ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<benchmark>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  mov    %esi,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %esi,%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L99
L16:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L16
L99:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L96
L72:
  mov    %rsp,%r13
  test   %ebx,%ebx
  jle    L41
  mov    %ebx,%eax
  mov    %r13,%r15
  lea    0x0(%r13,%rax,4),%r14
  nopl   0x0(%rax)
L23:
  xor    %eax,%eax
  add    $0x4,%r15
  call   <rand_float@plt>
  mulss  D14(%rip),%xmm0
  subss  D34(%rip),%xmm0
  movss  %xmm0,-0x4(%r15)
  cmp    %r14,%r15
  jne    L23
L41:
  xor    %eax,%eax
  call   <tic@plt>
  movsd  %xmm0,-0x48(%rbp)
  test   %r12d,%r12d
  jle    L48
  xor    %r14d,%r14d
  jmp    L85
  nopl   0x0(%rax,%rax,1)
L83:
  add    $0x1,%r14d
  cmp    %r14d,%r12d
  je     L48
L85:
  movss  D10(%rip),%xmm0
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    %ebx,%edi
  call   <stdlib_strided_dsvariancepn@plt>
  ucomisd %xmm0,%xmm0
  jnp    L83
  lea    D8(%rip),%r12
  mov    %r12,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  -0x48(%rbp),%xmm0
  mov    %r12,%rdi
  movsd  %xmm0,-0x48(%rbp)
  call   <printf@plt>
  movsd  -0x48(%rbp),%xmm0
L65:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax)
L48:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  -0x48(%rbp),%xmm0
  jmp    L65
  cs nopw 0x0(%rax,%rax,1)
L96:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L72
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D14
datatype: byte[4]
name: retrieve_data
arguments: data_label: D34
datatype: byte[4]
name: retrieve_data
arguments: data_label: D10
datatype: byte[4]
name: retrieve_data
arguments: data_label: D8
datatype: string

tool: {"D14": [0x00,0x40,0x9c,0x46]}

tool: {"D34": [0x00,0x40,0x1c,0x46]}

tool: {"D10": [0x00,0x00,0x80,0x3f]}

tool: D8: should not return NaN


assistant: ```c
double benchmark(int iterations, int len) {
  double elapsed;
  float x[len];
  double v;
  double t;
  int i;
  for (i = 0; i < len; i++) {
    x[i] = (rand_float() * 20000.0f) - 10000.0f;
  }
  v = 0.0;
  t = tic();
  for (i = 0; i < iterations; i++) {
    v = stdlib_strided_dsvariancepn(len, 1.0f, x, 1);
    if (v != v) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (v != v) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<build_url>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %ebx,%eax
  add    $0xe,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L2
  mov    $0x0,%eax
  jmp    L17
L2:
  mov    -0x38(%rbp),%rdi
  mov    -0x2c(%rbp),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rdi,%r9
  mov    %ecx,%r8d
  mov    %rdx,%rcx
  lea    D8(%rip),%rdx
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    -0x18(%rbp),%rax
L17:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type_infered: string

tool: D8: http://%s:%d%s

assistant: ```c
char *build_url(const char *host, int port, const char *path) {
  char *url;
  int len;
  len = strlen(host) + strlen(path) + 14;
  url = malloc(len);
  if (url == NULL)
    return NULL;
  snprintf(url, len, "http://%s:%d%s", host, port, path);
  return url;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<build_url>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r13d
  mov    %rdx,%r12
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0xe(%rbx,%rax,1),%r14d
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L1
  mov    %r12,%r9
  mov    %r13d,%r8d
  mov    %rbp,%rcx
  lea    D4(%rip),%rdx
  mov    %r14d,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
L1:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

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

tool: D4: http://%s:%d%s

assistant: ```c
char *build_url(const char *host, int port, const char *path) {
  char *url;
  int len;
  len = strlen(host) + strlen(path) + 14;
  url = malloc(len);
  if (url == NULL)
    return NULL;
  snprintf(url, len, "http://%s:%d%s", host, port, path);
  return url;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<build_url>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  lea    0xe(%r12,%rax,1),%r14d
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L5
  mov    %rbx,%r9
  mov    %ebp,%r8d
  mov    %r13,%rcx
  mov    %r14d,%esi
  lea    D6(%rip),%rdx
  mov    %rax,%rdi
  call   <snprintf@plt>
L5:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

