ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getFmt>:
  endbr64
  cmpb   $0x30,(%rdi)
  lea    D14(%rip),%rax
  je     L10
L18:
  ret
  nopl   0x0(%rax)
L10:
  movzbl 0x1(%rdi),%edx
  lea    D35(%rip),%rax
  cmp    $0x64,%dl
  je     L18
  cmp    $0x78,%dl
  lea    D19(%rip),%rax
  lea    D14(%rip),%rdx
  cmove  %rdx,%rax
  ret
```

tool_calls: name: parse_data
arguments: data_label: D14
data_type: string
name: parse_data
arguments: data_label: D35
data_type: string
name: parse_data
arguments: data_label: D19
data_type: string

tool: D14: %x

tool: D35: %d

tool: D19: %o

assistant: ```c
const char *getFmt(const char *a) {
  const char *fmt = "%x";
  if (*a == '0') {
    switch (*++a) {
    case 'x':
      fmt = "%x";
      ++a;
      break;
    case 'd':
      fmt = "%d";
      ++a;
      break;
    default:
      fmt = "%o";
      break;
    }
  }
  return fmt;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readFloatFromByteArray>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movss  -0xc(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
float readFloatFromByteArray(const char *byteArray, const int offset) {
  float val;
  memcpy(&val, byteArray + offset, sizeof(float));
  return val;
}
```
user: Translate the below assembly code into C programming language:

```asm
<readFloatFromByteArray>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movslq %esi,%rsi
  add    %rdi,%rsi
  lea    0x4(%rsp),%rdi
  mov    $0x4,%edx
  call   <memcpy@plt>
  movss  0x4(%rsp),%xmm0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float readFloatFromByteArray(const char *byteArray, const int offset) {
  float val;
  memcpy(&val, byteArray + offset, sizeof(float));
  return val;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<readFloatFromByteArray>:
  endbr64
  sub    $0x18,%rsp
  movslq %esi,%rsi
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  add    %rdi,%rsi
  lea    0x4(%rsp),%rdi
  call   <memcpy@plt>
  movss  0x4(%rsp),%xmm0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float readFloatFromByteArray(const char *byteArray, const int offset) {
  float val;
  memcpy(&val, byteArray + offset, sizeof(float));
  return val;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<readFloatFromByteArray>:
  endbr64
  sub    $0x18,%rsp
  movslq %esi,%rsi
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  add    %rdi,%rsi
  lea    0x4(%rsp),%rdi
  call   <memcpy@plt>
  movss  0x4(%rsp),%xmm0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float readFloatFromByteArray(const char *byteArray, const int offset) {
  float val;
  memcpy(&val, byteArray + offset, sizeof(float));
  return val;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<isprime_ui>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L39:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,%eax
  test   %eax,%eax
  jne    L22
  mov    $0x0,%eax
  jmp    L36
L22:
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  imul   %edx,%eax
  cmp    %eax,-0x14(%rbp)
  jae    L39
  mov    $0x1,%eax
L36:
  pop    %rbp
  ret
```

