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

```asm
<findTwoNonRepeatElements>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,%r15d
  mov    %rdx,%r13
  mov    %rcx,%r14
  test   %esi,%esi
  jle    L66
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rdx
  mov    $0x0,%ebx
L40:
  xor    (%rax),%ebx
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L40
  mov    %ebx,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%r12d
  and    $0x1,%r12d
  jne    L1
L123:
  mov    $0x0,%ebp
L81:
  sar    %ebx
  add    $0x1,%ebp
  mov    %ebx,%r12d
  and    $0x1,%r12d
  je     L81
  mov    %ebp,%esi
  lea    D69(%rip),%rdi
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L84
L50:
  add    %r12d,%r12d
  sub    $0x1,%ebp
  jne    L50
L84:
  mov    %r12d,%esi
  lea    D71(%rip),%rdi
  call   <printf@plt>
  test   %r15d,%r15d
  jle    L117
L7:
  mov    0x8(%rsp),%rsi
  mov    %rsi,%rax
  lea    -0x1(%r15),%edx
  lea    0x4(%rsi,%rdx,4),%rcx
  jmp    L52
L18:
  xor    %edx,(%r14)
L89:
  add    $0x4,%rax
  cmp    %rcx,%rax
  je     L117
L52:
  mov    (%rax),%edx
  test   %r12d,%edx
  je     L18
  xor    %edx,0x0(%r13)
  jmp    L89
L117:
  mov    0x0(%r13),%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    (%r14),%esi
  lea    D50(%rip),%rdi
  call   <printf@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  mov    $0x0,%esi
  lea    D69(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D71(%rip),%rdi
  call   <printf@plt>
  jmp    L7
L66:
  mov    $0x0,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%ebx
  jmp    L123
```

tool_calls: name: read_data
arguments: data_label: D4
type_infered: string
name: read_data
arguments: data_label: D69
type_infered: string
name: read_data
arguments: data_label: D71
type_infered: string
name: read_data
arguments: data_label: D1
type_infered: string
name: read_data
arguments: data_label: D50
type_infered: string

