ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse_words>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  test   %eax,%eax
  jle    L17
  movslq %eax,%rdx
  cmpb   $0xa,-0x1(%rbp,%rdx,1)
  je     L20
  lea    -0x1(%rax),%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <reverse_string@plt>
L4:
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L64
  mov    $0x1,%ebx
  mov    $0x0,%esi
  cmp    $0xa,%al
  jne    L16
L64:
  mov    $0x0,%edx
  mov    $0x0,%esi
L81:
  sub    $0x1,%edx
  mov    %rbp,%rdi
  call   <reverse_string@plt>
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L20:
  lea    -0x2(%rax),%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <reverse_string@plt>
  jmp    L4
L84:
  mov    %ebx,%edx
  movzbl 0x0(%rbp,%rbx,1),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L81
  cmp    $0xa,%al
  je     L81
L16:
  cmp    $0x20,%al
  jne    L84
  lea    -0x2(%rbx),%edx
  mov    %rbp,%rdi
  call   <reverse_string@plt>
  mov    %ebx,%esi
  jmp    L84
```

assistant: ```c
void reverse_words(char line[]) {
  int len = strlen(line);
  if (len < 1) {
    return;
  }
  if (line[len - 1] == '\n') {
    reverse_string(line, 0, len - 2);
  } else {
    reverse_string(line, 0, len - 1);
  }
  int wordstart = 0, wordend = 0;
  for (; line[wordend] != '\0' && line[wordend] != '\n'; wordend++) {
    if (line[wordend] == ' ') {
      reverse_string(line, wordstart, wordend - 1);
      wordstart = wordend + 1;
    }
  }
  reverse_string(line, wordstart, wordend - 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<reverse_words>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L87
  movslq %eax,%rdx
  cmpb   $0xa,-0x1(%rbp,%rdx,1)
  je     L98
  lea    -0x1(%rax),%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <reverse_string@plt>
L48:
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L12
  mov    $0x1,%ebx
  xor    %esi,%esi
  cmp    $0xa,%al
  jne    L86
  jmp    L12
  nopl   0x0(%rax)
L108:
  movzbl 0x0(%rbp,%rbx,1),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L105
L0:
  cmp    $0xa,%al
  je     L105
L86:
  lea    -0x1(%rbx),%r12d
  cmp    $0x20,%al
  jne    L108
  lea    -0x2(%rbx),%edx
  mov    %rbp,%rdi
  call   <reverse_string@plt>
  movzbl 0x0(%rbp,%rbx,1),%eax
  mov    %ebx,%esi
  add    $0x1,%rbx
  test   %al,%al
  jne    L0
L105:
  pop    %rbx
  mov    %r12d,%edx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <reverse_string@plt>
  nopw   0x0(%rax,%rax,1)
L98:
  lea    -0x2(%rax),%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <reverse_string@plt>
  jmp    L48
  nop
L87:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L12:
  mov    $0xffffffff,%r12d
  xor    %esi,%esi
  jmp    L105
```

