ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<help>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <error@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
data_type: string

tool: D7: Usage:
    convert <option> <input> <output>
Options:
    -a, --ascii: convert to ascii format
    -b, --big-endian: convert to big-endian format
    -l, --little-endian: convert to little-endian format


assistant: ```c
void help(void) {
  error("Usage:\n"
        "    convert <option> <input> <output>\n"
        "Options:\n"
        "    -a, --ascii: convert to ascii format\n"
        "    -b, --big-endian: convert to big-endian format\n"
        "    -l, --little-endian: convert to little-endian format\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<help>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <error@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D0: Usage:
    convert <option> <input> <output>
Options:
    -a, --ascii: convert to ascii format
    -b, --big-endian: convert to big-endian format
    -l, --little-endian: convert to little-endian format


assistant: ```c
void help(void) {
  error("Usage:\n"
        "    convert <option> <input> <output>\n"
        "Options:\n"
        "    -a, --ascii: convert to ascii format\n"
        "    -b, --big-endian: convert to big-endian format\n"
        "    -l, --little-endian: convert to little-endian format\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<help>:
  endbr64
  lea    D4(%rip),%rdi
  jmp    1050 <error@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D4
type_infered: string

tool: D4: Usage:
    convert <option> <input> <output>
Options:
    -a, --ascii: convert to ascii format
    -b, --big-endian: convert to big-endian format
    -l, --little-endian: convert to little-endian format


assistant: ```c
void help(void) {
  error("Usage:\n"
        "    convert <option> <input> <output>\n"
        "Options:\n"
        "    -a, --ascii: convert to ascii format\n"
        "    -b, --big-endian: convert to big-endian format\n"
        "    -l, --little-endian: convert to little-endian format\n");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<help>:
  endbr64
  lea    D8(%rip),%rdi
  jmp    1050 <error@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D8
data_type: string

tool: D8: Usage:
    convert <option> <input> <output>
Options:
    -a, --ascii: convert to ascii format
    -b, --big-endian: convert to big-endian format
    -l, --little-endian: convert to little-endian format


assistant: ```c
void help(void) {
  error("Usage:\n"
        "    convert <option> <input> <output>\n"
        "Options:\n"
        "    -a, --ascii: convert to ascii format\n"
        "    -b, --big-endian: convert to big-endian format\n"
        "    -l, --little-endian: convert to little-endian format\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L9
L30:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0x989680,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L22
L5:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D29(%rip),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L22:
  cmpl   $0x2,-0x14(%rbp)
  jle    L5
  addl   $0x1,-0x18(%rbp)
L9:
  cmpl   $0x6,-0x18(%rbp)
  jle    L30
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D29
type: string
name: unpack_data
arguments: tag: D11
type: string
name: unpack_data
arguments: tag: D8
type: string

tool: D29: sasum

tool: D11: # c::%s:len=%d


tool: D8: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "sasum", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D11(%rip),%r15
  jmp    L28
L2:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L27
L28:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0x989680,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L1:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D29(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L1
  jmp    L2
L27:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_label: D11
type: string
name: read_data
arguments: data_label: D29
type: string
name: read_data
arguments: data_label: D21
type: string

tool: D11: sasum

tool: D29: # c::%s:len=%d


tool: D21: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "sasum", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D13(%rip),%r15
  push   %r14
  lea    D10(%rip),%r14
  push   %r13
  lea    D23(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L17:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0x989680,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L11:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L11
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L17
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D13
data_type: string
name: unpack_data
arguments: data_tag: D10
data_type: string
name: unpack_data
arguments: data_tag: D23
data_type: string

tool: D13: sasum

tool: D10: # c::%s:len=%d


tool: D23: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "sasum", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D22(%rip),%r15
  push   %r14
  lea    D10(%rip),%r14
  push   %r13
  lea    D5(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L0:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0x989680,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L4:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L4
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L0
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D22
data_type: string
name: extract_symbolic_data
arguments: tag: D10
data_type: string
name: extract_symbolic_data
arguments: tag: D5
data_type: string

tool: D22: sasum

tool: D10: # c::%s:len=%d


tool: D5: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "sasum", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strDelNL>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L3:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L28
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
L28:
  addq   $0x1,-0x18(%rbp)
L5:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int strDelNL(char *suchen) {
  int counter = 0;
  while (*suchen != '\0') {
    if (*suchen == '\n')
      *suchen = '\0';
    suchen++;
  }
  return counter;
}
```
user: What is the c source code of the assembly code below?