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

tool: D6: http://%s:%d%s

assistant: ```c
char *build_url(const char *host, int port, const char *path) {
  char *url;
  int len;
  len = strlen(host) + strlen(path) + 14;
  url = malloc(len);
  if (url == NULL)
    return NULL;
  snprintf(url, len, "http://%s:%d%s", host, port, path);
  return url;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<build_url>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  lea    0xe(%r12,%rax,1),%r14d
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L3
  mov    %rbx,%r9
  mov    %ebp,%r8d
  mov    %r13,%rcx
  mov    %r14d,%esi
  lea    D1(%rip),%rdx
  mov    %rax,%rdi
  call   <snprintf@plt>
L3:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
datatype: string

tool: D1: http://%s:%d%s

assistant: ```c
char *build_url(const char *host, int port, const char *path) {
  char *url;
  int len;
  len = strlen(host) + strlen(path) + 14;
  url = malloc(len);
  if (url == NULL)
    return NULL;
  snprintf(url, len, "http://%s:%d%s", host, port, path);
  return url;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<itoa>:
  endbr64
  mov    %edi,%ecx
  mov    %edx,%r9d
  lea    -0x2(%rsi),%eax
  cmp    $0xe,%eax
  ja     L36
  mov    $0x0,%r8d
  lea    D10(%rip),%rdi
  lea    D12(%rip),%r10
L8:
  sub    $0x1,%rdi
  mov    %ecx,%eax
  mov    $0x0,%edx
  div    %esi
  movzbl (%r10,%rdx,1),%edx
  mov    %dl,(%rdi)
  mov    %ecx,%edx
  mov    %eax,%ecx
  add    $0x1,%r8d
  cmp    %r9d,%r8d
  jb     L8
  cmp    %esi,%edx
  jae    L8
L7:
  mov    %rdi,%rax
  ret
L36:
  mov    $0x0,%edi
  jmp    L7
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
type_guessed: string

tool: D12: 0123456789abcdef

assistant: ```c
char *itoa(unsigned int n, unsigned int base, unsigned int width) {
  if (base < 2 || base > 16) {
    return NULL;
  }
  static const char *digits = "0123456789abcdef";
  static char buffer[65];
  char *s = &buffer[64];
  unsigned int count = 0;
  do {
    *--s = digits[n % base];
    n /= base;
    count++;
  } while (count < width || n != 0);
  return s;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<itoa>:
  endbr64
  lea    -0x2(%rsi),%eax
  mov    %edi,%ecx
  mov    %edx,%r9d
  cmp    $0xe,%eax
  ja     L21
  xor    %edi,%edi
  lea    D12(%rip),%r8
  lea    D14(%rip),%r10
  nopl   0x0(%rax)
L10:
  mov    %ecx,%eax
  xor    %edx,%edx
  sub    $0x1,%r8
  add    $0x1,%edi
  div    %esi
  movzbl (%r10,%rdx,1),%edx
  mov    %dl,(%r8)
  mov    %ecx,%edx
  mov    %eax,%ecx
  cmp    %r9d,%edi
  jb     L10
  cmp    %esi,%edx
  jae    L10
  mov    %r8,%rax
  ret
  xchg   %ax,%ax
L21:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D14
type: string

