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

```asm
<start_upstream>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x440,%rsp
  mov    %edi,-0x434(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  call   <fork@plt>
  mov    %eax,%esi
  mov    $0x66,%edi
  call   <_ensure@plt>
  mov    %eax,-0x420(%rbp)
  cmpl   $0x0,-0x420(%rbp)
  jne    L22
L58:
  lea    -0x410(%rbp),%rax
  mov    %rax,-0x418(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    -0x418(%rbp),%rcx
  mov    $0x400,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,%esi
  mov    $0x6b,%edi
  call   <_ensure@plt>
  mov    %eax,-0x424(%rbp)
  cmpl   $0x0,-0x424(%rbp)
  je     L2
  jmp    L30
L36:
  mov    -0x424(%rbp),%edx
  mov    -0x418(%rbp),%rcx
  mov    -0x434(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %eax,%esi
  mov    $0x6e,%edi
  call   <_ensure@plt>
  mov    %eax,-0x41c(%rbp)
  mov    -0x41c(%rbp),%eax
  cltq
  add    %rax,-0x418(%rbp)
  mov    -0x41c(%rbp),%eax
  sub    %eax,-0x424(%rbp)
L30:
  cmpl   $0x0,-0x424(%rbp)
  jg     L36
  jmp    L58
L2:
  nop
  mov    $0x0,%edi
  call   <exit@plt>
L22:
  mov    -0x420(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int start_upstream(int ptmaster) {
  int ret = _ensure(102, fork());
  if (ret == 0) {
    char buf[1024];
    while (1) {
      char *bufptr = buf;
      int size = _ensure(107, read(STDIN_FILENO, bufptr, sizeof(buf)));
      if (size == 0)
        break;
      while (size > 0) {
        int rc = _ensure(110, write(ptmaster, bufptr, size));
        bufptr += rc;
        size -= rc;
      }
    }
    exit(0);
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<start_upstream>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %edi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <fork@plt>
  mov    %eax,%esi
  mov    $0x66,%edi
  call   <_ensure@plt>
  mov    %eax,%r14d
  test   %eax,%eax
  jne    L14
  mov    %rsp,%r13
L54:
  mov    $0x400,%edx
  mov    %r13,%rsi
  mov    D6(%rip),%rax
  mov    (%rax),%edi
  call   <read@plt>
  mov    %eax,%esi
  mov    $0x6b,%edi
  call   <_ensure@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L12
  jle    L54
  mov    %rsp,%rbp
L58:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <write@plt>
  mov    %eax,%esi
  mov    $0x6e,%edi
  call   <_ensure@plt>
  movslq %eax,%rdx
  add    %rdx,%rbp
  sub    %eax,%ebx
  test   %ebx,%ebx
  jg     L58
  jmp    L54
L12:
  mov    $0x0,%edi
  call   <exit@plt>
L14:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  mov    %r14d,%eax
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L51:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int start_upstream(int ptmaster) {
  int ret = _ensure(102, fork());
  if (ret == 0) {
    char buf[1024];
    while (1) {
      char *bufptr = buf;
      int size = _ensure(107, read(STDIN_FILENO, bufptr, sizeof(buf)));
      if (size == 0)
        break;
      while (size > 0) {
        int rc = _ensure(110, write(ptmaster, bufptr, size));
        bufptr += rc;
        size -= rc;
      }
    }
    exit(0);
  }
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<start_upstream>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <fork@plt>
  mov    $0x66,%edi
  mov    %eax,%esi
  call   <_ensure@plt>
  mov    %eax,%r13d
  test   %eax,%eax
  jne    L51
  mov    D8(%rip),%r12
  mov    %rsp,%rbp
  nopl   0x0(%rax)
L37:
  mov    (%r12),%edi
  mov    $0x400,%edx
  mov    %rbp,%rsi
  call   <read@plt>
  mov    $0x6b,%edi
  mov    %eax,%esi
  call   <_ensure@plt>
  mov    %eax,%r14d
  test   %eax,%eax
  je     L49
  jle    L37
  mov    %rbp,%r15
  nopl   0x0(%rax)
L29:
  mov    %r14d,%edx
  mov    %r15,%rsi
  mov    %ebx,%edi
  call   <write@plt>
  mov    $0x6e,%edi
  mov    %eax,%esi
  call   <_ensure@plt>
  movslq %eax,%rdx
  sub    %eax,%r14d
  add    %rdx,%r15
  test   %r14d,%r14d
  jg     L29
  jmp    L37
  nopl   0x0(%rax)
L49:
  xor    %edi,%edi
  call   <exit@plt>
L51:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x418,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int start_upstream(int ptmaster) {
  int ret = _ensure(102, fork());
  if (ret == 0) {
    char buf[1024];
    while (1) {
      char *bufptr = buf;
      int size = _ensure(107, read(STDIN_FILENO, bufptr, sizeof(buf)));
      if (size == 0)
        break;
      while (size > 0) {
        int rc = _ensure(110, write(ptmaster, bufptr, size));
        bufptr += rc;
        size -= rc;
      }
    }
    exit(0);
  }
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<start_upstream>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <fork@plt>
  mov    $0x66,%edi
  mov    %eax,%esi
  call   <_ensure@plt>
  mov    %eax,%r13d
  test   %eax,%eax
  jne    L1
  mov    D7(%rip),%r12
  mov    %rsp,%rbp
  nopl   0x0(%rax)
L20:
  mov    (%r12),%edi
  mov    $0x400,%edx
  mov    %rbp,%rsi
  call   <read@plt>
  mov    $0x6b,%edi
  mov    %eax,%esi
  call   <_ensure@plt>
  mov    %eax,%r14d
  test   %eax,%eax
  je     L42
  jle    L20
  mov    %rbp,%r15
  nopl   0x0(%rax)
L15:
  mov    %r14d,%edx
  mov    %r15,%rsi
  mov    %ebx,%edi
  call   <write@plt>
  mov    $0x6e,%edi
  mov    %eax,%esi
  call   <_ensure@plt>
  movslq %eax,%rdx
  sub    %eax,%r14d
  add    %rdx,%r15
  test   %r14d,%r14d
  jg     L15
  jmp    L20
  nopl   0x0(%rax)
L42:
  xor    %edi,%edi
  call   <exit@plt>
L1:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x418,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int start_upstream(int ptmaster) {
  int ret = _ensure(102, fork());
  if (ret == 0) {
    char buf[1024];
    while (1) {
      char *bufptr = buf;
      int size = _ensure(107, read(STDIN_FILENO, bufptr, sizeof(buf)));
      if (size == 0)
        break;
      while (size > 0) {
        int rc = _ensure(110, write(ptmaster, bufptr, size));
        bufptr += rc;
        size -= rc;
      }
    }
    exit(0);
  }
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<format_date_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpl   $0x9,-0x14(%rbp)
  jle    L16
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  lea    D16(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  jmp    L2
L16:
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  lea    D9(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
L2:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D16
datatype: string
name: read_data
arguments: data_tag: D9
datatype: string

tool: D16: %d

tool: D9: 0%d

assistant: ```c
char *format_date_int(int val) {
  char *result = (char *)malloc(2 * sizeof(char));
  if (val > 9) {
    sprintf(result, "%d", val);
  } else {
    sprintf(result, "0%d", val);
  }
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<format_date_int>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  cmp    $0x9,%ebp
  jle    L9
  mov    %ebp,%edx
  lea    D11(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
L14:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  mov    %ebp,%edx
  lea    D14(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  jmp    L14
```

