ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<pvCaptureToBufferEx>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <notify_pvcall@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
datatype: string

tool: D7: f

assistant: ```c
int pvCaptureToBufferEx(unsigned long ulBoardNum, void **hEvent, unsigned long ulFlags, int nNumFrames, void *pBuffer, void *pBufferDirect, int ulLineIncrement, long long *pTimeStamp) {
  notify_pvcall("f");
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pvCaptureToBufferEx>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <notify_pvcall@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
data_type: string

tool: D2: f

assistant: ```c
int pvCaptureToBufferEx(unsigned long ulBoardNum, void **hEvent, unsigned long ulFlags, int nNumFrames, void *pBuffer, void *pBufferDirect, int ulLineIncrement, long long *pTimeStamp) {
  notify_pvcall("f");
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sleepms>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x0,-0x24(%rbp)
  jle    L28
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cltq
  mov    %rax,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  imul   $0xf4240,%edx,%eax
  cltq
  mov    %rax,-0x20(%rbp)
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <nanosleep@plt>
  jmp    L5
L28:
  nop
L5:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

assistant: ```c
extern void sleepms(int ms) {
  struct timespec ts;
  if (ms <= 0)
    return;
  ts.tv_sec = (time_t)(ms / 1000);
  ts.tv_nsec = (long)(ms % 1000 * 1000000);
  nanosleep(&ts, NULL);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sleepms>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L0
  movslq %edi,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  movslq %eax,%rdx
  mov    %rdx,0x8(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  movslq %edi,%rdi
  mov    %rdi,(%rsp)
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <nanosleep@plt>
L0:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x28,%rsp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
extern void sleepms(int ms) {
  struct timespec ts;
  if (ms <= 0)
    return;
  ts.tv_sec = (time_t)(ms / 1000);
  ts.tv_nsec = (long)(ms % 1000 * 1000000);
  nanosleep(&ts, NULL);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<sleepms>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L1
  movslq %edi,%rax
  mov    %edi,%edx
  xor    %esi,%esi
  imul   $0x10624dd3,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x26,%rax
  sub    %edx,%eax
  movslq %eax,%rdx
  imul   $0x3e8,%eax,%eax
  mov    %rdx,0x8(%rsp)
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  movslq %edi,%rdi
  mov    %rdi,(%rsp)
  mov    %rsp,%rdi
  call   <nanosleep@plt>
L1:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
extern void sleepms(int ms) {
  struct timespec ts;
  if (ms <= 0)
    return;
  ts.tv_sec = (time_t)(ms / 1000);
  ts.tv_nsec = (long)(ms % 1000 * 1000000);
  nanosleep(&ts, NULL);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sleepms>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L15
  movslq %edi,%rax
  mov    %edi,%edx
  xor    %esi,%esi
  imul   $0x10624dd3,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x26,%rax
  sub    %edx,%eax
  imul   $0x3e8,%eax,%edx
  cltq
  movq   %rax,%xmm1
  sub    %edx,%edi
  imul   $0xf4240,%edi,%edi
  movslq %edi,%rcx
  mov    %rsp,%rdi
  movq   %rcx,%xmm0
  punpcklqdq %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  call   <nanosleep@plt>
L15:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
extern void sleepms(int ms) {
  struct timespec ts;
  if (ms <= 0)
    return;
  ts.tv_sec = (time_t)(ms / 1000);
  ts.tv_nsec = (long)(ms % 1000 * 1000000);
  nanosleep(&ts, NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<f1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L7:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %eax,%esi
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    %esi,%eax
  lea    (%rcx,%rax,1),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L5:
  cmpl   $0x6,-0x4(%rbp)
  jle    L7
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void f1(int64_t *restrict dst, int16_t *src1, int8_t *src2, uint32_t *index) {
  for (int i = 0; i < 7; ++i)
    dst[i] += (int16_t)(src1[i] + src2[index[i]]);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<f1>:
  endbr64
  mov    $0x0,%eax
L2:
  mov    (%rcx,%rax,8),%r9
  mov    (%rdi,%rax,4),%r8d
  add    (%rsi,%rax,8),%r8d
  add    (%rdx,%r9,8),%r8d
  mov    %r8d,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    $0x7,%rax
  jne    L2
  ret
```

assistant: ```c
void f1(int64_t *restrict dst, int16_t *src1, int8_t *src2, uint32_t *index) {
  for (int i = 0; i < 7; ++i)
    dst[i] += (int16_t)(src1[i] + src2[index[i]]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<f1>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    (%rcx,%rax,8),%r9
  mov    (%rdi,%rax,4),%r8d
  add    (%rsi,%rax,8),%r8d
  add    (%rdx,%r9,8),%r8d
  mov    %r8d,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    $0x7,%rax
  jne    L7
  ret
```

