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

```asm
<characterCounter>:
  endbr64
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x1(%rsi),%edi
  mov    $0x4,%esi
  sub    $0x8,%rsp
  call   <calloc@plt>
  movzbl (%rbx),%ecx
  mov    %rax,%r8
  test   %cl,%cl
  je     L94
  mov    $0x1,%edx
  xor    %eax,%eax
  xor    %r9d,%r9d
  mov    $0x1,%r10d
  jmp    L82
  nopl   0x0(%rax)
L91:
  add    $0x1,%rdx
  movzbl -0x1(%rbx,%rdx,1),%ecx
  test   %cl,%cl
  je     L3
L82:
  mov    %edx,%edi
  cmp    %cl,%bpl
  jne    L91
  add    %rbx,%rax
  cmp    (%rax),%bpl
  je     L70
  mov    %r10d,%esi
  sub    %eax,%esi
  nopl   0x0(%rax)
L56:
  lea    (%rsi,%rax,1),%ecx
  add    $0x1,%rax
  cmp    %bpl,(%rax)
  jne    L56
  movslq %r9d,%rax
  add    $0x1,%r9d
  mov    %ecx,(%r8,%rax,4)
  movslq %edi,%rax
L27:
  add    $0x1,%rdx
  movzbl -0x1(%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L82
  nop
L3:
  add    $0x1,%eax
  movslq %r9d,%r9
  movslq %eax,%rdx
  lea    (%r8,%r9,4),%rsi
L65:
  cmpb   $0x0,(%rbx,%rdx,1)
  je     L83
  cltq
  mov    $0x1,%edx
  add    %rbx,%rax
  sub    %eax,%edx
  nop
L54:
  lea    (%rdx,%rax,1),%ecx
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L54
  mov    %ecx,(%rsi)
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L70:
  movslq %r9d,%rax
  xor    %ecx,%ecx
  add    $0x1,%r9d
  mov    %ecx,(%r8,%rax,4)
  movslq %edi,%rax
  jmp    L27
L83:
  xor    %ecx,%ecx
  mov    %r8,%rax
  mov    %ecx,(%rsi)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L94:
  mov    %rax,%rsi
  mov    $0x1,%edx
  mov    $0x1,%eax
  jmp    L65
```

assistant: ```c
int *characterCounter(char *stringIn, int numTokens, char delim) {
  int i, j, k = 0, x = 0, numChars = 0;
  int *numCharacters = (int *)calloc(numTokens + 1, sizeof(int));
  for (i = 0; stringIn[i] != '\0'; i++) {
    if (stringIn[i] == delim) {
      for (j = x; stringIn[j] != delim; j++) {
        numChars++;
      }
      numCharacters[k] = numChars;
      numChars = 0;
      x = i + 1;
      k++;
    }
  }
  for (j = x + 1; stringIn[j] != '\0'; j++) {
    numChars++;
  }
  numCharacters[k] = numChars;
  return numCharacters;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movabs $0x8be589eac7,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x2,-0x8(%rbp)
  jmp    L13
L17:
  mov    -0x10(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x8(%rbp)
  mov    %rax,-0x10(%rbp)
L18:
  mov    -0x10(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x8(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  je     L17
  addq   $0x1,-0x8(%rbp)
L13:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L18
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_guessed: string

tool: D2: %llu


assistant: ```c
int main(void) {
  unsigned long long n = 600851475143ULL;
  unsigned long long i;
  for (i = 2ULL; i < n; i++) {
    while (n % i == 0) {
      n /= i;
    }
  }
  printf("%llu\n", n);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%ecx
  movabs $0x8be589eac7,%rsi
  jmp    L32
L17:
  mov    %rsi,%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rax,%rsi
  mov    $0x0,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L17
L32:
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  jae    L45
  mov    %rsi,%rax
  mov    $0x0,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L17
  jmp    L32
L45:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D7
type: string