assistant: ```c
int isprime_ui(unsigned int p, unsigned int *primes) {
  int i;
  for (i = 0; primes[i] * primes[i] <= p; i++)
    if (p % primes[i] == 0)
      return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isprime_ui>:
  endbr64
  mov    (%rsi),%ecx
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %eax,%edi
  jb     L26
  add    $0x4,%rsi
L20:
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %ecx
  test   %edx,%edx
  je     L19
  mov    (%rsi),%ecx
  add    $0x4,%rsi
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %edi,%eax
  jbe    L20
  mov    $0x1,%eax
  ret
L26:
  mov    $0x1,%eax
  ret
L19:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int isprime_ui(unsigned int p, unsigned int *primes) {
  int i;
  for (i = 0; primes[i] * primes[i] <= p; i++)
    if (p % primes[i] == 0)
      return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isprime_ui>:
  endbr64
  mov    (%rsi),%ecx
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %eax,%edi
  jb     L32
  add    $0x4,%rsi
  jmp    L39
  nopl   (%rax)
L30:
  mov    (%rsi),%ecx
  add    $0x4,%rsi
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %edi,%eax
  ja     L32
L39:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  test   %edx,%edx
  jne    L30
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L32:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int isprime_ui(unsigned int p, unsigned int *primes) {
  int i;
  for (i = 0; primes[i] * primes[i] <= p; i++)
    if (p % primes[i] == 0)
      return 0;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isprime_ui>:
  endbr64
  mov    (%rsi),%ecx
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %eax,%edi
  jb     L10
  add    $0x4,%rsi
  jmp    L34
  nopl   (%rax)
L35:
  mov    (%rsi),%ecx
  add    $0x4,%rsi
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %edi,%eax
  ja     L10
L34:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  test   %edx,%edx
  jne    L35
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L10:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int isprime_ui(unsigned int p, unsigned int *primes) {
  int i;
  for (i = 0; primes[i] * primes[i] <= p; i++)
    if (p % primes[i] == 0)
      return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dt_strlcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  cmp    %rax,-0x28(%rbp)
  ja     L18
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x8(%rbp),%rdx
  add    %rdx,%rax
  jmp    L10
L18:
  mov    -0x28(%rbp),%rax
  sub    -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  add    %rax,%rcx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <dt_strlcpy@plt>
  mov    -0x8(%rbp),%rdx
  add    %rdx,%rax
L10:
  leave
  ret
```

assistant: ```c
size_t dt_strlcat(char *dst, const char *src, size_t size) {
  size_t len = strlen(dst);
  if (size <= len + 1) {
    return len + strlen(src);
  }
  return len + dt_strlcpy(dst + len, src, size - len);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dt_strlcat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r13
  mov    %rdx,%r12
  call   <strlen@plt>
  mov    %rax,%rbx
  lea    0x1(%rax),%rax
  cmp    %r12,%rax
  jb     L11
  mov    %r13,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  mov    %r12,%rdx
  sub    %rbx,%rdx
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r13,%rsi
  call   <dt_strlcpy@plt>
  add    %rbx,%rax
  jmp    L16
```

assistant: ```c
size_t dt_strlcat(char *dst, const char *src, size_t size) {
  size_t len = strlen(dst);
  if (size <= len + 1) {
    return len + strlen(src);
  }
  return len + dt_strlcpy(dst + len, src, size - len);
}
```
user: What is the c source code of the assembly code below:

```asm
<dt_strlcat>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rax,%rbx
  add    $0x1,%rax
  cmp    %r12,%rax
  jb     L8
  mov    %r13,%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  add    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L8:
  mov    %r12,%rdx
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r13,%rsi
  sub    %rbx,%rdx
  call   <dt_strlcpy@plt>
  add    $0x8,%rsp
  add    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
size_t dt_strlcat(char *dst, const char *src, size_t size) {
  size_t len = strlen(dst);
  if (size <= len + 1) {
    return len + strlen(src);
  }
  return len + dt_strlcpy(dst + len, src, size - len);
}
```
user: Translate the below assembly code into C programming language:

```asm
<dt_strlcat>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rax,%rbx
  add    $0x1,%rax
  cmp    %r12,%rax
  jb     L0
  mov    %r13,%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  add    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L0:
  mov    %r12,%rdx
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r13,%rsi
  sub    %rbx,%rdx
  call   <dt_strlcpy@plt>
  add    $0x8,%rsp
  add    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
size_t dt_strlcat(char *dst, const char *src, size_t size) {
  size_t len = strlen(dst);
  if (size <= len + 1) {
    return len + strlen(src);
  }
  return len + dt_strlcpy(dst + len, src, size - len);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<uniphier_board_late_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
  pop    %rbp
  ret
```

assistant: ```c
void uniphier_board_late_init(void) {
  do {
  } while (0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<uniphier_board_late_init>:
  endbr64
  ret
```

assistant: ```c
void uniphier_board_late_init(void) {
  do {
  } while (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<uniphier_board_late_init>:
  endbr64
  ret
```

assistant: ```c
void uniphier_board_late_init(void) {
  do {
  } while (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
<uniphier_board_late_init>:
  endbr64
  ret
```

