ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<I2CMasterInitExpClk>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  call   <I2CMasterEnable@plt>
  cmp    $0x1,%r12d
  mov    $0x1e8480,%eax
  lea    0xc(%rbx),%ecx
  mov    $0x7a1200,%esi
  movslq %ecx,%rcx
  cmovne %eax,%esi
  add    $0xfc0,%ebx
  movslq %ebx,%rbx
  lea    -0x1(%rsi,%rbp,1),%eax
  cltd
  idiv   %esi
  sub    $0x1,%eax
  cltq
  mov    %rax,(%rcx)
  mov    (%rbx),%rax
  test   $0x1,%al
  je     L5
  add    $0x137477f,%ebp
  movslq %ebp,%rax
  sar    $0x1f,%ebp
  imul   $0x6944d07d,%rax,%rax
  sar    $0x37,%rax
  sub    %ebp,%eax
  sub    $0x1,%eax
  or     $0x80,%al
  cltq
  mov    %rax,(%rcx)
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void I2CMasterInitExpClk(uint32_t ui32Base, uint32_t ui32I2CClk, bool bFast) {
  uint32_t ui32SCLFreq;
  uint32_t ui32TPR;
  ;
  I2CMasterEnable(ui32Base);
  if (bFast == 1) {
    ui32SCLFreq = 400000;
  } else {
    ui32SCLFreq = 100000;
  }
  ui32TPR = ((ui32I2CClk + (2 * 10 * ui32SCLFreq) - 1) / (2 * 10 * ui32SCLFreq)) - 1;
  (*((volatile unsigned long *)(ui32Base + 0x0000000C))) = ui32TPR;
  if ((*((volatile unsigned long *)(ui32Base + 0x00000FC0))) & 0x00000001) {
    ui32TPR = ((ui32I2CClk + (2 * 3 * 3400000) - 1) / (2 * 3 * 3400000)) - 1;
    (*((volatile unsigned long *)(ui32Base + 0x0000000C))) = 0x00000080 | ui32TPR;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<has_suffix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L0
  mov    -0x8(%rbp),%eax
  cltq
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  sub    %rdx,%rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  lea    (%rcx,%rax,1),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L0
  mov    $0x1,%eax
  jmp    L15
L0:
  mov    $0x0,%eax
L15:
  leave
  ret
```

assistant: ```c
int has_suffix(char *filename, char *suffix) {
  int flen = strlen(filename);
  int slen = strlen(suffix);
  return flen > slen && strcmp(filename + flen - slen, suffix) == 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<has_suffix>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    $0x0,%edx
  cmp    %eax,%ebx
  jg     L16
L9:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  movslq %ebx,%rbx
  cltq
  sub    %rax,%rbx
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r12,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %dl
  movzbl %dl,%edx
  jmp    L9
```