tool_calls: name: read_data
arguments: data_tag: D11
data_type: string
name: read_data
arguments: data_tag: D14
data_type: string

tool: D11: %d

tool: D14: 0%d

assistant: ```c
char *format_date_int(int val) {
  char *result = (char *)malloc(2 * sizeof(char));
  if (val > 9) {
    sprintf(result, "%d", val);
  } else {
    sprintf(result, "0%d", val);
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<format_date_int>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  mov    $0x2,%edi
  push   %r12
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13d,%edx
  mov    %rax,%r12
  cmp    $0x9,%r13d
  jle    L2
  lea    D15(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
L2:
  lea    D6(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
data_type: string
name: extract_symbolic_data
arguments: tag: D6
data_type: string

tool: D15: %d

tool: D6: 0%d

assistant: ```c
char *format_date_int(int val) {
  char *result = (char *)malloc(2 * sizeof(char));
  if (val > 9) {
    sprintf(result, "%d", val);
  } else {
    sprintf(result, "0%d", val);
  }
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<format_date_int>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  mov    $0x2,%edi
  push   %r12
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13d,%edx
  mov    %rax,%r12
  cmp    $0x9,%r13d
  jle    L3
  lea    D2(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
L3:
  lea    D16(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
type_guessed: string
name: read_data
arguments: data_label: D16
type_guessed: string

tool: D2: %d

tool: D16: 0%d

assistant: ```c
char *format_date_int(int val) {
  char *result = (char *)malloc(2 * sizeof(char));
  if (val > 9) {
    sprintf(result, "%d", val);
  } else {
    sprintf(result, "0%d", val);
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<__esubm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  addq   $0x10,-0x18(%rbp)
  addq   $0x10,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x2,-0x10(%rbp)
  jmp    L20
L10:
  mov    -0x20(%rbp),%rax
  movzwl (%rax),%eax
  movzwl %ax,%edx
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  movzwl %ax,%ecx
  mov    %rdx,%rax
  sub    %rcx,%rax
  mov    -0xc(%rbp),%edx
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  and    $0x10000,%eax
  test   %rax,%rax
  je     L19
  movl   $0x1,-0xc(%rbp)
  jmp    L18
L19:
  movl   $0x0,-0xc(%rbp)
L18:
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  subq   $0x2,-0x18(%rbp)
  subq   $0x2,-0x20(%rbp)
  addl   $0x1,-0x10(%rbp)
L20:
  cmpl   $0x8,-0x10(%rbp)
  jle    L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void __esubm(const short unsigned int *x, short unsigned int *y) {
  unsigned long a;
  int i;
  unsigned int carry;
  x += (6 + 3) - 1;
  y += (6 + 3) - 1;
  carry = 0;
  for (i = 2; i < (6 + 3); i++) {
    a = (unsigned long)(*y) - (unsigned long)(*x) - carry;
    if (a & 0x10000)
      carry = 1;
    else
      carry = 0;
    *y = (unsigned short)a;
    --x;
    --y;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<__esubm>:
  endbr64
  mov    $0x10,%ecx
  mov    $0x0,%edx
L1:
  movzwl (%rsi,%rcx,1),%eax
  mov    %edx,%edx
  sub    %rdx,%rax
  movzwl (%rdi,%rcx,1),%edx
  sub    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x10,%rdx
  and    $0x1,%edx
  mov    %ax,(%rsi,%rcx,1)
  sub    $0x2,%rcx
  cmp    $0x2,%rcx
  jne    L1
  ret
```

assistant: ```c
void __esubm(const short unsigned int *x, short unsigned int *y) {
  unsigned long a;
  int i;
  unsigned int carry;
  x += (6 + 3) - 1;
  y += (6 + 3) - 1;
  carry = 0;
  for (i = 2; i < (6 + 3); i++) {
    a = (unsigned long)(*y) - (unsigned long)(*x) - carry;
    if (a & 0x10000)
      carry = 1;
    else
      carry = 0;
    *y = (unsigned short)a;
    --x;
    --y;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<__esubm>:
  endbr64
  mov    $0x10,%ecx
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L1:
  movzwl (%rsi,%rcx,1),%eax
  sub    %rdx,%rax
  movzwl (%rdi,%rcx,1),%edx
  sub    %rdx,%rax
  mov    %rax,%rdx
  mov    %ax,(%rsi,%rcx,1)
  sub    $0x2,%rcx
  shr    $0x10,%rdx
  and    $0x1,%edx
  cmp    $0x2,%rcx
  jne    L1
  ret
```