assistant: ```c
void uniphier_board_late_init(void) {
  do {
  } while (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x12345678,-0x38(%rbp)
  movb   $0x9a,-0x3b(%rbp)
  movss  D14(%rip),%xmm0
  movss  %xmm0,-0x34(%rbp)
  movw   $0xbcde,-0x3a(%rbp)
  movl   $0xf1234567,-0x30(%rbp)
  lea    -0x17(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  lea    -0x38(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <copyByte@plt>
  mov    %rax,-0x28(%rbp)
  lea    -0x3b(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <copyByte@plt>
  mov    %rax,-0x28(%rbp)
  lea    -0x34(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <copyByte@plt>
  mov    %rax,-0x28(%rbp)
  lea    -0x3a(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <copyByte@plt>
  mov    %rax,-0x28(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <copyByte@plt>
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L0
L7:
  mov    -0x2c(%rbp),%eax
  cltq
  movzbl -0x17(%rbp,%rax,1),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x2c(%rbp)
L0:
  cmpl   $0xe,-0x2c(%rbp)
  jle    L7
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D14
type_infered: f32
name: retrieve_data
arguments: data_tag: D5
type_infered: string

tool: D14: 1234.5678

tool: D5: array = %x


assistant: ```c
main(void) {
  int i;
  int iVal = 0x12345678;
  char cVal = 0x9a;
  float fVal = 1234.5678;
  short sVal = 0xbcde;
  int iVal1 = 0xf1234567;
  unsigned char message[15];
  unsigned char *src, *dest;
  dest = &message[0];
  src = (char *)&iVal;
  dest = copyByte(src, dest, 4);
  src = (char *)&cVal;
  dest = copyByte(src, dest, 1);
  src = (char *)&fVal;
  dest = copyByte(src, dest, 4);
  src = (char *)&sVal;
  dest = copyByte(src, dest, 2);
  src = (char *)&iVal1;
  dest = copyByte(src, dest, 4);
  for (i = 0; i < 15; i++)
    printf("array = %x\n", message[i]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x12345678,0xc(%rsp)
  movb   $0x9a,0x9(%rsp)
  movl   $0x449a522b,0x10(%rsp)
  movw   $0xbcde,0xa(%rsp)
  movl   $0xf1234567,0x14(%rsp)
  lea    0x19(%rsp),%rbx
  lea    0xc(%rsp),%rdi
  mov    $0x4,%edx
  mov    %rbx,%rsi
  call   <copyByte@plt>
  mov    %rax,%rsi
  lea    0x9(%rsp),%rdi
  mov    $0x1,%edx
  call   <copyByte@plt>
  mov    %rax,%rsi
  lea    0x10(%rsp),%rdi
  mov    $0x4,%edx
  call   <copyByte@plt>
  mov    %rax,%rsi
  lea    0xa(%rsp),%rdi
  mov    $0x2,%edx
  call   <copyByte@plt>
  mov    %rax,%rsi
  lea    0x14(%rsp),%rdi
  mov    $0x4,%edx
  call   <copyByte@plt>
  lea    0x28(%rsp),%r12
  lea    D5(%rip),%rbp
L11:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L11
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D5
type: string