tool: D7: %llu


assistant: ```c
int main(void) {
  unsigned long long n = 600851475143ULL;
  unsigned long long i;
  for (i = 2ULL; i < n; i++) {
    while (n % i == 0) {
      n /= i;
    }
  }
  printf("%llu\n", n);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%ecx
  movabs $0x8be589eac7,%rsi
L25:
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  jbe    L18
L23:
  mov    %rsi,%rax
  xor    %edx,%edx
  div    %rcx
  test   %rdx,%rdx
  jne    L25
  nopl   (%rax)
L37:
  mov    %rsi,%rax
  xor    %edx,%edx
  div    %rcx
  xor    %edx,%edx
  mov    %rax,%rsi
  div    %rcx
  test   %rdx,%rdx
  je     L37
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  ja     L23
L18:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D7: %llu


assistant: ```c
int main(void) {
  unsigned long long n = 600851475143ULL;
  unsigned long long i;
  for (i = 2ULL; i < n; i++) {
    while (n % i == 0) {
      n /= i;
    }
  }
  printf("%llu\n", n);
  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    $0x3,%ecx
  movabs $0x8be589eac7,%rsi
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %rsi,%rax
  xor    %edx,%edx
  div    %rcx
  test   %rdx,%rdx
  jne    L23
  nopl   (%rax)
L16:
  mov    %rsi,%rax
  xor    %edx,%edx
  div    %rcx
  xor    %edx,%edx
  mov    %rax,%rsi
  div    %rcx
  test   %rdx,%rdx
  je     L16
L23:
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  jb     L14
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type: string

tool: D8: %llu


assistant: ```c
int main(void) {
  unsigned long long n = 600851475143ULL;
  unsigned long long i;
  for (i = 2ULL; i < n; i++) {
    while (n % i == 0) {
      n /= i;
    }
  }
  printf("%llu\n", n);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x7a120,-0x4(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jle    L29
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
L29:
  mov    $0xb,%edi
  call   <srand@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L17
L14:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  lea    0x3(%rax),%ecx
  test   %eax,%eax
  cmovs  %ecx,%eax
  sar    $0x2,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  imul   $0x1033c4d7,%eax,%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L17:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L14
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type_infered: string

tool: D1: %u


assistant: ```c
int main(int argc, char *argv[]) {
  int i, n = 500000;
  if (argc > 1)
    n = atoi(argv[1]);
  srand(11);
  for (i = 0; i < n; ++i)
    printf("%u\n", (unsigned)((rand() % (n / 4)) * 271828183u));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jg     L0
  mov    $0xb,%edi
  call   <srand@plt>
  mov    $0x7a120,%r12d
L31:
  lea    0x3(%r12),%ebp
  test   %r12d,%r12d
  cmovns %r12d,%ebp
  sar    $0x2,%ebp
  mov    $0x0,%ebx
  lea    D4(%rip),%r13
L29:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idiv   %ebp
  imul   $0x1033c4d7,%edx,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %r12d,%ebx
  jl     L29
L13:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%r12d
  mov    $0xb,%edi
  call   <srand@plt>
  test   %r12d,%r12d
  jg     L31
  jmp    L13
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
data_type: string

