ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inOutQuint>:
  endbr64
  divss  %xmm3,%xmm0
  sub    $0x18,%rsp
  movss  %xmm1,0xc(%rsp)
  mulss  D12(%rip),%xmm2
  movss  D7(%rip),%xmm1
  movss  %xmm2,0x8(%rsp)
  addss  %xmm0,%xmm0
  comiss %xmm0,%xmm1
  ja     L7
  subss  D6(%rip),%xmm0
  mov    $0x5,%edi
  call   <pow@plt>
  pxor   %xmm0,%xmm0
  add    $0x2,%eax
  cvtsi2ss %eax,%xmm0
  mulss  0x8(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  add    $0x18,%rsp
  ret
  nop
L7:
  mov    $0x5,%edi
  call   <pow@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  mulss  0x8(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D12
type_guessed: byte[4]
name: read_data
arguments: data_tag: D7
type_guessed: byte[4]
name: read_data
arguments: data_tag: D6
type_guessed: byte[4]

tool: {"D12": [0x00,0x00,0x00,0x3f]}

tool: {"D7": [0x00,0x00,0x80,0x3f]}

tool: {"D6": [0x00,0x00,0x00,0x40]}

assistant: ```c
float inOutQuint(float t, float b, float c, float d) {
  t = t / d * 2;
  if (t < 1)
    return c / 2 * pow(t, 5) + b;
  return c / 2 * (pow(t - 2, 5) + 2) + b;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<inOutQuint>:
  endbr64
  divss  %xmm3,%xmm0
  sub    $0x18,%rsp
  movss  %xmm1,0xc(%rsp)
  mulss  D7(%rip),%xmm2
  movss  D18(%rip),%xmm1
  movss  %xmm2,0x8(%rsp)
  addss  %xmm0,%xmm0
  comiss %xmm0,%xmm1
  ja     L6
  subss  D28(%rip),%xmm0
  mov    $0x5,%edi
  call   <pow@plt>
  pxor   %xmm0,%xmm0
  add    $0x2,%eax
  cvtsi2ss %eax,%xmm0
  mulss  0x8(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  add    $0x18,%rsp
  ret
  nop
L6:
  mov    $0x5,%edi
  call   <pow@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  mulss  0x8(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D7
datatype: byte[4]
name: parse_data
arguments: tag: D18
datatype: byte[4]
name: parse_data
arguments: tag: D28
datatype: byte[4]

tool: {"D7": [0x00,0x00,0x00,0x3f]}

tool: {"D18": [0x00,0x00,0x80,0x3f]}

tool: {"D28": [0x00,0x00,0x00,0x40]}

assistant: ```c
float inOutQuint(float t, float b, float c, float d) {
  t = t / d * 2;
  if (t < 1)
    return c / 2 * pow(t, 5) + b;
  return c / 2 * (pow(t - 2, 5) + 2) + b;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<print_boilerplate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
data_type: string

tool: D6: wildmac-solver  Copyright (C) 2010  Stefan Guna
This program comes with ABSOLUTELY NO WARRANTY.
This is free software, and you are welcome to redistribute it
under certain conditions; see the accompanying license file
for details.



assistant: ```c
void print_boilerplate() {
  printf("wildmac-solver  Copyright (C) 2010  Stefan Guna\n"
         "This program comes with ABSOLUTELY NO WARRANTY.\n"
         "This is free software, and you are welcome to redistribute it\n"
         "under certain conditions; see the accompanying license file\n"
         "for details.\n\n");
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<print_boilerplate>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D1
type: string

tool: D1: wildmac-solver  Copyright (C) 2010  Stefan Guna
This program comes with ABSOLUTELY NO WARRANTY.
This is free software, and you are welcome to redistribute it
under certain conditions; see the accompanying license file
for details.



