ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<zlog_time>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L15
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L15:
  mov    %rsi,%r13
  mov    %rsp,%rdi
  xor    %esi,%esi
  mov    %edx,%ebp
  mov    %rcx,%r12
  call   <gettimeofday@plt>
  mov    0x4(%rsp),%r14d
  lea    0x4(%rsp),%rdi
  call   <localtime@plt>
  lea    D3(%rip),%rdx
  mov    $0x40,%esi
  mov    %r14d,%ecx
  lea    D28(%rip),%r14
  mov    %rax,%rbx
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <snprintf@plt>
  mov    0x8(%rbx),%r9d
  mov    0x4(%rbx),%r8d
  mov    %r14,%rdi
  mov    (%rbx),%ecx
  lea    D34(%rip),%rdx
  mov    $0x40,%esi
  xor    %eax,%eax
  call   <snprintf@plt>
  xor    %eax,%eax
  call   <zlog_get_buffer@plt>
  sub    $0x8,%rsp
  mov    %r13,%r9
  mov    %r14,%rcx
  push   %rbp
  mov    0x10(%rsp),%r8d
  lea    D15(%rip),%rdx
  mov    %rax,%rbx
  mov    $0x200,%esi
  mov    %rax,%rdi
  xor    %eax,%eax
  xor    %r13d,%r13d
  call   <snprintf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %r13d,%edi
  cltq
  lea    (%rbx,%rax,1),%rbp
  call   <va_start@plt>
  mov    %r12,%rdx
  mov    %r13d,%ecx
  mov    $0x200,%esi
  mov    %rbp,%rdi
  call   <vsnprintf@plt>
  xor    %eax,%eax
  call   <zlog_finish_buffer@plt>
  mov    %r13d,%edi
  call   <va_end@plt>
  pop    %rax
  pop    %rdx
  jmp    L19
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D3
type: string
name: read_data
arguments: data_label: D34
type: string
name: read_data
arguments: data_label: D15
type: string

tool: D3: %ld

tool: D34: %d:%02d:%d

tool: D15: [%s.%06lds] [@%s:%d] 

assistant: ```c
void zlog_time(int msg_level, char *filename, int line, char const *fmt, ...) {
  if (msg_level <= 0) {
    static char timebuf[64];
    struct timeval tv;
    time_t curtime;
    char *buffer = NULL;
    va_list va;
    gettimeofday(&tv, NULL);
    curtime = tv.tv_sec;
    struct tm *tm_struct = localtime(&tv.tv_sec);
    snprintf(timebuf, 64, "%ld", curtime);
    snprintf(timebuf, 64, "%d:%02d:%d", tm_struct->tm_hour, tm_struct->tm_min, tm_struct->tm_sec);
    buffer = zlog_get_buffer();
    snprintf(buffer, 512, "[%s.%06lds] [@%s:%d] ", timebuf, tv.tv_usec, filename, line);
    buffer += strlen(buffer);
    va_start(va, fmt);
    vsnprintf(buffer, 512, fmt, va);
    zlog_finish_buffer();
    va_end(va);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<zlog_time>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L12
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L12:
  mov    %rsi,%r13
  mov    %rsp,%rdi
  xor    %esi,%esi
  mov    %edx,%ebp
  mov    %rcx,%r12
  call   <gettimeofday@plt>
  mov    0x4(%rsp),%r14d
  lea    0x4(%rsp),%rdi
  call   <localtime@plt>
  lea    D31(%rip),%rdx
  mov    $0x40,%esi
  mov    %r14d,%ecx
  lea    D21(%rip),%r14
  mov    %rax,%rbx
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <snprintf@plt>
  mov    0x8(%rbx),%r9d
  mov    0x4(%rbx),%r8d
  mov    %r14,%rdi
  mov    (%rbx),%ecx
  lea    D30(%rip),%rdx
  mov    $0x40,%esi
  xor    %eax,%eax
  call   <snprintf@plt>
  xor    %eax,%eax
  call   <zlog_get_buffer@plt>
  sub    $0x8,%rsp
  mov    %r13,%r9
  mov    %r14,%rcx
  push   %rbp
  mov    0x10(%rsp),%r8d
  lea    D8(%rip),%rdx
  mov    %rax,%rbx
  mov    $0x200,%esi
  mov    %rax,%rdi
  xor    %eax,%eax
  xor    %r13d,%r13d
  call   <snprintf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %r13d,%edi
  cltq
  lea    (%rbx,%rax,1),%rbp
  call   <va_start@plt>
  mov    %r12,%rdx
  mov    %r13d,%ecx
  mov    $0x200,%esi
  mov    %rbp,%rdi
  call   <vsnprintf@plt>
  xor    %eax,%eax
  call   <zlog_finish_buffer@plt>
  mov    %r13d,%edi
  call   <va_end@plt>
  pop    %rax
  pop    %rdx
  jmp    L14
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D31
type_infered: string
name: unpack_data
arguments: label: D30
type_infered: string
name: unpack_data
arguments: label: D8
type_infered: string