tool: D4: Sum = %04X


tool: D69: First 1 from left (position from 0) = %d


tool: D71: Flag = %04X


tool: D1: X = %d


tool: D50: Y = %d


assistant: ```c
void findTwoNonRepeatElements(int *pdata, int size, int *pX, int *pY) {
  int i;
  int sum = 0;
  for (i = 0; i < size; i++) {
    sum ^= pdata[i];
  }
  printf("Sum = %04X\n", sum);
  i = 0;
  while ((sum & 1) == 0) {
    sum >>= 1;
    i++;
  }
  printf("First 1 from left (position from 0) = %d\n", i);
  int flag = 1;
  while (i > 0) {
    flag <<= 1;
    i--;
  }
  printf("Flag = %04X\n", flag);
  for (i = 0; i < size; i++) {
    if (pdata[i] & flag) {
      *pX ^= pdata[i];
    } else {
      *pY ^= pdata[i];
    }
  }
  printf("X = %d\n", *pX);
  printf("Y = %d\n", *pY);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<findTwoNonRepeatElements>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rcx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  test   %esi,%esi
  jle    L110
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi),%r8
  mov    %rdi,%rax
  xor    %ebx,%ebx
  mov    %rdx,%rbp
  lea    (%r8,%rdx,4),%rdx
  nopw   0x0(%rax,%rax,1)
L1:
  xor    (%rax),%ebx
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L1
  mov    %ebx,%esi
  lea    D25(%rip),%rdi
  mov    %ebx,%r12d
  mov    %r8,0x8(%rsp)
  call   <printf@plt>
  and    $0x1,%r12d
  mov    0x8(%rsp),%r8
  jne    L107
L43:
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L86:
  sar    %ebx
  add    $0x1,%ebp
  mov    %ebx,%r12d
  and    $0x1,%r12d
  je     L86
  mov    %ebp,%esi
  lea    D76(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L45:
  add    %r12d,%r12d
  sub    $0x1,%ebp
  jne    L45
  mov    %r12d,%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  test   %r14d,%r14d
  jle    L35
  mov    (%rsp),%rax
  lea    -0x1(%r14),%ebp
  lea    0x4(%rax),%r8
L82:
  mov    (%rsp),%rax
  lea    (%r8,%rbp,4),%rsi
  jmp    L54
  nopw   0x0(%rax,%rax,1)
L11:
  add    $0x4,%rax
  xor    %edx,(%r15)
  cmp    %rax,%rsi
  je     L35
L54:
  mov    (%rax),%edx
  test   %r12d,%edx
  jne    L11
  add    $0x4,%rax
  xor    %edx,0x0(%r13)
  cmp    %rax,%rsi
  jne    L54
L35:
  mov    (%r15),%esi
  lea    D43(%rip),%rdi
  call   <printf@plt>
  mov    0x0(%r13),%esi
  add    $0x18,%rsp
  lea    D63(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <printf@plt>
  nopl   (%rax)
L107:
  xor    %esi,%esi
  lea    D76(%rip),%rdi
  mov    %r8,0x8(%rsp)
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%r8
  jmp    L82
  xchg   %ax,%ax
L110:
  xor    %esi,%esi
  lea    D25(%rip),%rdi
  xor    %ebx,%ebx
  call   <printf@plt>
  jmp    L43
```

tool_calls: name: retrieve_data
arguments: label: D25
data_type: string
name: retrieve_data
arguments: label: D76
data_type: string
name: retrieve_data
arguments: label: D28
data_type: string
name: retrieve_data
arguments: label: D43
data_type: string
name: retrieve_data
arguments: label: D63
data_type: string

tool: D25: Sum = %04X


tool: D76: First 1 from left (position from 0) = %d


tool: D28: Flag = %04X


tool: D43: X = %d


tool: D63: Y = %d


assistant: ```c
void findTwoNonRepeatElements(int *pdata, int size, int *pX, int *pY) {
  int i;
  int sum = 0;
  for (i = 0; i < size; i++) {
    sum ^= pdata[i];
  }
  printf("Sum = %04X\n", sum);
  i = 0;
  while ((sum & 1) == 0) {
    sum >>= 1;
    i++;
  }
  printf("First 1 from left (position from 0) = %d\n", i);
  int flag = 1;
  while (i > 0) {
    flag <<= 1;
    i--;
  }
  printf("Flag = %04X\n", flag);
  for (i = 0; i < size; i++) {
    if (pdata[i] & flag) {
      *pX ^= pdata[i];
    } else {
      *pY ^= pdata[i];
    }
  }
  printf("X = %d\n", *pX);
  printf("Y = %d\n", *pY);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <sum@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x9,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  cmp    $0x2d,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xa,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  cmp    $0x37,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x64,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  cmp    $0x13ba,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x5,%esi
  mov    $0x3,%edi
  call   <sum@plt>
  cmp    $0xc,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  assert(0 == sum(0, 0));
  assert(1 == sum(1, 1));
  assert(45 == sum(1, 9));
  assert(55 == sum(1, 10));
  assert(5050 == sum(1, 100));
  assert(12 == sum(3, 5));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <sum@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  cmp    $0x1,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x9,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  cmp    $0x2d,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0xa,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  cmp    $0x37,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x64,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  cmp    $0x13ba,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x5,%esi
  mov    $0x3,%edi
  call   <sum@plt>
  cmp    $0xc,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  assert(0 == sum(0, 0));
  assert(1 == sum(1, 1));
  assert(45 == sum(1, 9));
  assert(55 == sum(1, 10));
  assert(5050 == sum(1, 100));
  assert(12 == sum(3, 5));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  call   <sum@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0x1,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x9,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0x2d,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0xa,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0x37,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x64,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0x13ba,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x5,%esi
  mov    $0x3,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0xc,%eax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  assert(0 == sum(0, 0));
  assert(1 == sum(1, 1));
  assert(45 == sum(1, 9));
  assert(55 == sum(1, 10));
  assert(5050 == sum(1, 100));
  assert(12 == sum(3, 5));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  call   <sum@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0x1,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x9,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0x2d,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0xa,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0x37,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x64,%esi
  mov    $0x1,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0x13ba,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x5,%esi
  mov    $0x3,%edi
  call   <sum@plt>
  xor    %edi,%edi
  cmp    $0xc,%eax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  assert(0 == sum(0, 0));
  assert(1 == sum(1, 1));
  assert(45 == sum(1, 9));
  assert(55 == sum(1, 10));
  assert(5050 == sum(1, 100));
  assert(12 == sum(3, 5));
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<allocate_row>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L11
L19:
  mov    -0x24(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L17
  mov    -0x24(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x1,%esi
  mov    $0xa1,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
L17:
  addl   $0x1,-0x14(%rbp)
L11:
  cmpl   $0x2,-0x14(%rbp)
  jle    L19
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void allocate_row(int row, char **output) {
  for (int i = 0; i < 3; i++) {
    if (output[row - i] == NULL) {
      output[row - i] = calloc(40 * 4 + 1, sizeof(char));
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<allocate_row>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movslq %edi,%rdi
  shl    $0x3,%rdi
  lea    (%rsi,%rdi,1),%rbx
  lea    -0x18(%rsi,%rdi,1),%rbp
L18:
  cmpq   $0x0,(%rbx)
  je     L17
L22:
  sub    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L18
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  mov    $0x1,%esi
  mov    $0xa1,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
  jmp    L22
```