assistant: ```c
void print_boilerplate() {
  printf("wildmac-solver  Copyright (C) 2010  Stefan Guna\n"
         "This program comes with ABSOLUTELY NO WARRANTY.\n"
         "This is free software, and you are welcome to redistribute it\n"
         "under certain conditions; see the accompanying license file\n"
         "for details.\n\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<print_boilerplate>:
  endbr64
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_label: D5
type_infered: string

tool: D5: wildmac-solver  Copyright (C) 2010  Stefan Guna
This program comes with ABSOLUTELY NO WARRANTY.
This is free software, and you are welcome to redistribute it
under certain conditions; see the accompanying license file
for details.



assistant: ```c
void print_boilerplate() {
  printf("wildmac-solver  Copyright (C) 2010  Stefan Guna\n"
         "This program comes with ABSOLUTELY NO WARRANTY.\n"
         "This is free software, and you are welcome to redistribute it\n"
         "under certain conditions; see the accompanying license file\n"
         "for details.\n\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

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

tool_calls: name: retrieve_data
arguments: data_tag: D4
data_type: string

tool: D4: wildmac-solver  Copyright (C) 2010  Stefan Guna
This program comes with ABSOLUTELY NO WARRANTY.
This is free software, and you are welcome to redistribute it
under certain conditions; see the accompanying license file
for details.



assistant: ```c
void print_boilerplate() {
  printf("wildmac-solver  Copyright (C) 2010  Stefan Guna\n"
         "This program comes with ABSOLUTELY NO WARRANTY.\n"
         "This is free software, and you are welcome to redistribute it\n"
         "under certain conditions; see the accompanying license file\n"
         "for details.\n\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printstraightupper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,D172(%rip)
  movl   $0x0,D179(%rip)
  jmp    L49
L28:
  mov    D179(%rip),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <islower@plt>
  test   %rax,%rax
  je     L32
  mov    D179(%rip),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    D179(%rip),%ebx
  mov    %eax,%edi
  call   <toupper@plt>
  movslq %ebx,%rdx
  lea    D106(%rip),%rcx
  mov    %al,(%rdx,%rcx,1)
  jmp    L3
L32:
  mov    D179(%rip),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    D179(%rip),%eax
  movzbl (%rdx),%edx
  cltq
  lea    D106(%rip),%rcx
  mov    %dl,(%rax,%rcx,1)
L3:
  mov    D179(%rip),%eax
  cltq
  lea    D106(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isxdigit@plt>
  test   %rax,%rax
  je     L48
  mov    D172(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D38(%rip)
L48:
  mov    D179(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D38(%rip)
L49:
  mov    D38(%rip),%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L28
  mov    D38(%rip),%eax
  cltq
  lea    D106(%rip),%rdx
  movb   $0x0,(%rax,%rdx,1)
  mov    D38(%rip),%edx
  mov    D38(%rip),%eax
  cmp    %eax,%edx
  je     L1
  mov    -0x18(%rbp),%rax
  lea    D106(%rip),%rdx
  lea    D38(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  nop
L1:
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D38
data_type: string