assistant: ```c
void __esubm(const short unsigned int *x, short unsigned int *y) {
  unsigned long a;
  int i;
  unsigned int carry;
  x += (6 + 3) - 1;
  y += (6 + 3) - 1;
  carry = 0;
  for (i = 2; i < (6 + 3); i++) {
    a = (unsigned long)(*y) - (unsigned long)(*x) - carry;
    if (a & 0x10000)
      carry = 1;
    else
      carry = 0;
    *y = (unsigned short)a;
    --x;
    --y;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__esubm>:
  endbr64
  movzwl 0x10(%rsi),%edx
  mov    %rsi,%rax
  movzwl 0x10(%rdi),%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %cx,0x10(%rax)
  movzwl %dx,%ecx
  movzwl 0xe(%rax),%edx
  sub    %rsi,%rcx
  movzwl 0xe(%rdi),%esi
  shr    $0x10,%rcx
  and    $0x1,%ecx
  sub    %rsi,%rdx
  sub    %rcx,%rdx
  mov    %dx,0xe(%rax)
  shr    $0x10,%rdx
  movzwl 0xc(%rdi),%esi
  mov    %rdx,%rcx
  movzwl 0xc(%rax),%edx
  and    $0x1,%ecx
  sub    %rsi,%rdx
  sub    %rcx,%rdx
  mov    %dx,0xc(%rax)
  shr    $0x10,%rdx
  movzwl 0xa(%rdi),%esi
  mov    %rdx,%rcx
  movzwl 0xa(%rax),%edx
  and    $0x1,%ecx
  sub    %rsi,%rdx
  sub    %rcx,%rdx
  mov    %dx,0xa(%rax)
  shr    $0x10,%rdx
  movzwl 0x8(%rdi),%esi
  mov    %rdx,%rcx
  movzwl 0x8(%rax),%edx
  and    $0x1,%ecx
  sub    %rsi,%rdx
  sub    %rcx,%rdx
  mov    %dx,0x8(%rax)
  shr    $0x10,%rdx
  movzwl 0x6(%rdi),%esi
  mov    %rdx,%rcx
  movzwl 0x6(%rax),%edx
  and    $0x1,%ecx
  sub    %rsi,%rdx
  sub    %rcx,%rdx
  movzwl 0x4(%rax),%ecx
  mov    %dx,0x6(%rax)
  movzwl 0x4(%rdi),%esi
  shr    $0x10,%rdx
  and    $0x1,%edx
  sub    %rsi,%rcx
  sub    %rdx,%rcx
  mov    %cx,0x4(%rax)
  ret
```

assistant: ```c
void __esubm(const short unsigned int *x, short unsigned int *y) {
  unsigned long a;
  int i;
  unsigned int carry;
  x += (6 + 3) - 1;
  y += (6 + 3) - 1;
  carry = 0;
  for (i = 2; i < (6 + 3); i++) {
    a = (unsigned long)(*y) - (unsigned long)(*x) - carry;
    if (a & 0x10000)
      carry = 1;
    else
      carry = 0;
    *y = (unsigned short)a;
    --x;
    --y;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strtrim>:
  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)
  jmp    L30
L3:
  subl   $0x1,-0x4(%rbp)
L30:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jg     L33
  cmpl   $0x0,-0x4(%rbp)
  jg     L3
L33:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L48
L40:
  addq   $0x1,-0x18(%rbp)
L48:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jg     L43
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L40
L43:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strtrim(char *str) {
  int tail = strlen(str);
  while (str[tail] <= ' ' && tail > 0)
    tail--;
  str[tail + 1] = 0;
  while (*str <= ' ' && *str != 0)
    str++;
  return str;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strtrim>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  movslq %eax,%rsi
  cmpb   $0x20,(%rbx,%rsi,1)
  jg     L66
  test   %eax,%eax
  jle    L66
  sub    $0x1,%eax
  cltq
L48:
  cmpb   $0x20,(%rbx,%rax,1)
  setle  %cl
  test   %eax,%eax
  setg   %dl
  mov    %rax,%rsi
  sub    $0x1,%rax
  test   %dl,%cl
  jne    L48
L66:
  movb   $0x0,0x1(%rbx,%rsi,1)
  movzbl (%rbx),%eax
  cmp    $0x20,%al
  jg     L64
  test   %al,%al
  je     L64
L30:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  cmp    $0x20,%al