assistant: ```c
int has_suffix(char *filename, char *suffix) {
  int flen = strlen(filename);
  int slen = strlen(suffix);
  return flen > slen && strcmp(filename + flen - slen, suffix) == 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<has_suffix>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  movslq %eax,%rbx
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    %eax,%ebx
  jg     L0
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L0:
  cltq
  mov    %r12,%rsi
  sub    %rax,%rbx
  lea    0x0(%rbp,%rbx,1),%rdi
  call   <strcmp@plt>
  xor    %r8d,%r8d
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  sete   %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int has_suffix(char *filename, char *suffix) {
  int flen = strlen(filename);
  int slen = strlen(suffix);
  return flen > slen && strcmp(filename + flen - slen, suffix) == 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<has_suffix>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  movslq %eax,%rbx
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    %eax,%ebx
  jg     L1
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L1:
  cltq
  mov    %r12,%rsi
  sub    %rax,%rbx
  lea    0x0(%rbp,%rbx,1),%rdi
  call   <strcmp@plt>
  xor    %r8d,%r8d
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  sete   %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int has_suffix(char *filename, char *suffix) {
  int flen = strlen(filename);
  int slen = strlen(suffix);
  return flen > slen && strcmp(filename + flen - slen, suffix) == 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<int_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x30,-0x4(%rbp)
  jmp    L25
L23:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  subl   $0x1,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L25:
  cmpl   $0x0,-0x4(%rbp)
  js     L8
  cmpl   $0x0,-0x14(%rbp)
  jne    L23
L8:
  mov    -0x20(%rbp),%rax
  add    $0x31,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  pop    %rbp
  ret
```

assistant: ```c
char *int_str(int num, char *str) {
  int i;
  for (i = 48; i >= 0 && num; i--, num /= 10)
    str[i] = num % 10 + 48;
  str[49] = '\0';
  return &(str[i + 1]);
}
```
user: What is the c source code of the assembly code below?

```asm
<int_str>:
  endbr64
  mov    %edi,%ecx
  mov    %rsi,%rax
  test   %edi,%edi
  je     L25
  mov    $0x2f,%edi
L14:
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,0x1(%rax,%rdi,1)
  mov    %edi,%r8d
  mov    %edx,%ecx
  mov    %edi,%esi
  not    %esi
  shr    $0x1f,%esi
  test   %edx,%edx
  setne  %dl
  sub    $0x1,%rdi
  test   %sil,%dl
  jne    L14
L5:
  movb   $0x0,0x31(%rax)
  add    $0x1,%r8d
  movslq %r8d,%r8
  add    %r8,%rax
  ret
L25:
  mov    $0x30,%r8d
  jmp    L5
```

assistant: ```c
char *int_str(int num, char *str) {
  int i;
  for (i = 48; i >= 0 && num; i--, num /= 10)
    str[i] = num % 10 + 48;
  str[49] = '\0';
  return &(str[i + 1]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<int_str>:
  endbr64
  mov    %edi,%edx
  mov    %rsi,%r9
  test   %edi,%edi
  je     L18
  mov    %esi,%edi
  lea    0x30(%rsi),%rcx
  not    %edi
  nopl   (%rax)
L13:
  movslq %edx,%rax
  mov    %edx,%esi
  mov    %rcx,%r8
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,(%rcx)
  mov    %eax,%edx
  lea    (%rdi,%rcx,1),%eax
  not    %eax
  shr    $0x1f,%eax
  test   %edx,%edx
  setne  %sil
  sub    $0x1,%rcx
  test   %al,%sil
  jne    L13
  movb   $0x0,0x31(%r9)
  mov    %r8,%rax
  ret
  xchg   %ax,%ax
L18:
  lea    0x31(%rsi),%r8
  movb   $0x0,0x31(%r9)
  mov    %r8,%rax
  ret
```

assistant: ```c
char *int_str(int num, char *str) {
  int i;
  for (i = 48; i >= 0 && num; i--, num /= 10)
    str[i] = num % 10 + 48;
  str[49] = '\0';
  return &(str[i + 1]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<int_str>:
  endbr64
  mov    %edi,%edx
  mov    %rsi,%r9
  test   %edi,%edi
  je     L4
  mov    %esi,%edi
  lea    0x30(%rsi),%rcx
  not    %edi
  nopl   (%rax)
L10:
  movslq %edx,%rax
  mov    %edx,%esi
  mov    %rcx,%r8
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,(%rcx)
  mov    %eax,%edx
  lea    (%rdi,%rcx,1),%eax
  not    %eax
  shr    $0x1f,%eax
  test   %edx,%edx
  setne  %sil
  sub    $0x1,%rcx
  test   %al,%sil
  jne    L10
  movb   $0x0,0x31(%r9)
  mov    %r8,%rax
  ret
  xchg   %ax,%ax
L4:
  lea    0x31(%rsi),%r8
  movb   $0x0,0x31(%r9)
  mov    %r8,%rax
  ret
```