tool: D38: %s


assistant: ```c
void printstraightupper(FILE *fh_out, int len, char *linein) {
  static int p, px;
  static char lineout[128] = {};
  px = 0;
  for (p = 0; p < len; p++) {
    if (islower(linein[p])) {
      lineout[p] = toupper(linein[p]);
    } else {
      lineout[p] = linein[p];
    }
    if (isxdigit(lineout[p])) {
      px++;
    }
  }
  lineout[p] = 0;
  if (p != px) {
    fprintf(fh_out, "%s\n", lineout);
  }
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<printstraightupper>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  movl   $0x0,D52(%rip)
  movl   $0x0,D109(%rip)
  test   %esi,%esi
  jle    L31
  mov    %esi,%r12d
  mov    %rdx,%rbx
  mov    $0x0,%eax
  lea    D28(%rip),%rbp
  jmp    L59
L20:
  movslq D109(%rip),%r14
  movsbl (%rbx,%r14,1),%edi
  call   <toupper@plt>
  mov    %al,0x0(%rbp,%r14,1)
  jmp    L33
L60:
  mov    D109(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D68(%rip)
  cmp    %r12d,%eax
  jge    L79
L59:
  cltq
  movsbl (%rbx,%rax,1),%edi
  call   <islower@plt>
  test   %rax,%rax
  jne    L20
  movslq D68(%rip),%rax
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,0x0(%rbp,%rax,1)
L33:
  movslq D68(%rip),%rax
  movsbl 0x0(%rbp,%rax,1),%edi
  call   <isxdigit@plt>
  test   %rax,%rax
  je     L60
  addl   $0x1,D52(%rip)
  jmp    L60
L31:
  mov    $0x0,%eax
L79:
  movslq %eax,%rdx
  lea    D28(%rip),%rcx
  movb   $0x0,(%rcx,%rdx,1)
  cmp    %eax,D68(%rip)
  jne    L69
L23:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L69:
  mov    %rcx,%rdx
  lea    D68(%rip),%rsi
  mov    %r13,%rdi
  call   <fprintf@plt>
  jmp    L23
```

tool_calls: name: retrieve_data
arguments: label: D68
type_infered: string

tool: D68: %s


assistant: ```c
void printstraightupper(FILE *fh_out, int len, char *linein) {
  static int p, px;
  static char lineout[128] = {};
  px = 0;
  for (p = 0; p < len; p++) {
    if (islower(linein[p])) {
      lineout[p] = toupper(linein[p]);
    } else {
      lineout[p] = linein[p];
    }
    if (isxdigit(lineout[p])) {
      px++;
    }
  }
  lineout[p] = 0;
  if (p != px) {
    fprintf(fh_out, "%s\n", lineout);
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<inverseStrToBCD>:
  endbr64
  mov    %rdi,%r8
  mov    %esi,%r9d
  mov    %rdx,%rsi
  mov    $0x1,%eax
  test   $0x1,%r9b
  jne    L23
  mov    %r9d,%ecx
  shr    %cx
  cmp    $0x1,%r9w
  jbe    L37
  mov    $0x0,%eax
  mov    $0x0,%r10d
  jmp    L79
L78:
  lea    -0x41(%rdi),%edx
  cmp    $0x5,%dl
  ja     L80
  shl    $0x4,%edi
  lea    -0x70(%rdi),%edx
  jmp    L70
L80:
  lea    -0x61(%rdi),%r11d
  mov    %r10d,%edx
  shl    $0x4,%edi
  sub    $0x70,%edi
  cmp    $0x6,%r11b
  cmovb  %edi,%edx
  jmp    L70
L109:
  lea    -0x61(%r11),%edi
  sub    $0x57,%r11d
  cmp    $0x6,%dil
  mov    %r11d,%edi
  cmovae %r10d,%edi
L48:
  or     %edi,%edx
  mov    %dl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %ax,%cx
  jbe    L52
L79:
  movzbl (%r8,%rax,2),%edi
  lea    -0x30(%rdi),%edx
  cmp    $0x9,%dl
  ja     L78
  shl    $0x4,%edi
  mov    %edi,%edx
L70:
  movzbl 0x1(%r8,%rax,2),%r11d
  lea    -0x30(%r11),%edi
  cmp    $0x9,%dil
  jbe    L48
  lea    -0x41(%r11),%edi
  cmp    $0x5,%dil
  ja     L109
  lea    -0x37(%r11),%edi
  jmp    L48
L52:
  cmp    $0x1,%r9w
  mov    $0x1,%eax
  cmovbe %eax,%ecx
  mov    %ecx,%edi
  shr    %di
  cmp    $0x1,%cx
  jbe    L3
  movzwl %cx,%ecx
  add    %rsi,%rcx
  mov    $0x0,%edx
L2:
  movzbl (%rsi,%rdx,1),%eax
  xor    -0x1(%rcx),%al
  mov    %al,(%rsi,%rdx,1)
  xor    -0x1(%rcx),%al
  mov    %al,-0x1(%rcx)
  xor    %al,(%rsi,%rdx,1)
  add    $0x1,%rdx
  sub    $0x1,%rcx
  cmp    %dx,%di
  ja     L2
  mov    $0x0,%eax
  ret
L37:
  mov    $0x0,%eax
  ret
L3:
  mov    $0x0,%eax
L23:
  ret
```