assistant: ```c
void reverse_words(char line[]) {
  int len = strlen(line);
  if (len < 1) {
    return;
  }
  if (line[len - 1] == '\n') {
    reverse_string(line, 0, len - 2);
  } else {
    reverse_string(line, 0, len - 1);
  }
  int wordstart = 0, wordend = 0;
  for (; line[wordend] != '\0' && line[wordend] != '\n'; wordend++) {
    if (line[wordend] == ' ') {
      reverse_string(line, wordstart, wordend - 1);
      wordstart = wordend + 1;
    }
  }
  reverse_string(line, wordstart, wordend - 1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<reverse_words>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L13
  movslq %eax,%rdx
  cmpb   $0xa,-0x1(%rbp,%rdx,1)
  je     L83
  lea    -0x1(%rax),%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <reverse_string@plt>
L49:
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L121
  mov    $0x1,%ebx
  xor    %esi,%esi
  cmp    $0xa,%al
  je     L121
  xchg   %ax,%ax
L43:
  mov    %ebx,%r12d
  lea    -0x1(%rbx),%r13d
  mov    %ebx,%r14d
  cmp    $0x20,%al
  je     L75
  movzbl 0x0(%rbp,%rbx,1),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L120
  cmp    $0xa,%al
  jne    L43
L120:
  pop    %rbx
  mov    %r13d,%edx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <reverse_string@plt>
  nopw   0x0(%rax,%rax,1)
L83:
  lea    -0x2(%rax),%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <reverse_string@plt>
  jmp    L49
  nopw   0x0(%rax,%rax,1)
L75:
  lea    -0x2(%rbx),%edx
  mov    %rbp,%rdi
  call   <reverse_string@plt>
  movzbl 0x0(%rbp,%rbx,1),%eax
  add    $0x1,%rbx
  cmp    $0xa,%al
  je     L31
  test   %al,%al
  je     L31
  mov    %r12d,%esi
  jmp    L43
  nopl   0x0(%rax)
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L31:
  mov    %r14d,%esi
  jmp    L120
  nopl   (%rax)
L121:
  mov    $0xffffffff,%r13d
  xor    %esi,%esi
  jmp    L120
```

assistant: ```c
void reverse_words(char line[]) {
  int len = strlen(line);
  if (len < 1) {
    return;
  }
  if (line[len - 1] == '\n') {
    reverse_string(line, 0, len - 2);
  } else {
    reverse_string(line, 0, len - 1);
  }
  int wordstart = 0, wordend = 0;
  for (; line[wordend] != '\0' && line[wordend] != '\n'; wordend++) {
    if (line[wordend] == ' ') {
      reverse_string(line, wordstart, wordend - 1);
      wordstart = wordend + 1;
    }
  }
  reverse_string(line, wordstart, wordend - 1);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<transform_to_log_area_ratios>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L67
L61:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <saturated_abs16@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  not    %eax
  shr    $0x1f,%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpl   $0x5665,-0x8(%rbp)
  jg     L27
  sarl   -0x8(%rbp)
  jmp    L73
L27:
  cmpl   $0x7999,-0x8(%rbp)
  jg     L8
  cmpl   $0x2b32,-0x8(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  subl   $0x2b33,-0x8(%rbp)
  jmp    L73
L8:
  cmpl   $0x65ff,-0x8(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  subl   $0x6600,-0x8(%rbp)
  shll   $0x2,-0x8(%rbp)
L73:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jns    L40
  mov    -0x8(%rbp),%eax
  neg    %eax
  jmp    L52
L40:
  mov    -0x8(%rbp),%eax
L52:
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  addl   $0x1,-0x4(%rbp)
  addq   $0x4,-0x18(%rbp)
L67:
  cmpl   $0x8,-0x4(%rbp)
  jle    L61
  nop
  nop
  leave
  ret
```