assistant: ```c
void allocate_row(int row, char **output) {
  for (int i = 0; i < 3; i++) {
    if (output[row - i] == NULL) {
      output[row - i] = calloc(40 * 4 + 1, sizeof(char));
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<allocate_row>:
  endbr64
  push   %rbp
  movslq %edi,%rdi
  push   %rbx
  shl    $0x3,%rdi
  lea    (%rsi,%rdi,1),%rbx
  lea    -0x18(%rsi,%rdi,1),%rbp
  sub    $0x8,%rsp
L13:
  cmpq   $0x0,(%rbx)
  je     L11
L27:
  sub    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L13
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L11:
  mov    $0x1,%esi
  mov    $0xa1,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
  jmp    L27
```

assistant: ```c
void allocate_row(int row, char **output) {
  for (int i = 0; i < 3; i++) {
    if (output[row - i] == NULL) {
      output[row - i] = calloc(40 * 4 + 1, sizeof(char));
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<allocate_row>:
  endbr64
  movslq %edi,%rdi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  lea    0x0(,%rdi,8),%rbx
  lea    (%rsi,%rbx,1),%r12
  cmpq   $0x0,(%r12)
  je     L3
  lea    -0x8(%rbp,%rbx,1),%r12
  cmpq   $0x0,(%r12)
  je     L17
L37:
  lea    -0x10(%rbp,%rbx,1),%rbx
  cmpq   $0x0,(%rbx)
  je     L22
L49:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L3:
  mov    $0x1,%esi
  mov    $0xa1,%edi
  call   <calloc@plt>
  mov    %rax,(%r12)
  lea    -0x8(%rbp,%rbx,1),%r12
  cmpq   $0x0,(%r12)
  jne    L37
  nop
L17:
  mov    $0x1,%esi
  mov    $0xa1,%edi
  lea    -0x10(%rbp,%rbx,1),%rbx
  call   <calloc@plt>
  mov    %rax,(%r12)
  cmpq   $0x0,(%rbx)
  jne    L49
L22:
  mov    $0x1,%esi
  mov    $0xa1,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void allocate_row(int row, char **output) {
  for (int i = 0; i < 3; i++) {
    if (output[row - i] == NULL) {
      output[row - i] = calloc(40 * 4 + 1, sizeof(char));
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<FrenchToSdn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L32
  cmpl   $0xe,-0x4(%rbp)
  jg     L32
  cmpl   $0x0,-0x8(%rbp)
  jle    L32
  cmpl   $0xd,-0x8(%rbp)
  jg     L32
  cmpl   $0x0,-0xc(%rbp)
  jle    L32
  cmpl   $0x1e,-0xc(%rbp)
  jle    L24
L32:
  mov    $0x0,%eax
  jmp    L34
L24:
  mov    -0x4(%rbp),%eax
  imul   $0x5b5,%eax,%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  imul   $0x1e,%eax,%eax
  add    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  add    $0x243f32,%eax
  cltq
L34:
  pop    %rbp
  ret
```