  jg     L64
  test   %al,%al
  jne    L30
L64:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *strtrim(char *str) {
  int tail = strlen(str);
  while (str[tail] <= ' ' && tail > 0)
    tail--;
  str[tail + 1] = 0;
  while (*str <= ' ' && *str != 0)
    str++;
  return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strtrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <strlen@plt>
  movslq %eax,%rsi
  cmpb   $0x20,(%r12,%rsi,1)
  jg     L62
  test   %eax,%eax
  jle    L62
  sub    $0x1,%eax
  cltq
  nopl   0x0(%rax)
L58:
  cmpb   $0x20,(%r12,%rax,1)
  mov    %rax,%rsi
  setle  %cl
  test   %eax,%eax
  setg   %dl
  sub    $0x1,%rax
  test   %dl,%cl
  jne    L58
L62:
  movb   $0x0,0x1(%r12,%rsi,1)
  movzbl (%r12),%eax
  cmp    $0x20,%al
  jle    L50
  jmp    L42
  nopl   0x0(%rax)
L45:
  movzbl 0x1(%r12),%eax
  add    $0x1,%r12
  cmp    $0x20,%al
  jg     L42
L50:
  test   %al,%al
  jne    L45
L42:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *strtrim(char *str) {
  int tail = strlen(str);
  while (str[tail] <= ' ' && tail > 0)
    tail--;
  str[tail + 1] = 0;
  while (*str <= ' ' && *str != 0)
    str++;
  return str;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strtrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <strlen@plt>
  movslq %eax,%rsi
  cmpb   $0x20,(%r12,%rsi,1)
  jg     L11
  test   %eax,%eax
  jle    L11
  sub    $0x1,%eax
  cltq
  nopl   0x0(%rax)
L54:
  cmpb   $0x20,(%r12,%rax,1)
  mov    %rax,%rsi
  setle  %cl
  test   %eax,%eax
  setg   %dl
  sub    $0x1,%rax
  test   %dl,%cl
  jne    L54
L11:
  movb   $0x0,0x1(%r12,%rsi,1)
  movzbl (%r12),%eax
  cmp    $0x20,%al
  jle    L25
  jmp    L38
  nopl   0x0(%rax)
L67:
  movzbl 0x1(%r12),%eax
  add    $0x1,%r12
  cmp    $0x20,%al
  jg     L38
L25:
  test   %al,%al
  jne    L67
L38:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *strtrim(char *str) {
  int tail = strlen(str);
  while (str[tail] <= ' ' && tail > 0)
    tail--;
  str[tail + 1] = 0;
  while (*str <= ' ' && *str != 0)
    str++;
  return str;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<memmove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jae    L1
  mov    -0x14(%rbp),%eax
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rdi
  mov    %eax,%ecx
  mov    %rdx,%rsi
  cld
  rep movsb %ds:(%rsi),%es:(%rdi)
  jmp    L17
L1:
  mov    -0x14(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rcx
  mov    -0x8(%rbp),%rax
  lea    (%rcx,%rax,1),%rdi
  mov    -0x14(%rbp),%eax
  mov    %eax,%ecx
  mov    %rdx,%rsi
  std
  rep movsb %ds:(%rsi),%es:(%rdi)
L17:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memmove(void *dest, const void *src, int n) {
  if (dest < src)
    __asm__("cld\n\t"
            "rep\n\t"
            "movsb" ::"c"(n),
            "S"(src), "D"(dest));
  else
    __asm__("std\n\t"
            "rep\n\t"
            "movsb" ::"c"(n),
            "S"((char *)src + n - 1), "D"((char *)dest + n - 1));
  return dest;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<memmove>:
  endbr64
  mov    %rdi,%rax
  mov    %edx,%ecx
  cmp    %rsi,%rdi
  jae    L1
  cld
  rep movsb %ds:(%rsi),%es:(%rdi)
  ret
L1:
  movslq %edx,%rdi
  sub    $0x1,%rdi
  add    %rdi,%rsi
  add    %rax,%rdi
  std
  rep movsb %ds:(%rsi),%es:(%rdi)
  ret
```