tool: D31: %ld

tool: D30: %d:%02d:%d

tool: D8: [%s.%06lds] [@%s:%d] 

assistant: ```c
void zlog_time(int msg_level, char *filename, int line, char const *fmt, ...) {
  if (msg_level <= 0) {
    static char timebuf[64];
    struct timeval tv;
    time_t curtime;
    char *buffer = NULL;
    va_list va;
    gettimeofday(&tv, NULL);
    curtime = tv.tv_sec;
    struct tm *tm_struct = localtime(&tv.tv_sec);
    snprintf(timebuf, 64, "%ld", curtime);
    snprintf(timebuf, 64, "%d:%02d:%d", tm_struct->tm_hour, tm_struct->tm_min, tm_struct->tm_sec);
    buffer = zlog_get_buffer();
    snprintf(buffer, 512, "[%s.%06lds] [@%s:%d] ", timebuf, tv.tv_usec, filename, line);
    buffer += strlen(buffer);
    va_start(va, fmt);
    vsnprintf(buffer, 512, fmt, va);
    zlog_finish_buffer();
    va_end(va);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<deriv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9d,-0x3c(%rbp)
  mov    -0x3c(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x18(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L5
L19:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  sub    $0x8,%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  movsd  -0x8(%rbp),%xmm0
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  sub    $0x8,%rax
  add    %rdx,%rax
  divsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L5:
  mov    -0x3c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L19
  nop
  nop
  leave
  ret
```

assistant: ```c
void deriv(double **dx, double **dy, int *dn, const double *x, const double *y, int n) {
  int i;
  *dn = n - 2;
  *dx = malloc(*dn * sizeof(double));
  *dy = malloc(*dn * sizeof(double));
  double delta = x[1] - x[0];
  for (i = 1; i < n - 1; i++) {
    (*dx)[i - 1] = x[i];
    (*dy)[i - 1] = (y[i + 1] - y[i - 1]) / (2 * delta);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<deriv>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%r15
  mov    %rcx,%rbp
  mov    %r8,%rbx
  mov    %r9d,%r14d
  lea    -0x2(%r9),%edi
  mov    %edi,(%rdx)
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
  mov    (%r15),%edi
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  movsd  0x8(%rbp),%xmm1
  subsd  0x0(%rbp),%xmm1
  cmp    $0x2,%r14d
  jle    L5
  addsd  %xmm1,%xmm1
  lea    -0x3(%r14),%eax
  lea    0x8(,%rax,8),%rcx
  mov    $0x0,%eax
L13:
  movsd  0x8(%rbp,%rax,1),%xmm0
  mov    (%r12),%rdx
  movsd  %xmm0,(%rdx,%rax,1)
  mov    0x0(%r13),%rdx
  movsd  0x10(%rbx,%rax,1),%xmm0
  subsd  (%rbx,%rax,1),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx,%rax,1)
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L13
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void deriv(double **dx, double **dy, int *dn, const double *x, const double *y, int n) {
  int i;
  *dn = n - 2;
  *dx = malloc(*dn * sizeof(double));
  *dy = malloc(*dn * sizeof(double));
  double delta = x[1] - x[0];
  for (i = 1; i < n - 1; i++) {
    (*dx)[i - 1] = x[i];
    (*dy)[i - 1] = (y[i + 1] - y[i - 1]) / (2 * delta);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<deriv>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rdi,%r13
  lea    -0x2(%r9),%edi
  push   %r12
  mov    %r9d,%r12d
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x8,%rsp
  mov    %edi,(%rdx)
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    (%r14),%edi
  mov    %rax,0x0(%r13)
  shl    $0x3,%edi
  call   <malloc@plt>
  movsd  0x8(%rbp),%xmm1
  subsd  0x0(%rbp),%xmm1
