ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<streq>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L26
  mov    $0x1,%eax
L36:
  cmp    %dl,-0x1(%rdi,%rax,1)
  jne    L33
  mov    %eax,%ecx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L36
  jmp    L48
L26:
  mov    $0x0,%ecx
  jmp    L48
L33:
  mov    $0x0,%ecx
L48:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int streq(char str1[], char str2[]) {
  int k;
  k = 0;
  while (str2[k]) {
    if ((str1[k] != str2[k]))
      return (0);
    k++;
  }
  return (k);
}
```
user: Translate the below assembly code into C programming language:

```asm
<streq>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L1
  mov    $0x1,%eax
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L32:
  mov    %eax,%r8d
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L16
L6:
  cmp    %dl,-0x1(%rdi,%rax,1)
  je     L32
L1:
  xor    %r8d,%r8d
L16:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int streq(char str1[], char str2[]) {
  int k;
  k = 0;
  while (str2[k]) {
    if ((str1[k] != str2[k]))
      return (0);
    k++;
  }
  return (k);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<streq>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L6
  mov    $0x1,%eax
  jmp    L25
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %eax,%r8d
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L14
L25:
  cmp    %dl,-0x1(%rdi,%rax,1)
  je     L5
L6:
  xor    %r8d,%r8d
L14:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int streq(char str1[], char str2[]) {
  int k;
  k = 0;
  while (str2[k]) {
    if ((str1[k] != str2[k]))
      return (0);
    k++;
  }
  return (k);
}
```
user: What is the c source code of the assembly code below:

```asm
<load16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rcx
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
uint16_t load16(const void *ptr) {
  uint16_t v;
  memcpy(&v, ptr, sizeof v);
  return v;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<load16>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  mov    $0x4,%edx
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint16_t load16(const void *ptr) {
  uint16_t v;
  memcpy(&v, ptr, sizeof v);
  return v;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<load16>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rsi
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint16_t load16(const void *ptr) {
  uint16_t v;
  memcpy(&v, ptr, sizeof v);
  return v;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<load16>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rsi
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint16_t load16(const void *ptr) {
  uint16_t v;
  memcpy(&v, ptr, sizeof v);
  return v;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_cmsMalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x1f400000,-0x8(%rbp)
  jbe    L11
  mov    $0x0,%eax
  jmp    L9
L11:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
L9:
  leave
  ret
```

assistant: ```c
void *_cmsMalloc(size_t size) {
  if (size > ((size_t)1024 * 1024 * 500))
    return NULL;
  if (size < 0)
    return NULL;
  return (void *)malloc(size);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_cmsMalloc>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x1f400000,%rdi
  ja     L6
  sub    $0x8,%rsp
  call   <malloc@plt>
  add    $0x8,%rsp
  ret
L6:
  ret
```

assistant: ```c
void *_cmsMalloc(size_t size) {
  if (size > ((size_t)1024 * 1024 * 500))
    return NULL;
  if (size < 0)
    return NULL;
  return (void *)malloc(size);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_cmsMalloc>:
  endbr64
  cmp    $0x1f400000,%rdi
  ja     L9
  jmp    1050 <malloc@plt>
  nopw   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *_cmsMalloc(size_t size) {
  if (size > ((size_t)1024 * 1024 * 500))
    return NULL;
  if (size < 0)
    return NULL;
  return (void *)malloc(size);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_cmsMalloc>:
  endbr64
  cmp    $0x1f400000,%rdi
  ja     L7
  jmp    1050 <malloc@plt>
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *_cmsMalloc(size_t size) {
  if (size > ((size_t)1024 * 1024 * 500))
    return NULL;
  if (size < 0)
    return NULL;
  return (void *)malloc(size);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<join_threads>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L3:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <pthread_join@plt>
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L3
  nop
  nop
  leave
  ret
```

assistant: ```c
void join_threads(pthread_t *threads, int threads_amount) {
  int counter = 0;
  while (counter < threads_amount) {
    pthread_join(threads[counter], NULL);
    counter++;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<join_threads>:
  endbr64
  test   %esi,%esi
  jle    L19
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rbp
L18:
  mov    $0x0,%esi
  mov    (%rbx),%edi
  call   <pthread_join@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L18
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  ret
```