```asm
<strDelNL>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L12
L4:
  mov    $0x0,%eax
  ret
L20:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L4
L12:
  cmp    $0xa,%al
  jne    L20
  movb   $0x0,(%rdi)
  jmp    L20
```

assistant: ```c
int strDelNL(char *suchen) {
  int counter = 0;
  while (*suchen != '\0') {
    if (*suchen == '\n')
      *suchen = '\0';
    suchen++;
  }
  return counter;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strDelNL>:
  endbr64
  jmp    L0
  cs nopw 0x0(%rax,%rax,1)
L16:
  cmp    $0xa,%al
  jne    L23
  movb   $0x0,(%rdi)
L23:
  add    $0x1,%rdi
L0:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L16
  xor    %eax,%eax
  ret
```

assistant: ```c
int strDelNL(char *suchen) {
  int counter = 0;
  while (*suchen != '\0') {
    if (*suchen == '\n')
      *suchen = '\0';
    suchen++;
  }
  return counter;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strDelNL>:
  endbr64
  jmp    L36
  cs nopw 0x0(%rax,%rax,1)
L15:
  add    $0x1,%rdi
L36:
  movzbl (%rdi),%eax
  test   %al,%al
  je     L39
L12:
  cmp    $0xa,%al
  jne    L15
  movb   $0x0,(%rdi)
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L12
L39:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strDelNL(char *suchen) {
  int counter = 0;
  while (*suchen != '\0') {
    if (*suchen == '\n')
      *suchen = '\0';
    suchen++;
  }
  return counter;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strsplit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  movl   $0x0,-0x24(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x48(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,(%rbx)
  jmp    L0
L14:
  mov    -0x48(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L7
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,(%rbx)
L0:
  cmpq   $0x0,-0x20(%rbp)
  jne    L14
  jmp    L23
L7:
  nop
L23:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x24(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int strsplit(const char *str, char *parts[], const char *delimiter) {
  char *pch;
  int i = 0;
  char *tmp = strdup(str);
  pch = strtok(tmp, delimiter);
  parts[i++] = strdup(pch);
  while (pch) {
    pch = strtok(NULL, delimiter);
    if (NULL == pch)
      break;
    parts[i++] = strdup(pch);
  }
  free(tmp);
  free(pch);
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strsplit>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  call   <strdup@plt>
  mov    %rax,%r14
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,(%r12)
  mov    $0x1,%ebx
  jmp    L16
L0:
  call   <strdup@plt>
  mov    %rax,(%r12,%rbx,8)
  add    $0x1,%rbx
L16:
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L0
  mov    %r14,%rdi
  call   <free@plt>
  mov    $0x0,%edi
  call   <free@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int strsplit(const char *str, char *parts[], const char *delimiter) {
  char *pch;
  int i = 0;
  char *tmp = strdup(str);
  pch = strtok(tmp, delimiter);
  parts[i++] = strdup(pch);
  while (pch) {
    pch = strtok(NULL, delimiter);
    if (NULL == pch)
      break;
    parts[i++] = strdup(pch);
  }
  free(tmp);
  free(pch);
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strsplit>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <strdup@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r14
  call   <strtok@plt>
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,(%r12)
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L2:
  call   <strdup@plt>
  mov    %rax,(%r12,%rbx,8)
  add    $0x1,%rbx
L4:
  xor    %edi,%edi
  mov    %rbp,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L2
  mov    %r14,%rdi
  call   <free@plt>
  xor    %edi,%edi
  call   <free@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int strsplit(const char *str, char *parts[], const char *delimiter) {
  char *pch;
  int i = 0;
  char *tmp = strdup(str);
  pch = strtok(tmp, delimiter);
  parts[i++] = strdup(pch);
  while (pch) {
    pch = strtok(NULL, delimiter);
    if (NULL == pch)
      break;
    parts[i++] = strdup(pch);
  }
  free(tmp);
  free(pch);
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strsplit>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <strdup@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r14
  call   <strtok@plt>
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,(%r12)
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L11:
  call   <strdup@plt>
  mov    %rax,(%r12,%rbx,8)
  add    $0x1,%rbx
L2:
  xor    %edi,%edi
  mov    %rbp,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L11
  mov    %r14,%rdi
  call   <free@plt>
  xor    %edi,%edi
  call   <free@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int strsplit(const char *str, char *parts[], const char *delimiter) {
  char *pch;
  int i = 0;
  char *tmp = strdup(str);
  pch = strtok(tmp, delimiter);
  parts[i++] = strdup(pch);
  while (pch) {
    pch = strtok(NULL, delimiter);
    if (NULL == pch)
      break;
    parts[i++] = strdup(pch);
  }
  free(tmp);
  free(pch);
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Double2TStg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  call   <Double2Stg@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <RTrimDecimal@plt>
  nop
  leave
  ret
```

assistant: ```c
void Double2TStg(char *lps, double factor) {
  Double2Stg(lps, factor);
  RTrimDecimal(lps);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Double2TStg>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <Double2Stg@plt>
  mov    %rbx,%rdi
  call   <RTrimDecimal@plt>
  pop    %rbx
  ret
```