assistant: ```c
void transform_to_log_area_ratios(int16_t r[8]) {
  int16_t temp;
  int i;
  for (i = 1; i <= 8; i++, r++) {
    temp = saturated_abs16(*r);
    assert(temp >= 0);
    if (temp < 22118) {
      temp >>= 1;
    } else if (temp < 31130) {
      assert(temp >= 11059);
      temp -= 11059;
    } else {
      assert(temp >= 26112);
      temp -= 26112;
      temp <<= 2;
    }
    *r = (*r < 0) ? -temp : temp;
    assert(*r != INT16_MIN);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<transform_to_log_area_ratios>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  lea    0x20(%rdi),%r12
  jmp    L51
L18:
  cmp    $0x7999,%ebx
  jg     L56
  mov    $0x1,%edi
  call   <assert@plt>
  lea    -0x2b33(%rbx),%eax
L7:
  mov    %eax,%edx
  neg    %edx
  cmpl   $0x0,0x0(%rbp)
  cmovs  %edx,%eax
  mov    %eax,0x0(%rbp)
  mov    D0(%rip),%rdx
  cmp    %eax,(%rdx)
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x4,%rbp
  cmp    %rbp,%r12
  je     L8
L51:
  mov    0x0(%rbp),%edi
  call   <saturated_abs16@plt>
  mov    %eax,%ebx
  mov    %eax,%edi
  not    %edi
  shr    $0x1f,%edi
  call   <assert@plt>
  cmp    $0x5665,%ebx
  jg     L18
  mov    %ebx,%eax
  sar    %eax
  jmp    L7
L56:
  mov    $0x1,%edi
  call   <assert@plt>
  lea    -0x19800(,%rbx,4),%eax
  jmp    L7
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void transform_to_log_area_ratios(int16_t r[8]) {
  int16_t temp;
  int i;
  for (i = 1; i <= 8; i++, r++) {
    temp = saturated_abs16(*r);
    assert(temp >= 0);
    if (temp < 22118) {
      temp >>= 1;
    } else if (temp < 31130) {
      assert(temp >= 11059);
      temp -= 11059;
    } else {
      assert(temp >= 26112);
      temp -= 26112;
      temp <<= 2;
    }
    *r = (*r < 0) ? -temp : temp;
    assert(*r != INT16_MIN);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<transform_to_log_area_ratios>:
  endbr64
  push   %r13
  lea    0x20(%rdi),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D3(%rip),%r12
  jmp    L51
  xchg   %ax,%ax
L27:
  mov    %ebx,%eax
  sar    %eax
L14:
  mov    0x0(%rbp),%ecx
  mov    %eax,%edx
  neg    %edx
  test   %ecx,%ecx
  cmovs  %edx,%eax
  xor    %edi,%edi
  mov    %eax,0x0(%rbp)
  cmp    %eax,(%r12)
  setne  %dil
  add    $0x4,%rbp
  call   <assert@plt>
  cmp    %rbp,%r13
  je     L0
L51:
  mov    0x0(%rbp),%edi
  call   <saturated_abs16@plt>
  mov    %eax,%edi
  mov    %eax,%ebx
  not    %edi
  shr    $0x1f,%edi
  call   <assert@plt>
  cmp    $0x5665,%ebx
  jle    L27
  mov    $0x1,%edi
  cmp    $0x7999,%ebx
  jg     L42
  call   <assert@plt>
  lea    -0x2b33(%rbx),%eax
  jmp    L14
  nopl   0x0(%rax,%rax,1)
L42:
  call   <assert@plt>
  lea    -0x19800(,%rbx,4),%eax
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void transform_to_log_area_ratios(int16_t r[8]) {
  int16_t temp;
  int i;
  for (i = 1; i <= 8; i++, r++) {
    temp = saturated_abs16(*r);
    assert(temp >= 0);
    if (temp < 22118) {
      temp >>= 1;
    } else if (temp < 31130) {
      assert(temp >= 11059);
      temp -= 11059;
    } else {
      assert(temp >= 26112);
      temp -= 26112;
      temp <<= 2;
    }
    *r = (*r < 0) ? -temp : temp;
    assert(*r != INT16_MIN);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<transform_to_log_area_ratios>:
  endbr64
  push   %r13
  lea    0x20(%rdi),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D1(%rip),%r12
  jmp    L55
  xchg   %ax,%ax
L28:
  mov    %ebx,%eax
  sar    %eax
L7:
  mov    0x0(%rbp),%ecx
  mov    %eax,%edx
  neg    %edx
  test   %ecx,%ecx
  cmovs  %edx,%eax
  xor    %edi,%edi
  mov    %eax,0x0(%rbp)
  cmp    %eax,(%r12)
  setne  %dil
  add    $0x4,%rbp
  call   <assert@plt>
  cmp    %r13,%rbp
  je     L51
L55:
  mov    0x0(%rbp),%edi
  call   <saturated_abs16@plt>
  mov    %eax,%edi
  mov    %eax,%ebx
  not    %edi
  shr    $0x1f,%edi
  call   <assert@plt>
  cmp    $0x5665,%ebx
  jle    L28
  mov    $0x1,%edi
  cmp    $0x7999,%ebx
  jg     L21
  call   <assert@plt>
  lea    -0x2b33(%rbx),%eax
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L21:
  call   <assert@plt>
  lea    -0x19800(,%rbx,4),%eax
  jmp    L7
  cs nopw 0x0(%rax,%rax,1)