assistant: ```c
void *memmove(void *dest, const void *src, int n) {
  if (dest < src)
    __asm__("cld\n\t"
            "rep\n\t"
            "movsb" ::"c"(n),
            "S"(src), "D"(dest));
  else
    __asm__("std\n\t"
            "rep\n\t"
            "movsb" ::"c"(n),
            "S"((char *)src + n - 1), "D"((char *)dest + n - 1));
  return dest;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<memmove>:
  endbr64
  mov    %rdi,%rax
  mov    %edx,%ecx
  cmp    %rsi,%rdi
  jae    L0
  cld
  rep movsb %ds:(%rsi),%es:(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  movslq %edx,%rdi
  sub    $0x1,%rdi
  add    %rdi,%rsi
  add    %rax,%rdi
  std
  rep movsb %ds:(%rsi),%es:(%rdi)
  ret
```

assistant: ```c
void *memmove(void *dest, const void *src, int n) {
  if (dest < src)
    __asm__("cld\n\t"
            "rep\n\t"
            "movsb" ::"c"(n),
            "S"(src), "D"(dest));
  else
    __asm__("std\n\t"
            "rep\n\t"
            "movsb" ::"c"(n),
            "S"((char *)src + n - 1), "D"((char *)dest + n - 1));
  return dest;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<memmove>:
  endbr64
  mov    %rdi,%rax
  mov    %edx,%ecx
  cmp    %rsi,%rdi
  jae    L6
  cld
  rep movsb %ds:(%rsi),%es:(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  movslq %edx,%rdi
  sub    $0x1,%rdi
  add    %rdi,%rsi
  add    %rax,%rdi
  std
  rep movsb %ds:(%rsi),%es:(%rdi)
  ret
```

