
L3:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <toupper@plt>
  movsbl %al,%edi
  sub    $0x41,%edi
  cmp    $0x20,%al
  cmove  %r14d,%edi
  mov    (%r12,%rbx,4),%esi
  call   <ModuloAdd@plt>
  lea    0x41(%rax),%edx
  cmp    $0x1a,%eax
  mov    $0x20,%eax
  cmovne %edx,%eax
  mov    %al,0x0(%r13,%rbx,1)
  add    $0x1,%rbx
L23:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L3
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *encrypt(char *key, char *file) {
  char F, K, N;
  int f, k, n, x;
  int i;
  int *keynum = malloc(sizeof(int) * strlen(key));
  char *newfile = malloc(sizeof(char) * strlen(file));
  memset(newfile, 0, sizeof(newfile));
  for (i = 0; i < strlen(key); i++) {
    K = toupper(key[i]);
    if (K == ' ')
      k = 26;
    else
      k = (int)K - 65;
    keynum[i] = k;
  }
  for (i = 0; i < strlen(file); i++) {
    F = toupper(file[i]);
    if (F == ' ')
      f = 26;
    else
      f = (int)F - 65;
    x = ModuloAdd(f, keynum[i]);
    if (0)
      printf("%c+%c = %d + %d = %d\n", F, K, f, keynum[i], x);
    if (x == 26)
      newfile[i] = ' ';
    else
      newfile[i] = (char)(x + 65);
  }
  return (newfile);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<encrypt>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    $0x1a,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    $0x8,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memset@plt>
  jmp    L8
  nopl   0x0(%rax,%rax,1)
L11:
  movsbl (%rbx,%r15,1),%edi
  call   <toupper@plt>
  movsbl %al,%edx
  sub    $0x41,%edx
  cmp    $0x20,%al
  mov    %edx,%eax
  cmove  %r14d,%eax
  mov    %eax,(%r12,%r15,4)
  add    $0x1,%r15
L8:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    %r15d,%eax
  jg     L11
  xor    %ebx,%ebx
  mov    $0x1a,%r14d
  jmp    L30
  nopw   0x0(%rax,%rax,1)
L34:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <toupper@plt>
  mov    (%r12,%rbx,4),%esi
  movsbl %al,%edi
  sub    $0x41,%edi
  cmp    $0x20,%al
  cmove  %r14d,%edi
  call   <ModuloAdd@plt>
  cmp    $0x1a,%eax
  lea    0x41(%rax),%edx
  mov    $0x20,%eax
  cmovne %edx,%eax
  mov    %al,0x0(%r13,%rbx,1)
  add    $0x1,%rbx
L30:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L34
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *encrypt(char *key, char *file) {
  char F, K, N;
  int f, k, n, x;
  int i;
  int *keynum = malloc(sizeof(int) * strlen(key));
  char *newfile = malloc(sizeof(char) * strlen(file));
  memset(newfile, 0, sizeof(newfile));
  for (i = 0; i < strlen(key); i++) {
    K = toupper(key[i]);
    if (K == ' ')
      k = 26;
    else
      k = (int)K - 65;
    keynum[i] = k;
  }
  for (i = 0; i < strlen(file); i++) {
    F = toupper(file[i]);
    if (F == ' ')
      f = 26;
    else
      f = (int)F - 65;
    x = ModuloAdd(f, keynum[i]);
    if (0)
      printf("%c+%c = %d + %d = %d\n", F, K, f, keynum[i], x);
    if (x == 26)
      newfile[i] = ' ';
    else
      newfile[i] = (char)(x + 65);
  }
  return (newfile);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<encrypt>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    $0x1a,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    $0x8,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memset@plt>
  jmp    L22
  nopl   0x0(%rax,%rax,1)
L13:
  movsbl (%rbx,%r15,1),%edi
  call   <toupper@plt>
  movsbl %al,%edx
  sub    $0x41,%edx
  cmp    $0x20,%al
  mov    %edx,%eax
  cmove  %r14d,%eax
  mov    %eax,(%r12,%r15,4)
  add    $0x1,%r15
L22:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    %r15d,%eax
  jg     L13
  xor    %ebx,%ebx
  mov    $0x1a,%r14d
  jmp    L37
  nopw   0x0(%rax,%rax,1)
L12:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <toupper@plt>
  mov    (%r12,%rbx,4),%esi
  movsbl %al,%edi
  sub    $0x41,%edi
  cmp    $0x20,%al
  cmove  %r14d,%edi
  call   <ModuloAdd@plt>
  cmp    $0x1a,%eax
  lea    0x41(%rax),%edx
  mov    $0x20,%eax
  cmovne %edx,%eax
  mov    %al,0x0(%r13,%rbx,1)
  add    $0x1,%rbx
L37:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L12
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *encrypt(char *key, char *file) {
  char F, K, N;
  int f, k, n, x;
  int i;
  int *keynum = malloc(sizeof(int) * strlen(key));
  char *newfile = malloc(sizeof(char) * strlen(file));
  memset(newfile, 0, sizeof(newfile));
  for (i = 0; i < strlen(key); i++) {
    K = toupper(key[i]);
    if (K == ' ')
      k = 26;
    else
      k = (int)K - 65;
    keynum[i] = k;
  }
  for (i = 0; i < strlen(file); i++) {
    F = toupper(file[i]);
    if (F == ' ')
      f = 26;
    else
      f = (int)F - 65;
    x = ModuloAdd(f, keynum[i]);
    if (0)
      printf("%c+%c = %d + %d = %d\n", F, K, f, keynum[i], x);
    if (x == 26)
      newfile[i] = ' ';
    else