assistant: ```c
void join_threads(pthread_t *threads, int threads_amount) {
  int counter = 0;
  while (counter < threads_amount) {
    pthread_join(threads[counter], NULL);
    counter++;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<join_threads>:
  endbr64
  test   %esi,%esi
  jle    L19
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x4(%rdi,%rax,4),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L12:
  mov    (%rbx),%edi
  xor    %esi,%esi
  add    $0x4,%rbx
  call   <pthread_join@plt>
  cmp    %rbp,%rbx
  jne    L12
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L19:
  ret
```

assistant: ```c
void join_threads(pthread_t *threads, int threads_amount) {
  int counter = 0;
  while (counter < threads_amount) {
    pthread_join(threads[counter], NULL);
    counter++;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<join_threads>:
  endbr64
  test   %esi,%esi
  jle    L4
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x4(%rdi,%rax,4),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L2:
  mov    (%rbx),%edi
  xor    %esi,%esi
  add    $0x4,%rbx
  call   <pthread_join@plt>
  cmp    %rbp,%rbx
  jne    L2
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L4:
  ret
```

assistant: ```c
void join_threads(pthread_t *threads, int threads_amount) {
  int counter = 0;
  while (counter < threads_amount) {
    pthread_join(threads[counter], NULL);
    counter++;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<nullpointerMemchr1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    $0x70,%esi
  mov    %rax,%rdi
  call   <memchr@plt>
  mov    %rax,-0x8(%rbp)
  nop
  leave
  ret
```

assistant: ```c
void nullpointerMemchr1(char *p, char *s) {
  p = memchr(s, 'p', strlen(s));
  (void)p;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<nullpointerMemchr1>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0x70,%esi
  mov    %rbx,%rdi
  call   <memchr@plt>
  pop    %rbx
  ret
```

assistant: ```c
void nullpointerMemchr1(char *p, char *s) {
  p = memchr(s, 'p', strlen(s));
  (void)p;
}
```
user: What is the c source code of the assembly code below:

```asm
<nullpointerMemchr1>:
  endbr64
  push   %rbp
  mov    %rsi,%rdi
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    $0x70,%esi
  pop    %rbp
  mov    %eax,%edx
  jmp    1070 <memchr@plt>
```

assistant: ```c
void nullpointerMemchr1(char *p, char *s) {
  p = memchr(s, 'p', strlen(s));
  (void)p;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<nullpointerMemchr1>:
  endbr64
  push   %rbp
  mov    %rsi,%rdi
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    $0x70,%esi
  pop    %rbp
  mov    %eax,%edx
  jmp    1070 <memchr@plt>
```

assistant: ```c
void nullpointerMemchr1(char *p, char *s) {
  p = memchr(s, 'p', strlen(s));
  (void)p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L14:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <isNumber@plt>
  test   %eax,%eax
  je     L14
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

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

tool: D3: %s

assistant: ```c
int main() {
  int i;
  char entered[100];
  do {
    scanf("%s", entered);
  } while (!isNumber(entered));
  return 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
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rbp
L6:
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <isNumber@plt>
  test   %eax,%eax
  je     L6
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D1
data_type: string

tool: D1: %s

assistant: ```c
int main() {
  int i;
  char entered[100];
  do {
    scanf("%s", entered);
  } while (!isNumber(entered));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L15:
  mov    %rbp,%rdi
  mov    %rbx,%rsi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <isNumber@plt>
  test   %eax,%eax
  je     L15
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

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

tool: D0: %s

assistant: ```c
int main() {
  int i;
  char entered[100];
  do {
    scanf("%s", entered);
  } while (!isNumber(entered));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L2:
  mov    %rbp,%rdi
  mov    %rbx,%rsi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <isNumber@plt>
  test   %eax,%eax
  je     L2
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