tool: D4: %u


assistant: ```c
int main(int argc, char *argv[]) {
  int i, n = 500000;
  if (argc > 1)
    n = atoi(argv[1]);
  srand(11);
  for (i = 0; i < n; ++i)
    printf("%u\n", (unsigned)((rand() % (n / 4)) * 271828183u));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jg     L24
  mov    $0xb,%edi
  mov    $0x7a120,%ebp
  call   <srand@plt>
L37:
  mov    %ebp,%ebx
  xor    %r13d,%r13d
  lea    D7(%rip),%r12
  sar    $0x2,%ebx
  nopl   0x0(%rax)
L39:
  xor    %eax,%eax
  add    $0x1,%r13d
  call   <rand@plt>
  mov    %r12,%rdi
  cltd
  idiv   %ebx
  imul   $0x1033c4d7,%edx,%esi
  call   <printf@plt>
  cmp    %ebp,%r13d
  jl     L39
L12:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    $0xb,%edi
  mov    %eax,%ebp
  call   <srand@plt>
  test   %ebp,%ebp
  jg     L37
  jmp    L12
```

tool_calls: name: retrieve_data
arguments: label: D7
type: string

tool: D7: %u


assistant: ```c
int main(int argc, char *argv[]) {
  int i, n = 500000;
  if (argc > 1)
    n = atoi(argv[1]);
  srand(11);
  for (i = 0; i < n; ++i)
    printf("%u\n", (unsigned)((rand() % (n / 4)) * 271828183u));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jg     L35
  mov    $0xb,%edi
  mov    $0x7a120,%ebp
  call   <srand@plt>
L24:
  mov    %ebp,%ebx
  xor    %r13d,%r13d
  lea    D6(%rip),%r12
  sar    $0x2,%ebx
  nopl   0x0(%rax)
L8:
  xor    %eax,%eax
  add    $0x1,%r13d
  call   <rand@plt>
  mov    %r12,%rdi
  cltd
  idiv   %ebx
  imul   $0x1033c4d7,%edx,%esi
  call   <printf@plt>
  cmp    %ebp,%r13d
  jl     L8
L2:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L35:
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    $0xb,%edi
  mov    %eax,%ebp
  call   <srand@plt>
  test   %ebp,%ebp
  jg     L24
  jmp    L2
```

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

tool: D6: %u


assistant: ```c
int main(int argc, char *argv[]) {
  int i, n = 500000;
  if (argc > 1)
    n = atoi(argv[1]);
  srand(11);
  for (i = 0; i < n; ++i)
    printf("%u\n", (unsigned)((rand() % (n / 4)) * 271828183u));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<UartReceiveByte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x4000c000,%edi
  call   <UARTCharGetNonBlocking@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0xffffffffffffffff,-0x8(%rbp)
  je     L14
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    $0x1,%eax
  jmp    L2
L14:
  mov    $0x0,%eax
L2:
  leave
  ret
```

assistant: ```c
unsigned char UartReceiveByte(unsigned char *data) {
  signed long result;
  result = UARTCharGetNonBlocking(0x4000C000);
  if (result != -1) {
    *data = (unsigned char)result;
    return 1;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<UartReceiveByte>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x4000c000,%edi
  call   <UARTCharGetNonBlocking@plt>
  mov    $0x0,%edx
  cmp    $0xffffffffffffffff,%rax
  je     L4
  mov    %al,(%rbx)
  mov    $0x1,%edx
L4:
  mov    %edx,%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned char UartReceiveByte(unsigned char *data) {
  signed long result;
  result = UARTCharGetNonBlocking(0x4000C000);
  if (result != -1) {
    *data = (unsigned char)result;
    return 1;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<UartReceiveByte>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x4000c000,%edi
  call   <UARTCharGetNonBlocking@plt>
  xor    %r8d,%r8d
  cmp    $0xffffffffffffffff,%rax
  je     L4
  mov    %al,(%rbx)
  mov    $0x1,%r8d
L4:
  mov    %r8d,%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned char UartReceiveByte(unsigned char *data) {
  signed long result;
  result = UARTCharGetNonBlocking(0x4000C000);
  if (result != -1) {
    *data = (unsigned char)result;
    return 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<UartReceiveByte>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x4000c000,%edi
  call   <UARTCharGetNonBlocking@plt>
  xor    %r8d,%r8d
  cmp    $0xffffffffffffffff,%rax
  je     L4
  mov    %al,(%rbx)
  mov    $0x1,%r8d
L4:
  mov    %r8d,%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned char UartReceiveByte(unsigned char *data) {
  signed long result;
  result = UARTCharGetNonBlocking(0x4000C000);
  if (result != -1) {
    *data = (unsigned char)result;
    return 1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L6
L19:
  addq   $0x1,-0x18(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L19
  mov    -0x18(%rbp),%rax
  sub    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long strlen(const char *str) {
  const char *start = str;
  while (*str)
    str++;
  return str - start;
}
```
user: What is the c source code of the assembly code below:

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L9
  mov    %rdi,%rax
L14:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L14
L26:
  sub    %rdi,%rax
  ret
L9:
  mov    %rdi,%rax
  jmp    L26
```

assistant: ```c
long strlen(const char *str) {
  const char *start = str;
  while (*str)
    str++;
  return str - start;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L7
  mov    %rdi,%rax
  nopl   0x0(%rax)
L8:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L8
  sub    %rdi,%rax
  ret
  nopl   (%rax)
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
long strlen(const char *str) {
  const char *start = str;
  while (*str)
    str++;
  return str - start;
}
```
user: What is the c source code of the assembly code below:

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L7
  mov    %rdi,%rax
  nopl   0x0(%rax)
L17:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L17
  sub    %rdi,%rax
  ret
  nopl   (%rax)
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
long strlen(const char *str) {
  const char *start = str;
  while (*str)
    str++;
  return str - start;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_line>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xe30,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %edx,%eax
  mov    %r8d,%ebx
  mov    %fs:0x28,%rdx
  mov    %rdx,0xe28(%rsp)
  xor    %edx,%edx
  lea    0x3(%rsi,%rsi,2),%r13d
  mov    %r13d,%edx
  and    $0xfffffffc,%edx
  imul   %eax,%edx
  add    %ecx,%edx
  mov    $0x0,%ecx
  mov    $0x0,%esi
  call   <OS_file_seek@plt>
  test   %eax,%eax
  jne    L37
  mov    %ebx,%eax
  sub    %ebp,%eax
  mov    %eax,%ebp
  shr    $0x1f,%ebp
  add    %eax,%ebp
  sar    %ebp
  cmp    $0xffffffff,%eax
  jl     L13
  lea    0x5b0(%rsp),%r14
  mov    $0x870,%edx
  mov    $0x0,%esi
  mov    %r14,%rdi
  call   <OS_memset@plt>
  lea    -0x3(%r13),%edx
  lea    0x0(%rbp,%rbp,2),%eax
  cltq
  lea    (%r14,%rax,1),%rsi
  mov    %r12,%rdi
  call   <OS_file_read@plt>
L0:
  test   %eax,%eax
  je     L2
L37:
  mov    0xe28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L44
  add    $0xe30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  lea    0x0(,%rbp,4),%eax
  mov    %ebp,%edx
  sub    %eax,%edx
  mov    $0x1,%ecx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <OS_file_seek@plt>
  test   %eax,%eax
  jne    L37
  lea    (%rbx,%rbx,2),%edx
  lea    0x5b0(%rsp),%rsi
  mov    %r12,%rdi
  call   <OS_file_read@plt>
  jmp    L0
L2:
  lea    0x5b0(%rsp),%rsi
  lea    0x10(%rsp),%rbp
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <convert_bgr_to_yuv@plt>
  lea    (%rbx,%rbx,1),%esi
  mov    %rbp,%rdi
  call   <FS460_image_set_start_write@plt>
  test   %eax,%eax
  jne    L37
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rbx
L51:
  mov    %rbx,%rdi
  call   <FS460_image_is_transfer_completed@plt>
  mov    %eax,%ecx
  or     0xc(%rsp),%ecx
  je     L51
  jmp    L37
L44:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_line(unsigned long file_handle, int width, int file_y, int header_size, int scaler_width) {
  int err;
  int completed;
  unsigned char line_yuv[720 * 2];
  unsigned char line_bgr[(((720 * 3) + 3) & ~3)];
  int h_offset;
  {}
  h_offset = (scaler_width - width) / 2;
  err = OS_file_seek(file_handle, 0, header_size + ((((width * 3) + 3) & ~3) * file_y), 0);
  if (!err) {
    if (h_offset < 0) {
      err = OS_file_seek(file_handle, 0, 3 * (-h_offset), 1);