assistant: ```c
void f1(int64_t *restrict dst, int16_t *src1, int8_t *src2, uint32_t *index) {
  for (int i = 0; i < 7; ++i)
    dst[i] += (int16_t)(src1[i] + src2[index[i]]);
}
```
user: What is the c source code of the assembly code below?

```asm
<f1>:
  endbr64
  mov    (%rcx),%r8
  mov    %rdi,%rax
  mov    (%rdi),%edi
  add    (%rsi),%edi
  add    (%rdx,%r8,8),%edi
  mov    0x8(%rcx),%r8
  mov    %edi,(%rax)
  mov    0x4(%rax),%edi
  add    0x8(%rsi),%edi
  add    (%rdx,%r8,8),%edi
  mov    0x10(%rcx),%r8
  mov    %edi,0x4(%rax)
  mov    0x8(%rax),%edi
  add    0x10(%rsi),%edi
  add    (%rdx,%r8,8),%edi
  mov    0x18(%rcx),%r8
  mov    %edi,0x8(%rax)
  mov    0xc(%rax),%edi
  add    0x18(%rsi),%edi
  add    (%rdx,%r8,8),%edi
  mov    0x20(%rcx),%r8
  mov    %edi,0xc(%rax)
  mov    0x10(%rax),%edi
  add    0x20(%rsi),%edi
  add    (%rdx,%r8,8),%edi
  mov    0x28(%rcx),%r8
  mov    %edi,0x10(%rax)
  mov    0x14(%rax),%edi
  add    0x28(%rsi),%edi
  add    (%rdx,%r8,8),%edi
  mov    %edi,0x14(%rax)
  mov    0x30(%rcx),%rdi
  mov    0x18(%rax),%ecx
  add    0x30(%rsi),%ecx
  add    (%rdx,%rdi,8),%ecx
  mov    %ecx,0x18(%rax)
  ret
```

assistant: ```c
void f1(int64_t *restrict dst, int16_t *src1, int8_t *src2, uint32_t *index) {
  for (int i = 0; i < 7; ++i)
    dst[i] += (int16_t)(src1[i] + src2[index[i]]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x12(%rbp),%rax
  mov    $0xc,%ecx
  lea    D2(%rip),%rdx
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x12(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D2
type: string
name: read_data
arguments: label: D5
type: string

tool: D2: bonjour toto %f

tool: D5: size : %d 
 

assistant: ```c
int main(void) {
  char buf[10];
  int size = snprintf(buf, 10, "bonjour toto %f", 12.);
  printf("size : %d \n ", size);
  puts(buf);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbx
  mov    $0xc,%ecx
  lea    D9(%rip),%rdx
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type_guessed: string
name: extract_symbolic_data
arguments: tag: D2
type_guessed: string

tool: D9: bonjour toto %f

tool: D2: size : %d 
 

assistant: ```c
int main(void) {
  char buf[10];
  int size = snprintf(buf, 10, "bonjour toto %f", 12.);
  printf("size : %d \n ", size);
  puts(buf);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0xc,%ecx
  mov    $0xa,%esi
  lea    D2(%rip),%rdx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbp
  mov    %rbp,%rdi
  call   <snprintf@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
type_infered: string
name: unpack_data
arguments: data_label: D4
type_infered: string

tool: D2: bonjour toto %f

tool: D4: size : %d 
 

assistant: ```c
int main(void) {
  char buf[10];
  int size = snprintf(buf, 10, "bonjour toto %f", 12.);
  printf("size : %d \n ", size);
  puts(buf);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0xc,%ecx
  mov    $0xa,%esi
  lea    D0(%rip),%rdx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbp
  mov    %rbp,%rdi
  call   <snprintf@plt>
  lea    D17(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type: string
name: disassemble_data
arguments: data_tag: D17
type: string

tool: D0: bonjour toto %f

tool: D17: size : %d 
 

assistant: ```c
int main(void) {
  char buf[10];
  int size = snprintf(buf, 10, "bonjour toto %f", 12.);
  printf("size : %d \n ", size);
  puts(buf);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<compfilenames>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  leave
  ret
```