      newfile[i] = (char)(x + 65);
  }
  return (newfile);
}
```
user: What is the c source code of the assembly code below:

```asm
<set>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,%eax
  mov    %al,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <openit@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x8,%eax
  mov    %al,-0xa(%rbp)
  movzbl -0x18(%rbp),%eax
  mov    %al,-0x9(%rbp)
  lea    -0xb(%rbp),%rcx
  mov    -0x10(%rbp),%eax
  mov    $0x3,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
void set(int addr, unsigned char val) {
  unsigned char msg[3];
  int fd;
  fd = openit();
  msg[0] = addr;
  msg[1] = addr >> 8;
  msg[2] = val;
  write(fd, msg, 3);
  close(fd);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<set>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebp
  mov    %esi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <openit@plt>
  mov    %eax,%ebx
  mov    %bpl,0x5(%rsp)
  mov    %ebp,%eax
  movzbl %ah,%eax
  mov    %al,0x6(%rsp)
  mov    %r12b,0x7(%rsp)
  lea    0x5(%rsp),%rsi
  mov    $0x3,%edx
  mov    %ebx,%edi
  call   <write@plt>
  mov    %ebx,%edi
  call   <close@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void set(int addr, unsigned char val) {
  unsigned char msg[3];
  int fd;
  fd = openit();
  msg[0] = addr;
  msg[1] = addr >> 8;
  msg[2] = val;
  write(fd, msg, 3);
  close(fd);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<set>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <openit@plt>
  lea    0x5(%rsp),%rsi
  mov    $0x3,%edx
  mov    %r12w,0x5(%rsp)
  mov    %eax,%ebp
  mov    %eax,%edi
  mov    %bl,0x7(%rsp)
  call   <write@plt>
  mov    %ebp,%edi
  call   <close@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void set(int addr, unsigned char val) {
  unsigned char msg[3];
  int fd;
  fd = openit();
  msg[0] = addr;
  msg[1] = addr >> 8;
  msg[2] = val;
  write(fd, msg, 3);
  close(fd);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<set>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <openit@plt>
  lea    0x4(%rsp),%rsi
  mov    $0x3,%edx
  mov    %r12w,0x4(%rsp)
  mov    %eax,%ebp
  mov    %eax,%edi
  mov    %bl,0x6(%rsp)
  call   <write@plt>
  mov    %ebp,%edi
  call   <close@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void set(int addr, unsigned char val) {
  unsigned char msg[3];
  int fd;
  fd = openit();
  msg[0] = addr;
  msg[1] = addr >> 8;
  msg[2] = val;
  write(fd, msg, 3);
  close(fd);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<hex2number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L33
L39:
  shll   $0x4,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <hexdigit2number@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jns    L27
  mov    -0xc(%rbp),%eax
  jmp    L29
L27:
  mov    -0xc(%rbp),%eax
  add    %eax,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L33:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L39
  mov    -0x10(%rbp),%eax
L29:
  leave
  ret
```

assistant: ```c
int hex2number(const char *buffer, const size_t len) {
  int result = 0, tmp;
  size_t pos;
  for (pos = 0; pos < len; pos++) {
    result = result << 4;
    tmp = hexdigit2number(buffer[pos]);
    if (tmp < 0)
      return tmp;
    result += tmp;
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hex2number>:
  endbr64
  test   %rsi,%rsi
  je     L21
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    (%rdi,%rsi,1),%r12
  mov    $0x0,%eax
L6:
  shl    $0x4,%eax
  mov    %eax,%ebp
  movsbl (%rbx),%edi
  call   <hexdigit2number@plt>
  test   %eax,%eax
  js     L17
  add    %ebp,%eax
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L6
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int hex2number(const char *buffer, const size_t len) {
  int result = 0, tmp;
  size_t pos;
  for (pos = 0; pos < len; pos++) {
    result = result << 4;
    tmp = hexdigit2number(buffer[pos]);
    if (tmp < 0)
      return tmp;
    result += tmp;
  }
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hex2number>:
  endbr64
  test   %rsi,%rsi
  je     L18
  push   %r12
  xor    %eax,%eax
  lea    (%rdi,%rsi,1),%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L8:
  add    $0x1,%rbx
  add    %ebp,%eax
  cmp    %r12,%rbx
  je     L23
L33:
  movsbl (%rbx),%edi
  shl    $0x4,%eax
  mov    %eax,%ebp
  call   <hexdigit2number@plt>
  test   %eax,%eax
  jns    L8
L23:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  xor    %eax,%eax
  ret
```

assistant: ```c
int hex2number(const char *buffer, const size_t len) {
  int result = 0, tmp;
  size_t pos;
  for (pos = 0; pos < len; pos++) {
    result = result << 4;
    tmp = hexdigit2number(buffer[pos]);
    if (tmp < 0)
      return tmp;
    result += tmp;
  }
  return result;
}
```