tool: D14: 0123456789abcdef

assistant: ```c
char *itoa(unsigned int n, unsigned int base, unsigned int width) {
  if (base < 2 || base > 16) {
    return NULL;
  }
  static const char *digits = "0123456789abcdef";
  static char buffer[65];
  char *s = &buffer[64];
  unsigned int count = 0;
  do {
    *--s = digits[n % base];
    n /= base;
    count++;
  } while (count < width || n != 0);
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<itoa>:
  endbr64
  lea    -0x2(%rsi),%eax
  mov    %edi,%ecx
  mov    %edx,%r9d
  cmp    $0xe,%eax
  ja     L29
  xor    %edi,%edi
  lea    D10(%rip),%r8
  lea    D3(%rip),%r10
  nopl   0x0(%rax)
L26:
  mov    %ecx,%eax
  xor    %edx,%edx
  sub    $0x1,%r8
  add    $0x1,%edi
  div    %esi
  movzbl (%r10,%rdx,1),%edx
  mov    %dl,(%r8)
  mov    %ecx,%edx
  mov    %eax,%ecx
  cmp    %r9d,%edi
  jb     L26
  cmp    %esi,%edx
  jae    L26
  mov    %r8,%rax
  ret
  xchg   %ax,%ax
L29:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type: string

tool: D3: 0123456789abcdef

assistant: ```c
char *itoa(unsigned int n, unsigned int base, unsigned int width) {
  if (base < 2 || base > 16) {
    return NULL;
  }
  static const char *digits = "0123456789abcdef";
  static char buffer[65];
  char *s = &buffer[64];
  unsigned int count = 0;
  do {
    *--s = digits[n % base];
    n /= base;
    count++;
  } while (count < width || n != 0);
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<DUMP>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L1:
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  cmp    %rax,-0x20(%rbp)
  ja     L1
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type_infered: string

tool: D6: 0x%x 

assistant: ```c
void DUMP(const char *s, size_t n) {
  unsigned int i;
  for (i = 0; i < n; i++)
    printf("0x%x ", s[i]);
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<DUMP>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L5
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0x0,%ebx
  mov    $0x0,%eax
  lea    D0(%rip),%r13
L9:
  movsbl (%r12,%rax,1),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  mov    %ebx,%eax
  cmp    %rbp,%rax
  jb     L9
L5:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
data_type: string

tool: D0: 0x%x 

assistant: ```c
void DUMP(const char *s, size_t n) {
  unsigned int i;
  for (i = 0; i < n; i++)
    printf("0x%x ", s[i]);
  printf("\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<DUMP>:
  endbr64
  test   %rsi,%rsi
  je     L10
  push   %r13
  xor    %eax,%eax
  lea    D7(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L5:
  movsbl (%r12,%rax,1),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x1(%rbx),%eax
  mov    %rax,%rbx
  cmp    %rbp,%rax
  jb     L5
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: data_tag: D7
type_infered: string

tool: D7: 0x%x 

assistant: ```c
void DUMP(const char *s, size_t n) {
  unsigned int i;
  for (i = 0; i < n; i++)
    printf("0x%x ", s[i]);
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<DUMP>:
  endbr64
  test   %rsi,%rsi
  je     L9
  push   %r13
  xor    %eax,%eax
  lea    D8(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L19:
  movsbl (%r12,%rax,1),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x1(%rbx),%eax
  mov    %rax,%rbx
  cmp    %rbp,%rax
  jb     L19
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type: string

tool: D8: 0x%x 

assistant: ```c
void DUMP(const char *s, size_t n) {
  unsigned int i;
  for (i = 0; i < n; i++)
    printf("0x%x ", s[i]);
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<file_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
L34:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L2
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  test   %rax,%rax
  jle    L29
  addl   $0x1,-0x4(%rbp)
  jmp    L34
L2:
  nop
  jmp    L23
L29:
  nop
L23:
  nop
  leave
  ret
```

assistant: ```c
void file_write(int fd, char *data) {
  int i = 0;
  while (1) {
    if (!data[i])
      break;
    if (write(fd, &data[i], 1) <= 0) {
      break;
    }
    i++;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<file_write>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %rsi,%rbx
  cmpb   $0x0,(%rsi)
  je     L16
L3:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <write@plt>
  test   %rax,%rax
  jle    L16
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L3
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void file_write(int fd, char *data) {
  int i = 0;
  while (1) {
    if (!data[i])
      break;
    if (write(fd, &data[i], 1) <= 0) {
      break;
    }
    i++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<file_write>:
  endbr64
  cmpb   $0x0,(%rsi)
  je     L0
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  jmp    L21
  cs nopw 0x0(%rax,%rax,1)
L29:
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  je     L28
L21:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <write@plt>
  test   %rax,%rax
  jg     L29
L28:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  ret
```