assistant: ```c
unsigned char inverseStrToBCD(unsigned char *s, unsigned short sLen, unsigned char *t, unsigned short tLen) {
  unsigned short i = 0, j = 0;
  if (sLen % 2)
    return 0x01;
  for (i = 0; i < sLen / 2; i++, s += 2) {
    t[i] = (((s[0] >= '0' && s[0] <= '9') ? (s[0] - '0') : ((s[0] >= 'A' && s[0] <= 'F') ? (s[0] - 'A' + 10) : ((s[0] >= 'a' && s[0] <= 'f') ? (s[0] - 'a' + 10) : 0))) << 4 | ((s[1] >= '0' && s[1] <= '9') ? (s[1] - '0') : ((s[1] >= 'A' && s[1] <= 'F') ? (s[1] - 'A' + 10) : ((s[1] >= 'a' && s[1] <= 'f') ? (s[1] - 'a' + 10) : 0))));
  }
  for (j = 0; j < i / 2; j++) {
    t[j] = t[j] ^ t[i - j - 1];
    t[i - j - 1] = t[j] ^ t[i - j - 1];
    t[j] = t[j] ^ t[i - j - 1];
  }
  return 0x00;
}
```
user: Translate the below assembly code into C programming language:

```asm
<inverseStrToBCD>:
  endbr64
  mov    %esi,%r8d
  mov    %rdx,%rax
  mov    $0x1,%r9d
  test   $0x1,%sil
  jne    L135
  mov    %esi,%r9d
  xor    %edx,%edx
  shr    %r9w
  jne    L11
  jmp    L104
  nopw   0x0(%rax,%rax,1)
L95:
  or     %r10d,%esi
L51:
  mov    %sil,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %dx,%r9w
  jbe    L22
L11:
  movzbl (%rdi,%rdx,2),%ecx
  lea    -0x30(%rcx),%esi
  cmp    $0x9,%sil
  ja     L130
  mov    %ecx,%esi
  shl    $0x4,%esi
L26:
  movzbl 0x1(%rdi,%rdx,2),%ecx
  lea    -0x30(%rcx),%r10d
  cmp    $0x9,%r10b
  jbe    L95
  lea    -0x41(%rcx),%r10d
  cmp    $0x5,%r10b
  ja     L71
  sub    $0x37,%ecx
  or     %ecx,%esi
  mov    %sil,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %dx,%r9w
  ja     L11
L22:
  shr    $0x2,%r8w
  je     L104
  sub    $0x1,%r8d
  movzwl %r9w,%r9d
  movzwl %r8w,%r8d
  lea    -0x1(%rax,%r9,1),%rcx
  lea    0x1(%rax,%r8,1),%rsi
  nopl   0x0(%rax)
L70:
  movzbl (%rax),%edx
  xor    (%rcx),%dl
  sub    $0x1,%rcx
  mov    %dl,(%rax)
  xor    0x1(%rcx),%dl
  mov    %dl,0x1(%rcx)
  xor    %dl,(%rax)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L70
L104:
  xor    %r9d,%r9d
L135:
  mov    %r9d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L130:
  lea    -0x41(%rcx),%esi
  cmp    $0x5,%sil
  ja     L14
  shl    $0x4,%ecx
  lea    -0x70(%rcx),%esi
  jmp    L26
  nopl   0x0(%rax)
L71:
  lea    -0x61(%rcx),%r10d
  sub    $0x57,%ecx
  or     %esi,%ecx
  cmp    $0x6,%r10b
  cmovb  %ecx,%esi
  jmp    L51
  nopl   (%rax)
L14:
  lea    -0x61(%rcx),%r10d
  shl    $0x4,%ecx
  xor    %esi,%esi
  sub    $0x70,%ecx
  cmp    $0x6,%r10b
  cmovb  %ecx,%esi
  jmp    L26
```

assistant: ```c
unsigned char inverseStrToBCD(unsigned char *s, unsigned short sLen, unsigned char *t, unsigned short tLen) {
  unsigned short i = 0, j = 0;
  if (sLen % 2)
    return 0x01;
  for (i = 0; i < sLen / 2; i++, s += 2) {
    t[i] = (((s[0] >= '0' && s[0] <= '9') ? (s[0] - '0') : ((s[0] >= 'A' && s[0] <= 'F') ? (s[0] - 'A' + 10) : ((s[0] >= 'a' && s[0] <= 'f') ? (s[0] - 'a' + 10) : 0))) << 4 | ((s[1] >= '0' && s[1] <= '9') ? (s[1] - '0') : ((s[1] >= 'A' && s[1] <= 'F') ? (s[1] - 'A' + 10) : ((s[1] >= 'a' && s[1] <= 'f') ? (s[1] - 'a' + 10) : 0))));
  }
  for (j = 0; j < i / 2; j++) {
    t[j] = t[j] ^ t[i - j - 1];
    t[i - j - 1] = t[j] ^ t[i - j - 1];
    t[j] = t[j] ^ t[i - j - 1];
  }
  return 0x00;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strings_equal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L74
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L58
L74:
  cmpq   $0x0,-0x10(%rbp)
  je     L67
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L58
L67:
  mov    $0x1,%eax
  jmp    L64
L58:
  cmpq   $0x0,-0x8(%rbp)
  je     L40
  cmpq   $0x0,-0x10(%rbp)
  je     L40
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L40
  mov    $0x1,%eax
  jmp    L64
L40:
  mov    $0x0,%eax
L64:
  leave
  ret
```