      if (!err) {
        err = OS_file_read(file_handle, line_bgr, scaler_width * 3);
      }
    } else {
      OS_memset(line_bgr, 0, sizeof(line_bgr));
      err = OS_file_read(file_handle, line_bgr + (3 * h_offset), width * 3);
    }
    if (!err) {
      convert_bgr_to_yuv(line_yuv, line_bgr, scaler_width);
      err = FS460_image_set_start_write(line_yuv, scaler_width * 2);
      if (!err) {
        completed = 0;
        while (!completed && !err)
          err = FS460_image_is_transfer_completed(&completed);
      }
    }
  }
  return err;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_line>:
  endbr64
  push   %r14
  mov    %edx,%r9d
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x3(%rsi,%rsi,2),%ebp
  xor    %esi,%esi
  mov    %ebp,%edx
  push   %rbx
  mov    %r8d,%ebx
  and    $0xfffffffc,%edx
  imul   %r9d,%edx
  sub    $0xe30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe28(%rsp)
  xor    %eax,%eax
  add    %ecx,%edx
  xor    %ecx,%ecx
  call   <OS_file_seek@plt>
  test   %eax,%eax
  jne    L19
  mov    %ebx,%eax
  sub    %r13d,%eax
  mov    %eax,%r13d
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  cmp    $0xffffffff,%eax
  jl     L35
  lea    0x5b0(%rsp),%r14
  mov    $0x870,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <OS_memset@plt>
  lea    0x0(%r13,%r13,2),%esi
  lea    -0x3(%rbp),%edx
  mov    %r12,%rdi
  movslq %esi,%rsi
  add    %r14,%rsi
  call   <OS_file_read@plt>
L26:
  test   %eax,%eax
  je     L7
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    0xe28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L84
  add    $0xe30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L35:
  lea    0x0(,%r13,4),%eax
  mov    %r13d,%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  sub    %eax,%edx
  mov    $0x1,%ecx
  call   <OS_file_seek@plt>
  test   %eax,%eax
  jne    L19
  lea    0x5b0(%rsp),%r14
  lea    (%rbx,%rbx,2),%edx
  mov    %r12,%rdi
  mov    %r14,%rsi
  call   <OS_file_read@plt>
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L7:
  lea    0x10(%rsp),%rbp
  mov    %r14,%rsi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <convert_bgr_to_yuv@plt>
  lea    (%rbx,%rbx,1),%esi
  mov    %rbp,%rdi
  call   <FS460_image_set_start_write@plt>
  test   %eax,%eax
  jne    L19
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rbx
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %rbx,%rdi
  call   <FS460_image_is_transfer_completed@plt>
  mov    %eax,%ecx
  or     0xc(%rsp),%ecx
  je     L3
  jmp    L19
L84:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_line(unsigned long file_handle, int width, int file_y, int header_size, int scaler_width) {
  int err;
  int completed;
  unsigned char line_yuv[720 * 2];
  unsigned char line_bgr[(((720 * 3) + 3) & ~3)];
  int h_offset;
  {}
  h_offset = (scaler_width - width) / 2;
  err = OS_file_seek(file_handle, 0, header_size + ((((width * 3) + 3) & ~3) * file_y), 0);
  if (!err) {
    if (h_offset < 0) {
      err = OS_file_seek(file_handle, 0, 3 * (-h_offset), 1);
      if (!err) {
        err = OS_file_read(file_handle, line_bgr, scaler_width * 3);
      }
    } else {
      OS_memset(line_bgr, 0, sizeof(line_bgr));
      err = OS_file_read(file_handle, line_bgr + (3 * h_offset), width * 3);
    }
    if (!err) {
      convert_bgr_to_yuv(line_yuv, line_bgr, scaler_width);
      err = FS460_image_set_start_write(line_yuv, scaler_width * 2);