assistant: ```c
int compfilenames(const void *s1, const void *s2) {
  char *r1 = *(char *const *)s1;
  char *r2 = *(char *const *)s2;
  return strcmp(r1, r2);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<compfilenames>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  call   <strcmp@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int compfilenames(const void *s1, const void *s2) {
  char *r1 = *(char *const *)s1;
  char *r2 = *(char *const *)s2;
  return strcmp(r1, r2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<compfilenames>:
  endbr64
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  jmp    1050 <strcmp@plt>
```

assistant: ```c
int compfilenames(const void *s1, const void *s2) {
  char *r1 = *(char *const *)s1;
  char *r2 = *(char *const *)s2;
  return strcmp(r1, r2);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<compfilenames>:
  endbr64
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  jmp    1050 <strcmp@plt>
```

assistant: ```c
int compfilenames(const void *s1, const void *s2) {
  char *r1 = *(char *const *)s1;
  char *r2 = *(char *const *)s2;
  return strcmp(r1, r2);
}
```
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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L44
L31:
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jne    L16
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  jne    L16
  mov    $0x0,%eax
  jmp    L37
L16:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L44:
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L31
  mov    $0x0,%eax
L37:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_guessed: string
name: unpack_data
arguments: data_tag: D18
type_guessed: string

tool: D9: %d


tool: D18: %d %d

assistant: ```c
int main() {
  int a, b, sum;
  while (scanf("%d %d", &a, &b)) {
    if (a == 0 && b == 0)
      return 0;
    sum = a + b;
    printf("%d\n", sum);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  lea    D18(%rip),%rbx
  lea    D5(%rip),%r13
  jmp    L8
L36:
  add    %eax,%esi
  mov    %r13,%rdi
  call   <printf@plt>
L8:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L0
  mov    (%rsp),%esi
  mov    0x4(%rsp),%eax
  mov    %esi,%ecx
  or     %eax,%ecx
  jne    L36
L0:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D18
datatype: string
name: unpack_data
arguments: data_tag: D5
datatype: string

tool: D18: %d %d

tool: D5: %d


assistant: ```c
int main() {
  int a, b, sum;
  while (scanf("%d %d", &a, &b)) {
    if (a == 0 && b == 0)
      return 0;
    sum = a + b;
    printf("%d\n", sum);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D13(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D0(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  jmp    L34
  cs nopw 0x0(%rax,%rax,1)
L4:
  mov    (%rsp),%esi
  mov    0x4(%rsp),%eax
  mov    %esi,%ecx
  or     %eax,%ecx
  je     L25
  add    %eax,%esi
  mov    %r13,%rdi
  call   <printf@plt>
L34:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L4
L25:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D13
type: string
name: retrieve_data
arguments: tag: D0
type: string

tool: D13: %d


tool: D0: %d %d

assistant: ```c
int main() {
  int a, b, sum;
  while (scanf("%d %d", &a, &b)) {
    if (a == 0 && b == 0)
      return 0;
    sum = a + b;
    printf("%d\n", sum);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D7(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  jmp    L31