  mov    %rax,(%r15)
  cmp    $0x2,%r12d
  jle    L15
  addsd  %xmm1,%xmm1
  mov    0x0(%r13),%rdi
  lea    -0x3(%r12),%esi
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L5:
  movsd  0x8(%rbp,%rdx,8),%xmm0
  mov    %rdx,%rcx
  movsd  %xmm0,(%rdi,%rdx,8)
  movsd  0x10(%rbx,%rdx,8),%xmm0
  subsd  (%rbx,%rdx,8),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax,%rdx,8)
  add    $0x1,%rdx
  cmp    %rcx,%rsi
  jne    L5
L15:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void deriv(double **dx, double **dy, int *dn, const double *x, const double *y, int n) {
  int i;
  *dn = n - 2;
  *dx = malloc(*dn * sizeof(double));
  *dy = malloc(*dn * sizeof(double));
  double delta = x[1] - x[0];
  for (i = 1; i < n - 1; i++) {
    (*dx)[i - 1] = x[i];
    (*dy)[i - 1] = (y[i + 1] - y[i - 1]) / (2 * delta);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<deriv>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %r9d,%r13d
  push   %r12
  lea    -0x2(%r9),%r12d
  push   %rbp
  lea    0x0(,%r12,8),%edi
  mov    %rcx,%rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x18,%rsp
  mov    %r12d,(%rdx)
  mov    %rsi,0x8(%rsp)
  call   <malloc@plt>
  mov    (%r15),%edi
  mov    %rax,(%r14)
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    0x8(%rsp),%rsi
  movsd  0x8(%rbp),%xmm1
  subsd  0x0(%rbp),%xmm1
  mov    %rax,(%rsi)
  cmp    $0x2,%r13d
  jle    L41
  mov    (%r14),%rsi
  mov    %rax,%rcx
  lea    0x20(%rbx),%rdi
  lea    0x10(%rbp),%rdx
  addsd  %xmm1,%xmm1
  lea    0x10(%rsi),%rax
  cmp    %rax,%rbx
  setae  %al
  cmp    %rdi,%rsi
  setae  %dil
  or     %edi,%eax
  cmp    $0x3,%r13d
  setne  %dil
  and    %edi,%eax
  cmp    %rdx,%rsi
  setne  %dil
  and    %edi,%eax
  lea    0x8(%rsi),%rdi
  cmp    %rdi,%rcx
  setne  %dil
  and    %edi,%eax
  cmp    %rdx,%rcx
  setne  %dl
  test   %dl,%al
  je     L10
  lea    0x8(%rbx),%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  cmp    $0x10,%rax
  jbe    L10
  mov    %r12d,%edx
  movapd %xmm1,%xmm2
  xor    %eax,%eax
  shr    %edx
  unpcklpd %xmm2,%xmm2
  shl    $0x4,%rdx
  xchg   %ax,%ax
L44:
  movupd 0x8(%rbp,%rax,1),%xmm3
  movups %xmm3,(%rsi,%rax,1)
  movupd 0x10(%rbx,%rax,1),%xmm0
  movupd (%rbx,%rax,1),%xmm4
  subpd  %xmm4,%xmm0
  divpd  %xmm2,%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L44
  mov    %r12d,%edx
  or     $0x1,%edx
  and    $0x1,%r12d
  je     L41
  movslq %edx,%rdx
  movsd  0x0(%rbp,%rdx,8),%xmm0
  lea    0x0(,%rdx,8),%rax
  movsd  %xmm0,-0x8(%rsi,%rax,1)
  movsd  0x8(%rbx,%rax,1),%xmm0
  subsd  -0x8(%rbx,%rax,1),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rcx,%rax,1)
L41:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  lea    -0x3(%r13),%edi
  xor    %eax,%eax
  xchg   %ax,%ax
L22:
  movsd  0x8(%rbp,%rax,8),%xmm0
  mov    %rax,%rdx
  movsd  %xmm0,(%rsi,%rax,8)
  movsd  0x10(%rbx,%rax,8),%xmm0
  subsd  (%rbx,%rax,8),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rdi
  jne    L22
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void deriv(double **dx, double **dy, int *dn, const double *x, const double *y, int n) {
  int i;
  *dn = n - 2;
  *dx = malloc(*dn * sizeof(double));
  *dy = malloc(*dn * sizeof(double));
  double delta = x[1] - x[0];
  for (i = 1; i < n - 1; i++) {
    (*dx)[i - 1] = x[i];
    (*dy)[i - 1] = (y[i + 1] - y[i - 1]) / (2 * delta);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L6:
  mov    $0x3c,%edi
  call   <sleep@plt>
  jmp    L6
```