tool: D5: array = %x


assistant: ```c
main(void) {
  int i;
  int iVal = 0x12345678;
  char cVal = 0x9a;
  float fVal = 1234.5678;
  short sVal = 0xbcde;
  int iVal1 = 0xf1234567;
  unsigned char message[15];
  unsigned char *src, *dest;
  dest = &message[0];
  src = (char *)&iVal;
  dest = copyByte(src, dest, 4);
  src = (char *)&cVal;
  dest = copyByte(src, dest, 1);
  src = (char *)&fVal;
  dest = copyByte(src, dest, 4);
  src = (char *)&sVal;
  dest = copyByte(src, dest, 2);
  src = (char *)&iVal1;
  dest = copyByte(src, dest, 4);
  for (i = 0; i < 15; i++)
    printf("array = %x\n", message[i]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x4,%edx
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x19(%rsp),%rbx
  mov    $0xffffbcde,%eax
  lea    0xc(%rsp),%rdi
  mov    %rbx,%rsi
  movl   $0x12345678,0xc(%rsp)
  lea    0x28(%rsp),%r12
  movb   $0x9a,0x9(%rsp)
  movl   $0x449a522b,0x10(%rsp)
  mov    %ax,0xa(%rsp)
  movl   $0xf1234567,0x14(%rsp)
  call   <copyByte@plt>
  lea    0x9(%rsp),%rdi
  mov    $0x1,%edx
  mov    %rax,%rsi
  call   <copyByte@plt>
  lea    0x10(%rsp),%rdi
  mov    $0x4,%edx
  mov    %rax,%rsi
  call   <copyByte@plt>
  lea    0xa(%rsp),%rdi
  mov    $0x2,%edx
  mov    %rax,%rsi
  call   <copyByte@plt>
  lea    0x14(%rsp),%rdi
  mov    $0x4,%edx
  mov    %rax,%rsi
  call   <copyByte@plt>
  xchg   %ax,%ax
L16:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L16
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

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

tool: D7: array = %x


assistant: ```c
main(void) {
  int i;
  int iVal = 0x12345678;
  char cVal = 0x9a;
  float fVal = 1234.5678;
  short sVal = 0xbcde;
  int iVal1 = 0xf1234567;
  unsigned char message[15];
  unsigned char *src, *dest;
  dest = &message[0];
  src = (char *)&iVal;
  dest = copyByte(src, dest, 4);
  src = (char *)&cVal;
  dest = copyByte(src, dest, 1);
  src = (char *)&fVal;
  dest = copyByte(src, dest, 4);
  src = (char *)&sVal;
  dest = copyByte(src, dest, 2);
  src = (char *)&iVal1;
  dest = copyByte(src, dest, 4);
  for (i = 0; i < 15; i++)
    printf("array = %x\n", message[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x4,%edx
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x19(%rsp),%rbx
  mov    $0xffffbcde,%eax
  lea    0xc(%rsp),%rdi
  mov    %rbx,%rsi
  movl   $0x12345678,0xc(%rsp)
  lea    0x28(%rsp),%r12
  movb   $0x9a,0x9(%rsp)
  movl   $0x449a522b,0x10(%rsp)
  mov    %ax,0xa(%rsp)
  movl   $0xf1234567,0x14(%rsp)
  call   <copyByte@plt>
  lea    0x9(%rsp),%rdi
  mov    $0x1,%edx
  mov    %rax,%rsi
  call   <copyByte@plt>
  lea    0x10(%rsp),%rdi
  mov    $0x4,%edx
  mov    %rax,%rsi
  call   <copyByte@plt>
  lea    0xa(%rsp),%rdi
  mov    $0x2,%edx
  mov    %rax,%rsi
  call   <copyByte@plt>
  lea    0x14(%rsp),%rdi
  mov    $0x4,%edx
  mov    %rax,%rsi
  call   <copyByte@plt>
  xchg   %ax,%ax
L0:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L0
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D1
datatype: string

tool: D1: array = %x


assistant: ```c
main(void) {
  int i;
  int iVal = 0x12345678;
  char cVal = 0x9a;
  float fVal = 1234.5678;
  short sVal = 0xbcde;
  int iVal1 = 0xf1234567;
  unsigned char message[15];
  unsigned char *src, *dest;
  dest = &message[0];
  src = (char *)&iVal;
  dest = copyByte(src, dest, 4);
  src = (char *)&cVal;
  dest = copyByte(src, dest, 1);
  src = (char *)&fVal;
  dest = copyByte(src, dest, 4);
  src = (char *)&sVal;
  dest = copyByte(src, dest, 2);
  src = (char *)&iVal1;
  dest = copyByte(src, dest, 4);
  for (i = 0; i < 15; i++)
    printf("array = %x\n", message[i]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<drawP>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x15,%ecx
  mov    $0xc,%edx
  mov    $0x22,%esi
  mov    $0x4,%edi
  call   <nxtFillRect@plt>
  mov    $0x2f,%ecx
  mov    $0x15,%edx
  mov    $0x32,%esi
  mov    $0xa,%edi
  call   <nxtFillRect@plt>
  mov    $0x23,%ecx
  mov    $0x7,%edx
  mov    $0x2b,%esi
  mov    $0x4,%edi
  call   <nxtFillRect@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void drawP() {
  nxtFillRect(4, 34, 12, 21);
  nxtFillRect(10, 50, 21, 47);
  nxtFillRect(4, 43, 7, 35);
}
```
user: Translate the below assembly code into C programming language:

```asm
<drawP>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x15,%ecx
  mov    $0xc,%edx
  mov    $0x22,%esi
  mov    $0x4,%edi
  call   <nxtFillRect@plt>
  mov    $0x2f,%ecx
  mov    $0x15,%edx
  mov    $0x32,%esi
  mov    $0xa,%edi
  call   <nxtFillRect@plt>
  mov    $0x23,%ecx
  mov    $0x7,%edx
  mov    $0x2b,%esi
  mov    $0x4,%edi
  call   <nxtFillRect@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void drawP() {
  nxtFillRect(4, 34, 12, 21);
  nxtFillRect(10, 50, 21, 47);
  nxtFillRect(4, 43, 7, 35);
}
```
user: Translate the below assembly code into C programming language:

```asm
<drawP>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x15,%ecx
  mov    $0xc,%edx
  mov    $0x22,%esi
  mov    $0x4,%edi
  call   <nxtFillRect@plt>
  mov    $0x2f,%ecx
  mov    $0x15,%edx
  mov    $0x32,%esi
  mov    $0xa,%edi
  call   <nxtFillRect@plt>
  mov    $0x23,%ecx
  mov    $0x7,%edx
  mov    $0x2b,%esi
  mov    $0x4,%edi
  add    $0x8,%rsp
  jmp    1050 <nxtFillRect@plt>
```

assistant: ```c
void drawP() {
  nxtFillRect(4, 34, 12, 21);
  nxtFillRect(10, 50, 21, 47);
  nxtFillRect(4, 43, 7, 35);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<guess_output>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x54(%rbp)
  lea    -0x50(%rbp),%rdx
  mov    -0x68(%rbp),%rax
  mov    $0x0,%ecx
  lea    D4(%rip),%rsi
  mov    %rax,%rdi
  call   <symtab_lookup@plt>
  test   %rax,%rax
  je     L48
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x50(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
L48:
  movl   $0x0,-0x58(%rbp)
  jmp    L0
L49:
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x70(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L23
  mov    -0x58(%rbp),%eax
  mov    %eax,-0x54(%rbp)
L23:
  addl   $0x1,-0x58(%rbp)
L0:
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x70(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L49
  cmpl   $0xffffffff,-0x54(%rbp)
  jne    L55
  mov    $0xffffffff,%eax
  jmp    L7
L55:
  mov    -0x54(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x70(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdx
  lea    D31(%rip),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  mov    $0x0,%eax
L7:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L65
  call   <__stack_chk_fail@plt>
L65:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D20
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D5
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D31
type_guessed: string

tool: D4: $outfmt

tool: D20: INFO: Unknown output format, defaulting to hex


tool: D5: hex

tool: D31: %s

assistant: ```c
int guess_output(struct SymTab **prgSyms, char *out) {
  int x, pIdx = -1;
  char fmtname[64];
  if (symtab_lookup(prgSyms, "$outfmt", fmtname, NULL) != 0) {
    printf("INFO: Unknown output format, defaulting to hex\n");
    sprintf(fmtname, "hex");
  }
  for (x = 0; out[x] != '\0'; x++) {
    if (out[x] == '.') {
      pIdx = x;
    }
  }
  if (pIdx == -1) {
    return -1;
  }
  sprintf(&out[pIdx + 1], "%s", fmtname);
  return 0;
}
```