L51:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void transform_to_log_area_ratios(int16_t r[8]) {
  int16_t temp;
  int i;
  for (i = 1; i <= 8; i++, r++) {
    temp = saturated_abs16(*r);
    assert(temp >= 0);
    if (temp < 22118) {
      temp >>= 1;
    } else if (temp < 31130) {
      assert(temp >= 11059);
      temp -= 11059;
    } else {
      assert(temp >= 26112);
      temp -= 26112;
      temp <<= 2;
    }
    *r = (*r < 0) ? -temp : temp;
    assert(*r != INT16_MIN);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<transmit_reqs_once>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x4000(%rsp),%r11
L23:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L23
  sub    $0xc0,%rsp
  mov    %edi,-0x4094(%rbp)
  mov    %rsi,-0x40a0(%rbp)
  mov    %rdx,-0x40a8(%rbp)
  mov    %ecx,-0x4098(%rbp)
  mov    %r8,-0x40b0(%rbp)
  mov    %r9d,-0x40b4(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x4084(%rbp)
  mov    -0x40a0(%rbp),%rax
  mov    %rax,-0x4078(%rbp)
L11:
  mov    -0x40b4(%rbp),%edi
  mov    -0x4084(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40a8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x40b0(%rbp),%rcx
  mov    -0x4078(%rbp),%rsi
  mov    -0x4098(%rbp),%eax
  mov    %edi,%r9d
  mov    %rcx,%r8
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <sendto@plt>
  mov    %eax,-0x4080(%rbp)
  mov    -0x4084(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40a8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,-0x4078(%rbp)
  lea    -0x40b4(%rbp),%rcx
  mov    -0x40b0(%rbp),%rdx
  lea    -0x4070(%rbp),%rsi
  mov    -0x4098(%rbp),%eax
  mov    %rcx,%r9
  mov    %rdx,%r8
  mov    $0x0,%ecx
  mov    $0x4000,%edx
  mov    %eax,%edi
  call   <recvfrom@plt>
  mov    %eax,-0x407c(%rbp)
  addl   $0x1,-0x4084(%rbp)
  mov    -0x4084(%rbp),%eax
  cmp    -0x4094(%rbp),%eax
  jl     L11
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
void transmit_reqs_once(int num_of_reqs, uint8_t *reqs, int *req_size_arr, int socket, struct sockaddr *si_other, socklen_t slen) {
  uint8_t *req_addr;
  uint8_t rec_data[4096];
  int sent_items, rec_items;
  int i;
  char req_str[100];
  i = 0;
  req_addr = reqs;
  do {
    sent_items = sendto(socket, req_addr, req_size_arr[i], 0, si_other, slen);
    req_addr += req_size_arr[i];
    rec_items = recvfrom(socket, &rec_data, sizeof(rec_data), 0, si_other, &slen);
    i++;
  } while (i < num_of_reqs);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<transmit_reqs_once>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L18:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L18
  sub    $0x28,%rsp
  mov    %edi,%r15d
  mov    %rsi,%rbp
  mov    %rdx,%r14
  mov    %ecx,%r13d
  mov    %r8,%r12
  mov    %r9d,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x4018(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebx
  lea    0xc(%rsp),%rax
  mov    %rax,(%rsp)
L28:
  mov    (%r14,%rbx,4),%edx
  mov    0xc(%rsp),%r9d
  mov    %r12,%r8
  mov    $0x0,%ecx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <sendto@plt>
  movslq (%r14,%rbx,4),%rax
  lea    0x0(%rbp,%rax,4),%rbp
  lea    0x10(%rsp),%rsi
  mov    (%rsp),%r9
  mov    %r12,%r8
  mov    $0x0,%ecx
  mov    $0x4000,%edx
  mov    %r13d,%edi
  call   <recvfrom@plt>
  add    $0x1,%rbx
  cmp    %ebx,%r15d
  jg     L28
  mov    0x4018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x4028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void transmit_reqs_once(int num_of_reqs, uint8_t *reqs, int *req_size_arr, int socket, struct sockaddr *si_other, socklen_t slen) {
  uint8_t *req_addr;
  uint8_t rec_data[4096];
  int sent_items, rec_items;
  int i;
  char req_str[100];
  i = 0;
  req_addr = reqs;
  do {
    sent_items = sendto(socket, req_addr, req_size_arr[i], 0, si_other, slen);
    req_addr += req_size_arr[i];
    rec_items = recvfrom(socket, &rec_data, sizeof(rec_data), 0, si_other, &slen);
    i++;
  } while (i < num_of_reqs);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<transmit_reqs_once>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L2:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L2
  sub    $0x28,%rsp
  xor    %r13d,%r13d