assistant: ```c
char *int_str(int num, char *str) {
  int i;
  for (i = 48; i >= 0 && num; i--, num /= 10)
    str[i] = num % 10 + 48;
  str[49] = '\0';
  return &(str[i + 1]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
L1:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0x71,%al
  jne    L1
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  char c;
  while ((getchar()) != 'q')
    ;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
L9:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0x71,%al
  jne    L9
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  char c;
  while ((getchar()) != 'q')
    ;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x71,%al
  jne    L4
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  char c;
  while ((getchar()) != 'q')
    ;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x71,%al
  jne    L3
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  char c;
  while ((getchar()) != 'q')
    ;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ispalindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L5:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edi
  mov    -0x8(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  lea    (%rdi,%rcx,1),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
L9:
  cmpl   $0x0,-0x8(%rbp)
  jg     L5
  mov    -0x14(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  sete   %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
bool ispalindrome(uint64_t x) {
  uint64_t t = x;
  uint64_t y = 0;
  while (t > 0) {
    y = (y * 10) + (t % 10);
    t = t / 10;
  }
  return x == y;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ispalindrome>:
  endbr64
  test   %edi,%edi
  jle    L1
  mov    %edi,%edx
  mov    $0x0,%ecx
L22:
  lea    (%rcx,%rcx,4),%r8d
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%esi
  sub    %ecx,%esi
  lea    (%rsi,%r8,2),%ecx
  mov    %edx,%esi
  mov    %eax,%edx
  cmp    $0x9,%esi
  jg     L22
L13:
  cmp    %ecx,%edi
  sete   %al
  movzbl %al,%eax
  ret
L1:
  mov    $0x0,%ecx
  jmp    L13
```

assistant: ```c
bool ispalindrome(uint64_t x) {
  uint64_t t = x;
  uint64_t y = 0;
  while (t > 0) {
    y = (y * 10) + (t % 10);
    t = t / 10;
  }
  return x == y;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ispalindrome>:
  endbr64
  test   %edi,%edi
  jle    L9
  mov    %edi,%edx
  xor    %ecx,%ecx
  mov    $0xcccccccd,%r9d
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %edx,%eax
  lea    (%rcx,%rcx,4),%r8d
  mov    %edx,%esi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%esi
  lea    (%rsi,%r8,2),%ecx
  mov    %edx,%esi
  mov    %eax,%edx
  cmp    $0x9,%esi
  jg     L13
  xor    %eax,%eax
  cmp    %ecx,%edi
  sete   %al
  ret
  nopl   0x0(%rax)
L9:
  xor    %ecx,%ecx
  xor    %eax,%eax
  cmp    %ecx,%edi
  sete   %al
  ret
```