assistant: ```c
long int FrenchToSdn(int year, int month, int day) {
  if (year < 1 || year > 14 || month < 1 || month > 13 || day < 1 || day > 30) {
    return (0);
  }
  return ((year * 1461) / 4 + (month - 1) * 30 + day + 2375474);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<FrenchToSdn>:
  endbr64
  lea    -0x1(%rdi),%ecx
  mov    $0x0,%eax
  cmp    $0xd,%ecx
  ja     L12
  lea    -0x1(%rsi),%ecx
  cmp    $0xc,%ecx
  ja     L12
  lea    -0x1(%rdx),%ecx
  cmp    $0x1d,%ecx
  ja     L12
  imul   $0x5b5,%edi,%edi
  lea    0x3(%rdi),%eax
  test   %edi,%edi
  cmovns %edi,%eax
  sar    $0x2,%eax
  sub    $0x1,%esi
  imul   $0x1e,%esi,%esi
  add    %esi,%eax
  lea    0x243f32(%rdx,%rax,1),%eax
  cltq
L12:
  ret
```

assistant: ```c
long int FrenchToSdn(int year, int month, int day) {
  if (year < 1 || year > 14 || month < 1 || month > 13 || day < 1 || day > 30) {
    return (0);
  }
  return ((year * 1461) / 4 + (month - 1) * 30 + day + 2375474);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<FrenchToSdn>:
  endbr64
  lea    -0x1(%rdi),%ecx
  xor    %eax,%eax
  cmp    $0xd,%ecx
  ja     L19
  sub    $0x1,%esi
  cmp    $0xc,%esi
  ja     L19
  lea    -0x1(%rdx),%ecx
  cmp    $0x1d,%ecx
  ja     L19
  imul   $0x5b5,%edi,%edi
  imul   $0x1e,%esi,%esi
  sar    $0x2,%edi
  add    %esi,%edi
  lea    0x243f32(%rdx,%rdi,1),%eax
  cltq
L19:
  ret
```

assistant: ```c
long int FrenchToSdn(int year, int month, int day) {
  if (year < 1 || year > 14 || month < 1 || month > 13 || day < 1 || day > 30) {
    return (0);
  }
  return ((year * 1461) / 4 + (month - 1) * 30 + day + 2375474);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<FrenchToSdn>:
  endbr64
  lea    -0x1(%rdi),%ecx
  xor    %eax,%eax
  cmp    $0xd,%ecx
  ja     L21
  sub    $0x1,%esi
  cmp    $0xc,%esi
  ja     L21
  lea    -0x1(%rdx),%ecx
  cmp    $0x1d,%ecx
  ja     L21
  imul   $0x5b5,%edi,%edi
  imul   $0x1e,%esi,%esi
  sar    $0x2,%edi
  add    %esi,%edi
  lea    0x243f32(%rdx,%rdi,1),%eax
  cltq
L21:
  ret
```

assistant: ```c
long int FrenchToSdn(int year, int month, int day) {
  if (year < 1 || year > 14 || month < 1 || month > 13 || day < 1 || day > 30) {
    return (0);
  }
  return ((year * 1461) / 4 + (month - 1) * 30 + day + 2375474);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<init_serial_comm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1c200,%edi
  call   <UART1_Init@plt>
  mov    $0xc8,%edi
  call   <Delay_ms@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void init_serial_comm() {
  UART1_Init(115200);
  Delay_ms(200);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<init_serial_comm>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1c200,%edi
  call   <UART1_Init@plt>
  mov    $0xc8,%edi
  call   <Delay_ms@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void init_serial_comm() {
  UART1_Init(115200);
  Delay_ms(200);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<init_serial_comm>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1c200,%edi
  call   <UART1_Init@plt>
  mov    $0xc8,%edi
  add    $0x8,%rsp
  jmp    1060 <Delay_ms@plt>
```