assistant: ```c
void Double2TStg(char *lps, double factor) {
  Double2Stg(lps, factor);
  RTrimDecimal(lps);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<Double2TStg>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <Double2Stg@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <RTrimDecimal@plt>
```

assistant: ```c
void Double2TStg(char *lps, double factor) {
  Double2Stg(lps, factor);
  RTrimDecimal(lps);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Double2TStg>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <Double2Stg@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <RTrimDecimal@plt>
```

assistant: ```c
void Double2TStg(char *lps, double factor) {
  Double2Stg(lps, factor);
  RTrimDecimal(lps);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<RSA_X931_hash_id>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x83,-0x4(%rbp)
  je     L11
  cmpl   $0x83,-0x4(%rbp)
  jg     L92
  cmpl   $0x82,-0x4(%rbp)
  je     L18
  cmpl   $0x82,-0x4(%rbp)
  jg     L92
  cmpl   $0x80,-0x4(%rbp)
  je     L98
  cmpl   $0x81,-0x4(%rbp)
  je     L0
  jmp    L92
L11:
  mov    $0x33,%eax
  jmp    L9
L18:
  mov    $0x34,%eax
  jmp    L9
L0:
  mov    $0x36,%eax
  jmp    L9
L98:
  mov    $0x35,%eax
  jmp    L9
L92:
  mov    $0xffffffff,%eax
L9:
  pop    %rbp
  ret
```

assistant: ```c
int RSA_X931_hash_id(int nid) {
  switch (nid) {
  case NID_sha1:
    return 0x33;
  case NID_sha256:
    return 0x34;
  case NID_sha384:
    return 0x36;
  case NID_sha512:
    return 0x35;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RSA_X931_hash_id>:
  endbr64
  cmp    $0x82,%edi
  je     L2
  jg     L29
  mov    $0x35,%eax
  cmp    $0x80,%edi
  je     L26
  cmp    $0x81,%edi
  mov    $0xffffffff,%eax
  mov    $0x36,%edx
  cmove  %edx,%eax
  ret
L29:
  cmp    $0x83,%edi
  mov    $0xffffffff,%eax
  mov    $0x33,%edx
  cmove  %edx,%eax
  ret
L2:
  mov    $0x34,%eax
L26:
  ret
```

assistant: ```c
int RSA_X931_hash_id(int nid) {
  switch (nid) {
  case NID_sha1:
    return 0x33;
  case NID_sha256:
    return 0x34;
  case NID_sha384:
    return 0x36;
  case NID_sha512:
    return 0x35;
  }
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<RSA_X931_hash_id>:
  endbr64
  add    $0xffffff80,%edi
  mov    $0xffffffff,%eax
  cmp    $0x3,%edi
  ja     L6
  lea    D0(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L6:
  ret
```

tool_calls: name: disassemble_data
arguments: label: D0
type_guessed: string

tool: D0: 5

assistant: ```c
int RSA_X931_hash_id(int nid) {
  switch (nid) {
  case NID_sha1:
    return 0x33;
  case NID_sha256:
    return 0x34;
  case NID_sha384:
    return 0x36;
  case NID_sha512:
    return 0x35;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RSA_X931_hash_id>:
  endbr64
  add    $0xffffff80,%edi
  mov    $0xffffffff,%eax
  cmp    $0x3,%edi
  ja     L0
  lea    D9(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L0:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_guessed: string

tool: D9: 5

assistant: ```c
int RSA_X931_hash_id(int nid) {
  switch (nid) {
  case NID_sha1:
    return 0x33;
  case NID_sha256:
    return 0x34;
  case NID_sha384:
    return 0x36;
  case NID_sha512:
    return 0x35;
  }
  return -1;
}
```