assistant: ```c
int main(void) {
  for (;;) {
    sleep(60);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
L4:
  mov    $0x3c,%edi
  call   <sleep@plt>
  jmp    L4
```

assistant: ```c
int main(void) {
  for (;;) {
    sleep(60);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L8:
  mov    $0x3c,%edi
  call   <sleep@plt>
  jmp    L8
```

assistant: ```c
int main(void) {
  for (;;) {
    sleep(60);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L6:
  mov    $0x3c,%edi
  call   <sleep@plt>
  jmp    L6
```

assistant: ```c
int main(void) {
  for (;;) {
    sleep(60);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<SetHandShake>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xee,%edi
  call   <SendChar@plt>
  mov    $0x4,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xfc,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void SetHandShake() {
  SendChar((0XEE) & 0xFF);
  SendChar((0x04) & 0xFF);
  SendChar((((0XFFFCFFFF) >> 16) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) >> 16) & 0xFF);
  SendChar((((0XFFFCFFFF) & 0xFFFF) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) & 0xFFFF) & 0xFF);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<SetHandShake>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xee,%edi
  call   <SendChar@plt>
  mov    $0x4,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xfc,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void SetHandShake() {
  SendChar((0XEE) & 0xFF);
  SendChar((0x04) & 0xFF);
  SendChar((((0XFFFCFFFF) >> 16) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) >> 16) & 0xFF);
  SendChar((((0XFFFCFFFF) & 0xFFFF) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) & 0xFFFF) & 0xFF);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<SetHandShake>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xee,%edi
  call   <SendChar@plt>
  mov    $0x4,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xfc,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  add    $0x8,%rsp
  jmp    1050 <SendChar@plt>
```

assistant: ```c
void SetHandShake() {
  SendChar((0XEE) & 0xFF);
  SendChar((0x04) & 0xFF);
  SendChar((((0XFFFCFFFF) >> 16) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) >> 16) & 0xFF);
  SendChar((((0XFFFCFFFF) & 0xFFFF) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) & 0xFFFF) & 0xFF);
}
```
user: What is the c source code of the assembly code below?