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

tool: D4: %s

assistant: ```c
int main() {
  int i;
  char entered[100];
  do {
    scanf("%s", entered);
  } while (!isNumber(entered));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<digit_multiple>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L13
L0:
  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)
L13:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  je     L27
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edi
  mov    %edi,%eax
  shl    $0x2,%eax
  add    %edi,%eax
  add    %eax,%eax
  mov    %ecx,%edx
  sub    %eax,%edx
  mov    %edx,%edi
  mov    -0x4(%rbp),%eax
  cltd
  idiv   %edi
  mov    %edx,%ecx
  mov    %ecx,%eax
  test   %eax,%eax
  jle    L0
L27:
  cmpl   $0x0,-0x14(%rbp)
  sete   %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int digit_multiple(int i) {
  for (int n = i; i % 10 && n % (i % 10) < 1; i /= 10)
    ;
  return !i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<digit_multiple>:
  endbr64
  mov    %edi,%r8d
  movslq %edi,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%eax
  add    %eax,%eax
  mov    %edi,%ecx
  sub    %eax,%ecx
  je     L57
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jg     L18
  mov    %edi,%esi
L6:
  movslq %esi,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  sar    $0x1f,%esi
  sub    %esi,%edx
  mov    %edx,%esi
  movslq %edx,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  mov    %edx,%eax
  sar    $0x1f,%eax
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%edi
  add    %edi,%edi
  mov    %edx,%ecx
  sub    %edi,%ecx
  je     L36
  mov    %r8d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jle    L6
L36:
  test   %esi,%esi
  sete   %al
  movzbl %al,%eax
  ret
L57:
  mov    %edi,%esi
  jmp    L36
L18:
  mov    %edi,%esi
  jmp    L36
```

assistant: ```c
int digit_multiple(int i) {
  for (int n = i; i % 10 && n % (i % 10) < 1; i /= 10)
    ;
  return !i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<digit_multiple>:
  endbr64
  movslq %edi,%rcx
  mov    %edi,%eax
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x22,%rcx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%eax
  mov    %edi,%ecx
  add    %eax,%eax
  sub    %eax,%ecx
  je     L14
  mov    %edi,%eax
  mov    %edi,%esi
  cltd
  idiv   %ecx
  test   %edx,%edx
  jle    L39
  jmp    L43
  nopl   0x0(%rax)
L8:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jg     L43
L39:
  movslq %esi,%rdx
  mov    %esi,%eax
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%eax
  sar    $0x22,%rdx
  mov    %edx,%esi
  sub    %eax,%esi
  movslq %esi,%rcx
  mov    %esi,%eax
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x22,%rcx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%edx
  mov    %esi,%ecx
  add    %edx,%edx
  sub    %edx,%ecx
  jne    L8
L43:
  xor    %eax,%eax
  test   %esi,%esi
  sete   %al
  ret
L14:
  mov    %edi,%esi
  jmp    L43
```

assistant: ```c
int digit_multiple(int i) {
  for (int n = i; i % 10 && n % (i % 10) < 1; i /= 10)
    ;
  return !i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<digit_multiple>:
  endbr64
  movslq %edi,%rcx
  mov    %edi,%eax
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x22,%rcx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%eax
  mov    %edi,%ecx
  add    %eax,%eax
  sub    %eax,%ecx
  je     L40
  mov    %edi,%eax
  mov    %edi,%esi
  cltd
  idiv   %ecx
  test   %edx,%edx
  jle    L47
  jmp    L51
  nopl   0x0(%rax)
L2:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jg     L51
L47:
  movslq %esi,%rdx
  mov    %esi,%eax
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%eax
  sar    $0x22,%rdx
  mov    %edx,%esi
  sub    %eax,%esi
  movslq %esi,%rcx
  mov    %esi,%eax
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x22,%rcx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%edx
  mov    %esi,%ecx
  add    %edx,%edx
  sub    %edx,%ecx
  jne    L2
L51:
  xor    %eax,%eax
  test   %esi,%esi
  sete   %al
  ret
L40:
  mov    %edi,%esi
  jmp    L51
```

assistant: ```c
int digit_multiple(int i) {
  for (int n = i; i % 10 && n % (i % 10) < 1; i /= 10)
    ;
  return !i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bump_up>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *bump_up(void *mtemp, int i_size) {
  unsigned char *bmptr;
  bmptr = (unsigned char *)mtemp;
  bmptr += i_size;
  return ((void *)bmptr);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<bump_up>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,1),%rax
  ret
```

assistant: ```c
void *bump_up(void *mtemp, int i_size) {
  unsigned char *bmptr;
  bmptr = (unsigned char *)mtemp;
  bmptr += i_size;
  return ((void *)bmptr);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bump_up>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,1),%rax
  ret
```

assistant: ```c
void *bump_up(void *mtemp, int i_size) {
  unsigned char *bmptr;
  bmptr = (unsigned char *)mtemp;
  bmptr += i_size;
  return ((void *)bmptr);
}
```
user: What is the c source code of the assembly code below:

```asm
<bump_up>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,1),%rax
  ret
```