assistant: ```c
void init_serial_comm() {
  UART1_Init(115200);
  Delay_ms(200);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<init_serial_comm>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1c200,%edi
  call   <UART1_Init@plt>
  mov    $0xc8,%edi
  add    $0x8,%rsp
  jmp    1060 <Delay_ms@plt>
```

assistant: ```c
void init_serial_comm() {
  UART1_Init(115200);
  Delay_ms(200);
}
```
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    $0x10,%rsp
  movl   $0x1,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L2
L4:
  mov    $0x0,%eax
  call   <unknown2@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x5,-0x4(%rbp)
  jle    L29
  addl   $0x1,-0x8(%rbp)
L29:
  mov    -0x8(%rbp),%eax
  add    %eax,-0xc(%rbp)
L2:
  mov    $0x0,%eax
  call   <unknown1@plt>
  test   %rax,%rax
  jne    L4
  cmpl   $0x0,-0xc(%rbp)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int k = 1;
  int w = 1;
  while (unknown1()) {
    int z = unknown2();
    if (z > 5)
      w++;
    k += w;
  }
  if (k <= 0) {
    goto ERROR;
  ERROR:;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  jmp    L9
L4:
  mov    $0x0,%eax
  call   <unknown2@plt>
L9:
  mov    $0x0,%eax
  call   <unknown1@plt>
  test   %rax,%rax
  jne    L4
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int k = 1;
  int w = 1;
  while (unknown1()) {
    int z = unknown2();
    if (z > 5)
      w++;
    k += w;
  }
  if (k <= 0) {
    goto ERROR;
  ERROR:;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  call   <unknown2@plt>
L13:
  xor    %eax,%eax
  call   <unknown1@plt>
  test   %rax,%rax
  jne    L6
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int k = 1;
  int w = 1;
  while (unknown1()) {
    int z = unknown2();
    if (z > 5)
      w++;
    k += w;
  }
  if (k <= 0) {
    goto ERROR;
  ERROR:;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  call   <unknown2@plt>
L10:
  xor    %eax,%eax
  call   <unknown1@plt>
  test   %rax,%rax
  jne    L6
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int k = 1;
  int w = 1;
  while (unknown1()) {
    int z = unknown2();
    if (z > 5)
      w++;
    k += w;
  }
  if (k <= 0) {
    goto ERROR;
  ERROR:;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<errorcheck_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x11(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L20
L26:
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L20:
  mov    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <contains_char_const@plt>
  test   %eax,%eax
  je     L26
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D36
type_guessed: string
name: retrieve_data
arguments: data_tag: D19
type_guessed: string
name: retrieve_data
arguments: data_tag: D24
type_guessed: string

tool: D36: %s

tool: D19:  %c

tool: D24: INVALID ENTRY

assistant: ```c
char *errorcheck_char(char *prompt, char *legal) {
  char container;
  char *response = &container;
  printf("%s", prompt);
  scanf(" %c", response);
  while (!contains_char_const(legal, response)) {
    printf("INVALID ENTRY\n");
    printf("%s", prompt);
    scanf(" %c", response);
  }
  return response;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<errorcheck_char>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rsi
  lea    D31(%rip),%rdi
  call   <printf@plt>
  lea    0x7(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  lea    0x7(%rsp),%rbx
  lea    D6(%rip),%r15
  lea    D31(%rip),%r14
  lea    D2(%rip),%r13
  jmp    L16
L3:
  mov    %r15,%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
L16:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <contains_char_const@plt>
  test   %eax,%eax
  je     L3
  lea    0x7(%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D31
data_type: string
name: disassemble_data
arguments: label: D2
data_type: string
name: disassemble_data
arguments: label: D6
data_type: string

tool: D31: %s

tool: D2:  %c

tool: D6: INVALID ENTRY

assistant: ```c
char *errorcheck_char(char *prompt, char *legal) {
  char container;
  char *response = &container;
  printf("%s", prompt);
  scanf(" %c", response);
  while (!contains_char_const(legal, response)) {
    printf("INVALID ENTRY\n");
    printf("%s", prompt);
    scanf(" %c", response);
  }
  return response;
}
```