      if (!err) {
        completed = 0;
        while (!completed && !err)
          err = FS460_image_is_transfer_completed(&completed);
      }
    }
  }
  return err;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_line>:
  endbr64
  push   %r14
  mov    %edx,%r9d
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x3(%rsi,%rsi,2),%ebp
  xor    %esi,%esi
  mov    %ebp,%edx
  push   %rbx
  mov    %r8d,%ebx
  and    $0xfffffffc,%edx
  imul   %r9d,%edx
  sub    $0xe30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe28(%rsp)
  xor    %eax,%eax
  add    %ecx,%edx
  xor    %ecx,%ecx
  call   <OS_file_seek@plt>
  test   %eax,%eax
  jne    L22
  mov    %ebx,%eax
  sub    %r13d,%eax
  mov    %eax,%r13d
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  cmp    $0xffffffff,%eax
  jl     L18
  lea    0x5b0(%rsp),%r14
  mov    $0x870,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <OS_memset@plt>
  lea    0x0(%r13,%r13,2),%esi
  lea    -0x3(%rbp),%edx
  mov    %r12,%rdi
  movslq %esi,%rsi
  add    %r14,%rsi
  call   <OS_file_read@plt>
L72:
  test   %eax,%eax
  je     L61
  cs nopw 0x0(%rax,%rax,1)
L22:
  mov    0xe28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0xe30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  lea    0x0(,%r13,4),%eax
  mov    %r13d,%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  sub    %eax,%edx
  mov    $0x1,%ecx
  call   <OS_file_seek@plt>
  test   %eax,%eax
  jne    L22
  lea    0x5b0(%rsp),%r14
  lea    (%rbx,%rbx,2),%edx
  mov    %r12,%rdi
  mov    %r14,%rsi
  call   <OS_file_read@plt>
  jmp    L72
  nopl   0x0(%rax,%rax,1)
L61:
  lea    0x10(%rsp),%rbp
  mov    %r14,%rsi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <convert_bgr_to_yuv@plt>
  lea    (%rbx,%rbx,1),%esi
  mov    %rbp,%rdi
  call   <FS460_image_set_start_write@plt>
  test   %eax,%eax
  jne    L22
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rbx
  nopw   0x0(%rax,%rax,1)
L60:
  mov    %rbx,%rdi
  call   <FS460_image_is_transfer_completed@plt>
  mov    %eax,%ecx
  or     0xc(%rsp),%ecx
  je     L60
  jmp    L22
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_line(unsigned long file_handle, int width, int file_y, int header_size, int scaler_width) {
  int err;
  int completed;
  unsigned char line_yuv[720 * 2];
  unsigned char line_bgr[(((720 * 3) + 3) & ~3)];
  int h_offset;
  {}
  h_offset = (scaler_width - width) / 2;
  err = OS_file_seek(file_handle, 0, header_size + ((((width * 3) + 3) & ~3) * file_y), 0);
  if (!err) {
    if (h_offset < 0) {
      err = OS_file_seek(file_handle, 0, 3 * (-h_offset), 1);
      if (!err) {
        err = OS_file_read(file_handle, line_bgr, scaler_width * 3);
      }
    } else {
      OS_memset(line_bgr, 0, sizeof(line_bgr));
      err = OS_file_read(file_handle, line_bgr + (3 * h_offset), width * 3);
    }
    if (!err) {
      convert_bgr_to_yuv(line_yuv, line_bgr, scaler_width);
      err = FS460_image_set_start_write(line_yuv, scaler_width * 2);
      if (!err) {
        completed = 0;
        while (!completed && !err)
          err = FS460_image_is_transfer_completed(&completed);
      }
    }
  }
  return err;
}
```
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
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
data_type: string

tool: D8: Hello, World.

assistant: ```c
int main(void) {
  puts("Hello, World.");
  return 0;
}
```