assistant: ```c
void *memmove(void *dest, const void *src, int n) {
  if (dest < src)
    __asm__("cld\n\t"
            "rep\n\t"
            "movsb" ::"c"(n),
            "S"(src), "D"(dest));
  else
    __asm__("std\n\t"
            "rep\n\t"
            "movsb" ::"c"(n),
            "S"((char *)src + n - 1), "D"((char *)dest + n - 1));
  return dest;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_uiRcMifClose>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x14,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x7d3,%edx
  lea    D17(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    $0x1,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <iDoRemoteFunc@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  cmpl   $0x0,-0xc(%rbp)
  je     L2
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <vWriteLogTxt@plt>
  mov    $0x1,%eax
  jmp    L17
L2:
  mov    $0x0,%edi
  call   <iGetParaInt@plt>
L17:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D17
type_guessed: string
name: disassemble_data
arguments: label: D19
type_guessed: string

tool: D17: %u()

tool: D19: _uiRcMifClose iDoRemoteFunc ret = %d

assistant: ```c
unsigned int _uiRcMifClose(void) {
  unsigned int uiDataLen, uiRet;
  char *psFunCall = malloc(20);
  uiDataLen = sprintf(psFunCall, "%u()", 2003);
  uiRet = iDoRemoteFunc((unsigned char *)psFunCall, uiDataLen, 1);
  free(psFunCall);
  if (uiRet) {
    vWriteLogTxt("_uiRcMifClose iDoRemoteFunc ret = %d", uiRet);
    return 1;
  }
  return iGetParaInt(0);
}
```