assistant: ```c
bool ispalindrome(uint64_t x) {
  uint64_t t = x;
  uint64_t y = 0;
  while (t > 0) {
    y = (y * 10) + (t % 10);
    t = t / 10;
  }
  return x == y;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ispalindrome>:
  endbr64
  test   %edi,%edi
  jle    L4
  mov    %edi,%edx
  xor    %ecx,%ecx
  mov    $0xcccccccd,%r9d
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %edx,%eax
  lea    (%rcx,%rcx,4),%r8d
  mov    %edx,%esi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%esi
  lea    (%rsi,%r8,2),%ecx
  mov    %edx,%esi
  mov    %eax,%edx
  cmp    $0x9,%esi
  jg     L0
  xor    %eax,%eax
  cmp    %ecx,%edi
  sete   %al
  ret
  nopl   0x0(%rax)
L4:
  xor    %ecx,%ecx
  xor    %eax,%eax
  cmp    %ecx,%edi
  sete   %al
  ret
```

assistant: ```c
bool ispalindrome(uint64_t x) {
  uint64_t t = x;
  uint64_t y = 0;
  while (t > 0) {
    y = (y * 10) + (t % 10);
    t = t / 10;
  }
  return x == y;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sspm_append_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsbl -0x1c(%rbp),%edx
  lea    -0xc(%rbp),%rax
  mov    %edx,%ecx
  lea    D3(%rip),%rdx
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sspm_append_string@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
type_guessed: string

tool: D3: =%02X

assistant: ```c
void sspm_append_hex(struct sspm_buffer *buf, char ch) {
  char tmp[4];
  snprintf(tmp, sizeof(tmp), "=%02X", ch);
  sspm_append_string(buf, tmp);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sspm_append_hex>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movsbl %sil,%ecx
  lea    0x4(%rsp),%rbp
  lea    D8(%rip),%rdx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <snprintf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sspm_append_string@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D8
data_type: string

tool: D8: =%02X

assistant: ```c
void sspm_append_hex(struct sspm_buffer *buf, char ch) {
  char tmp[4];
  snprintf(tmp, sizeof(tmp), "=%02X", ch);
  sspm_append_string(buf, tmp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sspm_append_hex>:
  endbr64
  push   %r12
  movsbl %sil,%ecx
  lea    D8(%rip),%rdx
  mov    $0x4,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdi
  call   <snprintf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <sspm_append_string@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D8
type_guessed: string

tool: D8: =%02X

assistant: ```c
void sspm_append_hex(struct sspm_buffer *buf, char ch) {
  char tmp[4];
  snprintf(tmp, sizeof(tmp), "=%02X", ch);
  sspm_append_string(buf, tmp);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sspm_append_hex>:
  endbr64
  push   %r12
  movsbl %sil,%ecx
  lea    D8(%rip),%rdx
  mov    $0x4,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdi
  call   <snprintf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <sspm_append_string@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

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

tool: D8: =%02X

assistant: ```c
void sspm_append_hex(struct sspm_buffer *buf, char ch) {
  char tmp[4];
  snprintf(tmp, sizeof(tmp), "=%02X", ch);
  sspm_append_string(buf, tmp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strtrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strspn@plt>
  add    %rax,-0x28(%rbp)
  jmp    L21
L2:
  mov    -0x28(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L21:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L12
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strspn@plt>
  mov    %rax,%rbx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %rax,%rbx
  jne    L2
L12:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *strtrim(char *str, char *ws) {
  char *s = str;
  str += strspn(str, ws);
  while (*str != '\0' && strspn(str, ws) != strlen(str))
    *(s++) = *(str++);
  *s = '\0';
  return (str);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strtrim>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%r14
  call   <strspn@plt>
  lea    (%r12,%rax,1),%rbx
  movzbl (%rbx),%r13d
  test   %r13b,%r13b
  je     L27
L23:
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <strspn@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    %rax,%rbp
  je     L27
  add    $0x1,%rbx
  add    $0x1,%r12
  mov    %r13b,-0x1(%r12)
  movzbl (%rbx),%r13d
  test   %r13b,%r13b
  jne    L23
L27:
  movb   $0x0,(%r12)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *strtrim(char *str, char *ws) {
  char *s = str;
  str += strspn(str, ws);
  while (*str != '\0' && strspn(str, ws) != strlen(str))
    *(s++) = *(str++);
  *s = '\0';
  return (str);
}
```
user: What is the c source code of the assembly code below?

```asm
<strtrim>:
  endbr64
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strspn@plt>
  lea    0x0(%rbp,%rax,1),%r12
  movzbl (%r12),%r14d
  test   %r14b,%r14b
  jne    L21
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L14:
  add    $0x1,%r12
  mov    %r14b,0x0(%rbp)
  add    $0x1,%rbp
  movzbl (%r12),%r14d
  test   %r14b,%r14b
  je     L10
L21:
  mov    %r12,%rdi
  mov    %r13,%rsi
  call   <strspn@plt>
  mov    %r12,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  cmp    %rax,%rbx
  jne    L14
L10:
  pop    %rbx
  mov    %r12,%rax
  movb   $0x0,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *strtrim(char *str, char *ws) {
  char *s = str;
  str += strspn(str, ws);
  while (*str != '\0' && strspn(str, ws) != strlen(str))
    *(s++) = *(str++);
  *s = '\0';
  return (str);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strtrim>:
  endbr64
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strspn@plt>
  lea    0x0(%rbp,%rax,1),%r12
  movzbl (%r12),%r14d
  test   %r14b,%r14b
  jne    L30
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L29:
  add    $0x1,%r12
  mov    %r14b,0x0(%rbp)
  add    $0x1,%rbp
  movzbl (%r12),%r14d
  test   %r14b,%r14b
  je     L5
L30:
  mov    %r12,%rdi
  mov    %r13,%rsi
  call   <strspn@plt>
  mov    %r12,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  cmp    %rax,%rbx
  jne    L29
L5:
  pop    %rbx
  mov    %r12,%rax
  movb   $0x0,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *strtrim(char *str, char *ws) {
  char *s = str;
  str += strspn(str, ws);
  while (*str != '\0' && strspn(str, ws) != strlen(str))
    *(s++) = *(str++);
  *s = '\0';
  return (str);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<check_center>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  jmp    L70
  nopl   (%rax)
L95:
  add    $0x1,%rbp
L70:
  movsbl 0x0(%rbp),%edi
  call   <iskuuhaku@plt>
  test   %rax,%rax
  jne    L95
  cmpb   $0x0,0x0(%rbp)
  jne    L106
L61:
  mov    $0x1,%eax
L86:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L106:
  mov    $0x4,%edx
  lea    D20(%rip),%rsi
  lea    0x4(%rbp),%rbx
  mov    %rbp,%rdi
  call   <str2ncmp@plt>
  test   %rax,%rax
  je     L0
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L60:
  add    $0x1,%rbx
L0:
  movsbl (%rbx),%edi
  call   <iskuuhaku@plt>
  test   %rax,%rax
  jne    L60
  cmpb   $0x0,(%rbx)
  je     L61
L34:
  lea    D24(%rip),%rsi
  mov    %rbp,%rdi
  call   <strst2@plt>
  lea    -0x1(%rax),%rbx
  test   %rax,%rax
  jne    L15
  jmp    L59
L129:
  sub    $0x1,%rbx
L15:
  movsbl (%rbx),%edi
  call   <iskuuhaku@plt>
  test   %rax,%rax
  jne    L129
L7:
  movzbl 0x1(%rbx),%r12d
  movb   $0x0,0x1(%rbx)
  lea    D15(%rip),%rsi
  mov    %rbp,%rdi
  call   <strst2@plt>
  test   %rax,%rax
  je     L17
  mov    %r12b,0x1(%rbx)
L155:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L59:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  jmp    L123
  nopw   0x0(%rax,%rax,1)
L46:
  sub    $0x1,%rbx
L123:
  movsbl (%rbx),%edi
  call   <iskuuhaku@plt>
  test   %rax,%rax
  jne    L46
  jmp    L7
L17:
  lea    D4(%rip),%rsi
  mov    %rbp,%rdi
  call   <strst2@plt>
  mov    %r12b,0x1(%rbx)
  test   %rax,%rax
  jne    L155
  mov    $0xffffffff,%eax
  jmp    L86
```