  mov    %edi,0x8(%rsp)
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    %ecx,%ebp
  mov    %r9d,0xc(%rsp)
  mov    %r8,%r14
  lea    0xc(%rsp),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x4018(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  mov    %rax,(%rsp)
  nopl   0x0(%rax,%rax,1)
L14:
  mov    (%r12,%r13,4),%edx
  mov    0xc(%rsp),%r9d
  mov    %rbx,%rsi
  mov    %r14,%r8
  xor    %ecx,%ecx
  mov    %ebp,%edi
  call   <sendto@plt>
  movslq (%r12,%r13,4),%rax
  mov    (%rsp),%rsi
  mov    %r15,%r9
  mov    %r14,%r8
  xor    %ecx,%ecx
  mov    $0x4000,%edx
  mov    %ebp,%edi
  lea    (%rbx,%rax,4),%rbx
  add    $0x1,%r13
  call   <recvfrom@plt>
  cmp    %r13d,0x8(%rsp)
  jg     L14
  mov    0x4018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x4028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void transmit_reqs_once(int num_of_reqs, uint8_t *reqs, int *req_size_arr, int socket, struct sockaddr *si_other, socklen_t slen) {
  uint8_t *req_addr;
  uint8_t rec_data[4096];
  int sent_items, rec_items;
  int i;
  char req_str[100];
  i = 0;
  req_addr = reqs;
  do {
    sent_items = sendto(socket, req_addr, req_size_arr[i], 0, si_other, slen);
    req_addr += req_size_arr[i];
    rec_items = recvfrom(socket, &rec_data, sizeof(rec_data), 0, si_other, &slen);
    i++;
  } while (i < num_of_reqs);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<transmit_reqs_once>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L24:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L24
  sub    $0x28,%rsp
  xor    %r13d,%r13d
  mov    %edi,0x8(%rsp)
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    %ecx,%ebp
  mov    %r9d,0xc(%rsp)
  mov    %r8,%r14
  lea    0xc(%rsp),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x4018(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  mov    %rax,(%rsp)
  nopl   0x0(%rax,%rax,1)
L0:
  mov    (%r12,%r13,4),%edx
  mov    0xc(%rsp),%r9d
  mov    %rbx,%rsi
  mov    %r14,%r8
  xor    %ecx,%ecx
  mov    %ebp,%edi
  call   <sendto@plt>
  movslq (%r12,%r13,4),%rax
  mov    (%rsp),%rsi
  mov    %r15,%r9
  mov    %r14,%r8
  xor    %ecx,%ecx
  mov    $0x4000,%edx
  mov    %ebp,%edi
  lea    (%rbx,%rax,4),%rbx
  add    $0x1,%r13
  call   <recvfrom@plt>
  cmp    %r13d,0x8(%rsp)
  jg     L0
  mov    0x4018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x4028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void transmit_reqs_once(int num_of_reqs, uint8_t *reqs, int *req_size_arr, int socket, struct sockaddr *si_other, socklen_t slen) {
  uint8_t *req_addr;
  uint8_t rec_data[4096];
  int sent_items, rec_items;
  int i;
  char req_str[100];
  i = 0;
  req_addr = reqs;
  do {
    sent_items = sendto(socket, req_addr, req_size_arr[i], 0, si_other, slen);
    req_addr += req_size_arr[i];
    rec_items = recvfrom(socket, &rec_data, sizeof(rec_data), 0, si_other, &slen);
    i++;
  } while (i < num_of_reqs);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<imgGamma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x450,%rsp
  mov    %rdi,-0x428(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %rdx,-0x438(%rbp)
  mov    %rcx,-0x440(%rbp)
  mov    %r8,-0x448(%rbp)
  mov    %r9,-0x450(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x414(%rbp)
  movl   $0x0,-0x414(%rbp)
  jmp    L1
L4:
  mov    -0x430(%rbp),%rax
  movsd  (%rax),%xmm1
  movsd  D9(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x414(%rbp),%xmm0
  movsd  D19(%rip),%xmm1
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <pow@plt>
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x8,%eax
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x414(%rbp),%eax
  cltq
  mov    %ecx,-0x410(%rbp,%rax,4)
  addl   $0x1,-0x414(%rbp)
L1:
  cmpl   $0xff,-0x414(%rbp)
  jle    L4
  lea    -0x410(%rbp),%rdi
  mov    -0x448(%rbp),%rcx
  mov    -0x440(%rbp),%rdx
  mov    -0x438(%rbp),%rsi
  mov    -0x428(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <lookUpTable@plt>
  mov    -0x450(%rbp),%rax
  movl   $0x1,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
data_type: f64
name: disassemble_data
arguments: label: D19
data_type: f64