assistant: ```c
void file_write(int fd, char *data) {
  int i = 0;
  while (1) {
    if (!data[i])
      break;
    if (write(fd, &data[i], 1) <= 0) {
      break;
    }
    i++;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<file_write>:
  endbr64
  cmpb   $0x0,(%rsi)
  je     L0
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  jmp    L35
  cs nopw 0x0(%rax,%rax,1)
L32:
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  je     L13
L35:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <write@plt>
  test   %rax,%rax
  jg     L32
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  ret
```

assistant: ```c
void file_write(int fd, char *data) {
  int i = 0;
  while (1) {
    if (!data[i])
      break;
    if (write(fd, &data[i], 1) <= 0) {
      break;
    }
    i++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x5c(%rbp)
  movl   $0x0,-0x58(%rbp)
  movl   $0x0,-0x5c(%rbp)
  jmp    L39
L61:
  movl   $0x0,-0x58(%rbp)
  jmp    L117
L41:
  lea    -0x50(%rbp),%rdx
  mov    -0x58(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x5c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rcx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x58(%rbp)
L117:
  cmpl   $0x3,-0x58(%rbp)
  jle    L41
  addl   $0x1,-0x5c(%rbp)
L39:
  cmpl   $0x3,-0x5c(%rbp)
  jle    L61
  movl   $0x1,-0x5c(%rbp)
  jmp    L97
L22:
  movl   $0x0,-0x58(%rbp)
  jmp    L77
L53:
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x5c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    -0x50(%rbp,%rax,4),%eax
  mov    %eax,-0x54(%rbp)
  mov    -0x5c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x58(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    -0x50(%rbp,%rax,4),%eax
  mov    -0x58(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x5c(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  mov    %eax,-0x50(%rbp,%rdx,4)
  mov    -0x5c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x58(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,%rdx
  mov    -0x54(%rbp),%eax
  mov    %eax,-0x50(%rbp,%rdx,4)
  addl   $0x1,-0x58(%rbp)
L77:
  mov    -0x58(%rbp),%eax
  cmp    -0x5c(%rbp),%eax
  jl     L53
  addl   $0x1,-0x5c(%rbp)
L97:
  cmpl   $0x3,-0x5c(%rbp)
  jle    L22
  movl   $0x0,-0x5c(%rbp)
  jmp    L64
L90:
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x58(%rbp)
  jmp    L49
L105:
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x5c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    -0x50(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x58(%rbp)
L49:
  cmpl   $0x3,-0x58(%rbp)
  jle    L105
  addl   $0x1,-0x5c(%rbp)
L64:
  cmpl   $0x3,-0x5c(%rbp)
  jle    L90
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L91
  call   <__stack_chk_fail@plt>
L91:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
data_type: string
name: disassemble_data
arguments: tag: D18
data_type: string

tool: D0: %d

tool: D18: %2d 

assistant: ```c
int main() {
  int a[4][4], i = 0, k = 0;
  for (i = 0; i < 4; i++) {
    for (k = 0; k < 4; k++) {
      scanf("%d", &a[i][k]);
    }
  }
  int h;
  for (i = 1; i < 4; i++) {
    for (k = 0; k < i; k++) {
      h = a[i][k];
      a[i][k] = a[k][i];
      a[k][i] = h;
    }
  }
  for (i = 0; i < 4; i++) {
    printf("\n");