assistant: ```c
int strings_equal(const char *expected, const char *received) {
  if ((expected == NULL || expected[0] == '\0') && (received == NULL || received[0] == '\0'))
    return 1;
  return (expected && received && strcmp(expected, received) == 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
<strings_equal>:
  endbr64
  test   %rdi,%rdi
  je     L10
  cmpb   $0x0,(%rdi)
  jne    L42
L10:
  test   %rsi,%rsi
  je     L48
  mov    $0x1,%eax
  cmpb   $0x0,(%rsi)
  je     L31
L42:
  test   %rdi,%rdi
  je     L51
  test   %rsi,%rsi
  jne    L56
L51:
  mov    $0x0,%eax
  ret
L56:
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
L48:
  mov    $0x1,%eax
  ret
L31:
  ret
```

assistant: ```c
int strings_equal(const char *expected, const char *received) {
  if ((expected == NULL || expected[0] == '\0') && (received == NULL || received[0] == '\0'))
    return 1;
  return (expected && received && strcmp(expected, received) == 0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strings_equal>:
  endbr64
  test   %rdi,%rdi
  je     L43
  cmpb   $0x0,(%rdi)
  jne    L29
L43:
  mov    $0x1,%eax
  test   %rsi,%rsi
  je     L10
  cmpb   $0x0,(%rsi)
  je     L53
L29:
  test   %rdi,%rdi
  je     L46
  test   %rsi,%rsi
  je     L46
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
  xchg   %ax,%ax
L53:
  ret
  nopl   0x0(%rax)
L46:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  ret
```