tool: D9: 1

tool: D19: 255

assistant: ```c
void imgGamma(int *image, double *gamma_value, int *width, int *height, int *depth, int *ret) {
  int i = 0;
  int lut[256];
  for (i = 0; i < 256; i++) {
    lut[i] = (int)(pow(i / 255.0, 1.0 / *gamma_value) * 255);
  }
  lookUpTable(image, width, height, depth, lut);
  *ret = 1;
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<imgGamma>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x428,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,%rbp
  mov    %rdx,%r13
  mov    %rcx,%r14
  mov    %r8,%r15
  mov    %r9,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebx
L4:
  movsd  D7(%rip),%xmm1
  divsd  0x0(%rbp),%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  divsd  D19(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  movapd %xmm1,%xmm0
  call   <pow@plt>
  mov    %eax,%edx
  shl    $0x8,%eax
  sub    %edx,%eax
  mov    %eax,0x10(%rsp,%rbx,4)
  add    $0x1,%rbx
  cmp    $0x100,%rbx
  jne    L4
  lea    0x10(%rsp),%r8
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    0x8(%rsp),%rdi
  call   <lookUpTable@plt>
  movl   $0x1,(%r12)
  mov    0x418(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x428,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
type: f64
name: read_data
arguments: tag: D19
type: f64

tool: D7: 1

tool: D19: 255

assistant: ```c
void imgGamma(int *image, double *gamma_value, int *width, int *height, int *depth, int *ret) {
  int i = 0;
  int lut[256];
  for (i = 0; i < 256; i++) {
    lut[i] = (int)(pow(i / 255.0, 1.0 / *gamma_value) * 255);
  }
  lookUpTable(image, width, height, depth, lut);
  *ret = 1;
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<imgGamma>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x428,%rsp
  mov    %rdi,(%rsp)
  lea    0x10(%rsp),%rbp
  mov    %r9,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  nopl   (%rax)
L17:
  pxor   %xmm0,%xmm0
  movsd  D12(%rip),%xmm1
  divsd  (%rbx),%xmm1
  cvtsi2sd %r12d,%xmm0
  divsd  D11(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  movapd %xmm1,%xmm0
  call   <pow@plt>
  mov    %eax,%r8d
  shl    $0x8,%eax
  sub    %r8d,%eax
  mov    %eax,0x0(%rbp,%r12,4)
  add    $0x1,%r12
  cmp    $0x100,%r12
  jne    L17
  mov    (%rsp),%rdi
  mov    %rbp,%r8
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  call   <lookUpTable@plt>
  mov    0x8(%rsp),%rax
  movl   $0x1,(%rax)
  mov    0x418(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x428,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D12
type_infered: f64
name: unpack_data
arguments: data_tag: D11
type_infered: f64

tool: D12: 1

tool: D11: 255

assistant: ```c
void imgGamma(int *image, double *gamma_value, int *width, int *height, int *depth, int *ret) {
  int i = 0;
  int lut[256];
  for (i = 0; i < 256; i++) {
    lut[i] = (int)(pow(i / 255.0, 1.0 / *gamma_value) * 255);
  }
  lookUpTable(image, width, height, depth, lut);
  *ret = 1;
  return;
}
```