    for (k = 0; k < 4; k++) {
      printf("%2d ", a[i][k]);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    0x10(%rsp),%rbp
  mov    %rbp,%r12
  mov    $0x0,%r15d
  lea    D1(%rip),%r14
L68:
  lea    -0x10(%r12),%rbx
L49:
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L49
  add    $0x4,%r15
  add    $0x10,%r12
  cmp    $0x10,%r15
  jne    L68
  lea    0x10(%rsp),%r10
  lea    0x4(%rsp),%r9
  lea    0x14(%r13),%rdi
  mov    $0x1,%r8d
  jmp    L26
L57:
  mov    (%rdx),%ecx
  mov    (%rax),%esi
  mov    %esi,(%rdx)
  mov    %ecx,(%rax)
  add    $0x4,%rdx
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L57
L40:
  add    $0x1,%r8d
  add    $0x10,%r10
  add    $0x4,%r9
  add    $0x14,%rdi
  cmp    $0x4,%r8d
  je     L91
L26:
  mov    %r9,%rax
  mov    %r10,%rdx
  test   %r8d,%r8d
  jg     L57
  jmp    L40
L91:
  add    $0x50,%r13
  lea    D10(%rip),%r12
L89:
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    -0x10(%rbp),%rbx
L96:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L96
  add    $0x10,%rbp
  cmp    %r13,%rbp
  jne    L89
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D1
datatype: string
name: retrieve_data
arguments: tag: D10
datatype: string

tool: D1: %d

tool: D10: %2d 

assistant: ```c
int main() {
  int a[4][4], i = 0, k = 0;
  for (i = 0; i < 4; i++) {
    for (k = 0; k < 4; k++) {
      scanf("%d", &a[i][k]);
    }
  }
  int h;
  for (i = 1; i < 4; i++) {
    for (k = 0; k < i; k++) {
      h = a[i][k];
      a[i][k] = a[k][i];
      a[k][i] = h;
    }
  }
  for (i = 0; i < 4; i++) {
    printf("\n");
    for (k = 0; k < 4; k++) {
      printf("%2d ", a[i][k]);
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  lea    0x20(%rsp),%r15
  mov    %rax,0x8(%rsp)
  mov    %r15,%rbp
  mov    %r15,%rbx
L27:
  lea    -0x10(%rbx),%r14
L42:
  mov    %r14,%rsi
  mov    %r12,%rdi
  add    $0x4,%r14
  call   <scanf@plt>
  cmp    %rbx,%r14
  jne    L42
  add    $0x4,%r13
  lea    0x10(%r14),%rbx
  cmp    $0x10,%r13
  jne    L27
  mov    0x8(%rsp),%rax
  mov    %r15,%r9
  lea    0x14(%rsp),%r8
  lea    0x14(%rax),%rdi
L17:
  mov    %r8,%rax
  mov    %r9,%rdx
L10:
  mov    (%rdx),%ecx
  mov    (%rax),%esi
  add    $0x10,%rax
  add    $0x4,%rdx
  mov    %esi,-0x4(%rdx)
  mov    %ecx,-0x10(%rax)
  cmp    %rdi,%rax
  jne    L10
  add    $0x4,%r8
  add    $0x10,%r9
  lea    0x14(%rax),%rdi
  cmp    %r15,%r8
  jne    L17
  mov    0x8(%rsp),%r14
  lea    D4(%rip),%r12
  add    $0x50,%r14
L47:
  mov    $0xa,%edi
  lea    -0x10(%rbp),%rbx
  call   <putchar@plt>
L50:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L50
  lea    0x10(%rbx),%rbp
  cmp    %r14,%rbp
  jne    L47
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D2
data_type: string
name: read_data
arguments: data_label: D4
data_type: string

tool: D2: %d

tool: D4: %2d 

assistant: ```c
int main() {
  int a[4][4], i = 0, k = 0;
  for (i = 0; i < 4; i++) {
    for (k = 0; k < 4; k++) {
      scanf("%d", &a[i][k]);
    }
  }
  int h;
  for (i = 1; i < 4; i++) {
    for (k = 0; k < i; k++) {
      h = a[i][k];
      a[i][k] = a[k][i];
      a[k][i] = h;
    }
  }
  for (i = 0; i < 4; i++) {
    printf("\n");
    for (k = 0; k < 4; k++) {
      printf("%2d ", a[i][k]);
    }
  }
}
```