tool_calls: name: retrieve_data
arguments: label: D20
type_guessed: string
name: retrieve_data
arguments: label: D24
type_guessed: string
name: retrieve_data
arguments: label: D15
type_guessed: string
name: retrieve_data
arguments: label: D4
type_guessed: string

tool: D20: void

tool: D24: ,

tool: D15:  

tool: D4: 	

assistant: ```c
int check_center(char *s) {
  char *b, *t;
  char bak;
  int naf;
  int len;
  naf = 0;
  while (iskuuhaku(*s)) {
    ++s;
  }
  if (*s == 0) {
    return 1;
  }
  if (str2ncmp(s, "void", 4) == 0) {
    t = s + 4;
    while (iskuuhaku(*t)) {
      ++t;
    }
    if (*t == 0) {
      return 1;
    }
  }
  if ((b = strst2(s, ",")) != 0) {
    t = b - 1;
    while (iskuuhaku(*t)) {
      --t;
    }
    bak = *(t + 1);
    *(t + 1) = (char)0;
    if (strst2(s, " ") || strst2(s, "\t")) {
    } else {
      naf = 1;
    }
    *(t + 1) = bak;
  } else {
    len = strlen(s);
    t = s + len - 1;
    while (iskuuhaku(*t)) {
      --t;
    }
    bak = *(t + 1);
    *(t + 1) = (char)0;
    if (strst2(s, " ") || strst2(s, "\t")) {
    } else {
      naf = 1;
    }
    *(t + 1) = bak;
  }
  if (naf) {
    return -1;
  }
  return 0;
}
```