assistant: ```c
int strings_equal(const char *expected, const char *received) {
  if ((expected == NULL || expected[0] == '\0') && (received == NULL || received[0] == '\0'))
    return 1;
  return (expected && received && strcmp(expected, received) == 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strings_equal>:
  endbr64
  test   %rdi,%rdi
  je     L55
  cmpb   $0x0,(%rdi)
  jne    L48
L55:
  mov    $0x1,%eax
  test   %rsi,%rsi
  je     L38
  cmpb   $0x0,(%rsi)
  je     L15
L48:
  test   %rdi,%rdi
  je     L14
  test   %rsi,%rsi
  je     L14
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
  xchg   %ax,%ax
L15:
  ret
  nopl   0x0(%rax)
L14:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  ret
```

assistant: ```c
int strings_equal(const char *expected, const char *received) {
  if ((expected == NULL || expected[0] == '\0') && (received == NULL || received[0] == '\0'))
    return 1;
  return (expected && received && strcmp(expected, received) == 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memrchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L17
L23:
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L17
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  jmp    L3
L17:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L23
  mov    $0x0,%eax
L3:
  pop    %rbp
  ret
```

assistant: ```c
void *memrchr(const void *m, int c, size_t n) {
  const uint8_t *s = m;
  c = (uint8_t)c;
  while (n--) {
    if (s[n] == c) {
      return (void *)(s + n);
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<memrchr>:
  endbr64
  lea    -0x4(%rdi,%rdx,4),%rax
L27:
  test   %rdx,%rdx
  je     L0
  mov    %rax,%rcx
  sub    $0x1,%rdx
  sub    $0x4,%rax
  cmp    %esi,(%rcx)
  jne    L27
  jmp    L25
L0:
  mov    $0x0,%ecx
L25:
  mov    %rcx,%rax
  ret
```

assistant: ```c
void *memrchr(const void *m, int c, size_t n) {
  const uint8_t *s = m;
  c = (uint8_t)c;
  while (n--) {
    if (s[n] == c) {
      return (void *)(s + n);
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<memrchr>:
  endbr64
  lea    -0x4(%rdi,%rdx,4),%rax
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L29:
  mov    %rax,%r8
  sub    $0x1,%rdx
  sub    $0x4,%rax
  cmp    %esi,(%r8)
  je     L22
L19:
  test   %rdx,%rdx
  jne    L29
  xor    %r8d,%r8d
L22:
  mov    %r8,%rax
  ret
```

assistant: ```c
void *memrchr(const void *m, int c, size_t n) {
  const uint8_t *s = m;
  c = (uint8_t)c;
  while (n--) {
    if (s[n] == c) {
      return (void *)(s + n);
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<memrchr>:
  endbr64
  lea    -0x4(%rdi,%rdx,4),%rax
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %rax,%r8
  sub    $0x1,%rdx
  sub    $0x4,%rax
  cmp    %esi,(%r8)
  je     L3
L2:
  test   %rdx,%rdx
  jne    L14
  xor    %r8d,%r8d
L3:
  mov    %r8,%rax
  ret
```

assistant: ```c
void *memrchr(const void *m, int c, size_t n) {
  const uint8_t *s = m;
  c = (uint8_t)c;
  while (n--) {
    if (s[n] == c) {
      return (void *)(s + n);
    }
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tarefa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L0
L17:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L17
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type: string

tool: D3:  %d  

assistant: ```c
void tarefa(int r) {
  for (int e = 1; e <= r; e++)
    printf(" %d  ", r);
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<tarefa>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L8
  mov    %edi,%ebp
  mov    $0x1,%ebx
  lea    D3(%rip),%r12
L17:
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jge    L17
L8:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type: string

tool: D3:  %d  

assistant: ```c
void tarefa(int r) {
  for (int e = 1; e <= r; e++)
    printf(" %d  ", r);
  printf("\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<tarefa>:
  endbr64
  test   %edi,%edi
  jle    L8
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jge    L1
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax)
L8:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type_infered: string

tool: D6:  %d  

assistant: ```c
void tarefa(int r) {
  for (int e = 1; e <= r; e++)
    printf(" %d  ", r);
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tarefa>:
  endbr64
  test   %edi,%edi
  jle    L18
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jge    L2
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax)
L18:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: parse_data
arguments: tag: D4
data_type: string

tool: D4:  %d  

assistant: ```c
void tarefa(int r) {
  for (int e = 1; e <= r; e++)
    printf(" %d  ", r);
  printf("\n");
}
```