assistant: ```c
void *bump_up(void *mtemp, int i_size) {
  unsigned char *bmptr;
  bmptr = (unsigned char *)mtemp;
  bmptr += i_size;
  return ((void *)bmptr);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<make_url>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  add    $0x26,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  test   %rax,%rax
  jne    L22
  lea    D6(%rip),%rax
L22:
  cmpq   $0x0,-0x38(%rbp)
  je     L20
  lea    D1(%rip),%rdx
  jmp    L27
L20:
  lea    D6(%rip),%rdx
L27:
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rsi
  mov    -0x18(%rbp),%rdi
  mov    %rax,%r9
  mov    %rdx,%r8
  mov    %rsi,%rdx
  lea    D27(%rip),%rax
  mov    %rax,%rsi
  call   <sprintf@plt>
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D6
type: string
name: parse_data
arguments: label: D1
type: string
name: parse_data
arguments: label: D27
type: string

tool: 

tool: D1: ?rev=

tool: D27: %s%s%s%s

assistant: ```c
char *make_url(const char *url_prefix, const char *path, const char *rev) {
  char *urlbuf = malloc(strlen(url_prefix) + strlen(path) + strlen("?rev=") + 32 + 1);
  sprintf(urlbuf, "%s%s%s%s", url_prefix, path, rev ? "?rev=" : "", rev ?: "");
  return urlbuf;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<make_url>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbx
  call   <strlen@plt>
  mov    %rax,%rbp
  mov    %r13,%rdi
  call   <strlen@plt>
  lea    0x26(%rbp,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  lea    D10(%rip),%r8
  test   %rbx,%rbx
  je     L13
L12:
  mov    %rbx,%r9
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D15(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  lea    D16(%rip),%rbx
  mov    %rbx,%r8
  jmp    L12
```

tool_calls: name: parse_data
arguments: data_tag: D10
data_type: string
name: parse_data
arguments: data_tag: D15
data_type: string
name: parse_data
arguments: data_tag: D16
data_type: string

tool: D10: ?rev=

tool: D15: %s%s%s%s

tool: 

assistant: ```c
char *make_url(const char *url_prefix, const char *path, const char *rev) {
  char *urlbuf = malloc(strlen(url_prefix) + strlen(path) + strlen("?rev=") + 32 + 1);
  sprintf(urlbuf, "%s%s%s%s", url_prefix, path, rev ? "?rev=" : "", rev ?: "");
  return urlbuf;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<make_url>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x26(%r12,%rax,1),%rdi
  call   <malloc@plt>
  lea    D5(%rip),%r8
  mov    %rax,%r12
  test   %rbx,%rbx
  je     L7
L6:
  mov    %rbx,%r9
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rdi
  lea    D17(%rip),%rsi
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  lea    D29(%rip),%rbx
  mov    %rbx,%r8
  jmp    L6
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
datatype: string
name: retrieve_data
arguments: data_tag: D17
datatype: string
name: retrieve_data
arguments: data_tag: D29
datatype: string

tool: D5: ?rev=

tool: D17: %s%s%s%s

tool: 

assistant: ```c
char *make_url(const char *url_prefix, const char *path, const char *rev) {
  char *urlbuf = malloc(strlen(url_prefix) + strlen(path) + strlen("?rev=") + 32 + 1);
  sprintf(urlbuf, "%s%s%s%s", url_prefix, path, rev ? "?rev=" : "", rev ?: "");
  return urlbuf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<make_url>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x26(%r12,%rax,1),%rdi
  call   <malloc@plt>
  lea    D7(%rip),%r8
  mov    %rax,%r12
  test   %rbx,%rbx
  je     L18
L10:
  mov    %rbx,%r9
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rdi
  lea    D28(%rip),%rsi
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  lea    D19(%rip),%rbx
  mov    %rbx,%r8
  jmp    L10
```

tool_calls: name: read_data
arguments: label: D7
datatype: string
name: read_data
arguments: label: D28
datatype: string
name: read_data
arguments: label: D19
datatype: string

tool: D7: ?rev=

tool: D28: %s%s%s%s

tool: 

assistant: ```c
char *make_url(const char *url_prefix, const char *path, const char *rev) {
  char *urlbuf = malloc(strlen(url_prefix) + strlen(path) + strlen("?rev=") + 32 + 1);
  sprintf(urlbuf, "%s%s%s%s", url_prefix, path, rev ? "?rev=" : "", rev ?: "");
  return urlbuf;
}
```