  cs nopw 0x0(%rax,%rax,1)
L39:
  mov    (%rsp),%esi
  mov    0x4(%rsp),%eax
  mov    %esi,%ecx
  or     %eax,%ecx
  je     L14
  add    %eax,%esi
  mov    %r13,%rdi
  call   <printf@plt>
L31:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L39
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D16
datatype: string
name: read_data
arguments: data_label: D7
datatype: string

tool: D16: %d


tool: D7: %d %d

assistant: ```c
int main() {
  int a, b, sum;
  while (scanf("%d %d", &a, &b)) {
    if (a == 0 && b == 0)
      return 0;
    sum = a + b;
    printf("%d\n", sum);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x168,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x158(%rsp)
  xor    %eax,%eax
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%ebx
L103:
  mov    %ebx,%ebp
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,0x10(%rsp,%rbx,4)
  cmp    $0xa,%eax
  je     L1
  add    $0x1,%rbx
  cmp    $0x50,%rbx
  jne    L103
  jmp    L44
L1:
  add    $0x1,%ebp
  movslq %ebp,%rbp
  movl   $0x0,0x10(%rsp,%rbp,4)
L44:
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  test   %eax,%eax
  je     L71
  mov    0xc(%rsp),%esi
  lea    0x10(%rsp),%rcx
  jmp    L19
L16:
  lea    -0x41(%rax),%edx
  cmp    $0x19,%edx
  ja     L94
  lea    -0x41(%rsi,%rax,1),%edx
  movslq %edx,%rax
  imul   $0x4ec4ec4f,%rax,%rax
  sar    $0x23,%rax
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  imul   $0x1a,%eax,%eax
  sub    %eax,%edx
  add    $0x41,%edx
  mov    %edx,(%rcx)
L94:
  add    $0x4,%rcx
  mov    (%rcx),%eax
  test   %eax,%eax
  je     L71
L19:
  lea    -0x61(%rax),%edx
  cmp    $0x19,%edx
  ja     L16
  lea    -0x61(%rsi,%rax,1),%edx
  movslq %edx,%rax
  imul   $0x4ec4ec4f,%rax,%rax
  sar    $0x23,%rax
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  imul   $0x1a,%eax,%eax
  sub    %eax,%edx
  add    $0x61,%edx
  mov    %edx,(%rcx)
  jmp    L94
L71:
  lea    D35(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%edi
  test   %edi,%edi
  je     L52
  lea    0x14(%rsp),%rbx
L55:
  call   <putchar@plt>
  add    $0x4,%rbx
  mov    -0x4(%rbx),%edi
  test   %edi,%edi
  jne    L55
L52:
  mov    0x158(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  mov    $0x0,%eax
  add    $0x168,%rsp
  pop    %rbx
  pop    %rbp
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D23
data_type: string
name: unpack_data
arguments: data_tag: D3
data_type: string
name: unpack_data
arguments: data_tag: D14
data_type: string
name: unpack_data
arguments: data_tag: D35
data_type: string

tool: D23: Enter message to be encrypted: 

tool: D3: Enter shift amount (1-25): 

tool: D14: %d

tool: D35: Encrypted message: 

assistant: ```c
int main(void) {
  int message[80];
  int n;
  printf("Enter message to be encrypted: ");
  int i = 0;
  while (i < 80) {
    message[i] = getchar();
    if (message[i] == '\n') {
      message[i + 1] = '\0';
      break;
    }
    i += 1;
  }
  printf("Enter shift amount (1-25): ");
  scanf("%d", &n);
  for (i = 0; message[i] != '\0'; i += 1) {
    if (message[i] >= 'a' && message[i] <= 'z') {
      message[i] = ((message[i] - 'a') + n) % 26 + 'a';
    } else if (message[i] >= 'A' && message[i] <= 'Z') {
      message[i] = ((message[i] - 'A') + n) % 26 + 'A';
    } else {
      continue;
    }
  }
  printf("Encrypted message: ");
  for (i = 0; message[i] != '\0'; i += 1) {
    printf("%c", message[i]);
  }
  return 0;
}
```