```asm
<SetHandShake>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xee,%edi
  call   <SendChar@plt>
  mov    $0x4,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xfc,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  add    $0x8,%rsp
  jmp    1050 <SendChar@plt>
```

assistant: ```c
void SetHandShake() {
  SendChar((0XEE) & 0xFF);
  SendChar((0x04) & 0xFF);
  SendChar((((0XFFFCFFFF) >> 16) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) >> 16) & 0xFF);
  SendChar((((0XFFFCFFFF) & 0xFFFF) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) & 0xFFFF) & 0xFF);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movss  -0x14(%rbp),%xmm0
  movss  -0x10(%rbp),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  movss  -0xc(%rbp),%xmm0
  divss  %xmm1,%xmm0
  cvtss2sd %xmm0,%xmm0
  mov    -0x18(%rbp),%edx
  pxor   %xmm2,%xmm2
  cvtss2sd -0xc(%rbp),%xmm2
  movq   %xmm2,%rax
  movapd %xmm0,%xmm1
  mov    %edx,%esi
  movq   %rax,%xmm0
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D10
type: string
name: retrieve_data
arguments: data_label: D18
type: string
name: retrieve_data
arguments: data_label: D58
type: string
name: retrieve_data
arguments: data_label: D14
type: string
name: retrieve_data
arguments: data_label: D52
type: string
name: retrieve_data
arguments: data_label: D41
type: string

tool: D10: Enter expected number of years. Ex. If 2 years, enter: 2 : 

tool: D18: %d

tool: D58: Enter the price value. Ex. 250.00 : 

tool: D14: %f

tool: D52: Enter the salvage value. Ex. 50.00 : 

tool: D41: DEPRECIATION IS: %.1f / %d = %.1f

assistant: ```c
int main() {
  int year;
  float num1, num2, vardep;
  printf("Enter expected number of years. Ex. If 2 years, enter: 2 : ");
  scanf("%d", &year);
  printf("Enter the price value. Ex. 250.00 : ");
  scanf("%f", &num1);
  printf("Enter the salvage value. Ex. 50.00 : ");
  scanf("%f", &num2);
  vardep = num1 - num2;
  printf("DEPRECIATION IS: %.1f / %d = %.1f", vardep, year, vardep / year);
  printf("\n");
}
```
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    D58(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D42(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D10(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D53(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movss  0x10(%rsp),%xmm0
  subss  0x14(%rsp),%xmm0
  mov    0xc(%rsp),%esi
  pxor   %xmm2,%xmm2
  cvtsi2ss %esi,%xmm2
  movaps %xmm0,%xmm1
  divss  %xmm2,%xmm1
  cvtss2sd %xmm0,%xmm0
  cvtss2sd %xmm1,%xmm1
  lea    D56(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@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: disassemble_data
arguments: label: D58
datatype: string
name: disassemble_data
arguments: label: D42
datatype: string
name: disassemble_data
arguments: label: D27
datatype: string
name: disassemble_data
arguments: label: D10
datatype: string
name: disassemble_data
arguments: label: D53
datatype: string
name: disassemble_data
arguments: label: D56
datatype: string

tool: D58: Enter expected number of years. Ex. If 2 years, enter: 2 : 

tool: D42: %d

tool: D27: Enter the price value. Ex. 250.00 : 

tool: D10: %f

tool: D53: Enter the salvage value. Ex. 50.00 : 

tool: D56: DEPRECIATION IS: %.1f / %d = %.1f

assistant: ```c
int main() {
  int year;
  float num1, num2, vardep;
  printf("Enter expected number of years. Ex. If 2 years, enter: 2 : ");
  scanf("%d", &year);
  printf("Enter the price value. Ex. 250.00 : ");
  scanf("%f", &num1);
  printf("Enter the salvage value. Ex. 50.00 : ");
  scanf("%f", &num2);
  vardep = num1 - num2;
  printf("DEPRECIATION IS: %.1f / %d = %.1f", vardep, year, vardep / year);
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D24(%rip),%rdi
  lea    D19(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  pxor   %xmm2,%xmm2
  movss  0x10(%rsp),%xmm0
  subss  0x14(%rsp),%xmm0
  lea    D50(%rip),%rdi
  mov    $0x2,%eax
  cvtsi2ss %esi,%xmm2
  movaps %xmm0,%xmm1
  cvtss2sd %xmm0,%xmm0
  divss  %xmm2,%xmm1
  cvtss2sd %xmm1,%xmm1
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D24
type_infered: string
name: extract_symbolic_data
arguments: tag: D19
type_infered: string
name: extract_symbolic_data
arguments: tag: D25
type_infered: string
name: extract_symbolic_data
arguments: tag: D27
type_infered: string
name: extract_symbolic_data
arguments: tag: D58
type_infered: string
name: extract_symbolic_data
arguments: tag: D50
type_infered: string

tool: D24: Enter expected number of years. Ex. If 2 years, enter: 2 : 

tool: D19: %f

tool: D25: %d

tool: D27: Enter the price value. Ex. 250.00 : 

tool: D58: Enter the salvage value. Ex. 50.00 : 

tool: D50: DEPRECIATION IS: %.1f / %d = %.1f

assistant: ```c
int main() {
  int year;
  float num1, num2, vardep;
  printf("Enter expected number of years. Ex. If 2 years, enter: 2 : ");
  scanf("%d", &year);
  printf("Enter the price value. Ex. 250.00 : ");
  scanf("%f", &num1);
  printf("Enter the salvage value. Ex. 50.00 : ");
  scanf("%f", &num2);
  vardep = num1 - num2;
  printf("DEPRECIATION IS: %.1f / %d = %.1f", vardep, year, vardep / year);
  printf("\n");
}
```