ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<write_memory>:
  endbr64
  mov    %rsi,%rax
  cmp    $0x2,%ecx
  je     L4
  cmp    $0x4,%ecx
  je     L47
  cmp    $0x1,%ecx
  je     L7
L82:
  ret
L47:
  lea    -0x4(%rdx),%ecx
  lea    0x4(%rsi,%rcx,1),%rcx
  test   %edx,%edx
  je     L12
L58:
  mov    %r8,(%rax)
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L58
  jmp    L82
L12:
  ret
L4:
  lea    -0x2(%rdx),%ecx
  lea    0x2(%rsi,%rcx,1),%rcx
  test   %edx,%edx
  je     L43
L94:
  mov    %r8w,(%rax)
  add    $0x2,%rax
  cmp    %rax,%rcx
  jne    L94
  ret
L43:
  ret
L7:
  test   %edx,%edx
  je     L82
  mov    %edx,%edx
  add    %rsi,%rdx
L37:
  mov    %r8b,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L37
  ret
```

assistant: ```c
void write_memory(unsigned long phys_addr, void *addr, int len, int iosize, unsigned long value) {
  switch (iosize) {
  case 1:
    while (len) {
      *(unsigned char *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  case 2:
    while (len) {
      *(unsigned short *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  case 4:
    while (len) {
      *(unsigned long *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<write_memory>:
  endbr64
  mov    %rsi,%rdi
  cmp    $0x2,%ecx
  je     L41
  cmp    $0x4,%ecx
  je     L75
  cmp    $0x1,%ecx
  je     L31
L6:
  ret
  nopw   0x0(%rax,%rax,1)
L31:
  test   %edx,%edx
  je     L6
  sub    $0x1,%edx
  movzbl %r8b,%esi
  add    $0x1,%rdx
  jmp    1050 <memset@plt>
  nopl   0x0(%rax)
L75:
  lea    -0x4(%rdx),%eax
  lea    0x4(%rsi,%rax,1),%rax
  test   %edx,%edx
  je     L0
  nopl   0x0(%rax)
L79:
  mov    %r8,(%rdi)
  add    $0x4,%rdi
  cmp    %rdi,%rax
  jne    L79
  ret
  nopl   (%rax)
L41:
  lea    -0x2(%rdx),%eax
  lea    0x2(%rsi,%rax,1),%rax
  test   %edx,%edx
  je     L22
  nopl   0x0(%rax)
L30:
  mov    %r8w,(%rdi)
  add    $0x2,%rdi
  cmp    %rax,%rdi
  jne    L30
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  ret
  nopl   0x0(%rax)
L22:
  ret
```

assistant: ```c
void write_memory(unsigned long phys_addr, void *addr, int len, int iosize, unsigned long value) {
  switch (iosize) {
  case 1:
    while (len) {
      *(unsigned char *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  case 2:
    while (len) {
      *(unsigned short *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  case 4:
    while (len) {
      *(unsigned long *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<write_memory>:
  endbr64
  mov    %rsi,%rdi
  cmp    $0x2,%ecx
  je     L73
  cmp    $0x4,%ecx
  je     L67
  cmp    $0x1,%ecx
  je     L72
L123:
  ret
  nopw   0x0(%rax,%rax,1)
L72:
  test   %edx,%edx
  je     L123
  sub    $0x1,%edx
  movzbl %r8b,%esi
  add    $0x1,%rdx
  jmp    1050 <memset@plt>
  nopl   0x0(%rax)
L67:
  lea    -0x4(%rdx),%eax
  lea    0x4(%rsi,%rax,1),%rax
  test   %edx,%edx
  je     L143
  nopl   0x0(%rax,%rax,1)
L44:
  mov    %r8,(%rdi)
  add    $0x4,%rdi
  cmp    %rax,%rdi
  jne    L44
  ret
  nopl   (%rax)
L73:
  test   %edx,%edx
  je     L12
  lea    -0x2(%rdx),%eax
  mov    %r8d,%esi
  mov    %eax,%r9d
  shr    %r9d
  lea    0x1(%r9),%r10d
  cmp    $0xc,%eax
  jbe    L138
  mov    %r10d,%ecx
  movd   %r8d,%xmm0
  mov    %rdi,%rax
  shr    $0x3,%ecx
  punpcklwd %xmm0,%xmm0
  sub    $0x1,%ecx
  pshufd $0x0,%xmm0,%xmm0
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax)
L118:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L118
  mov    %r10d,%eax
  and    $0xfffffff8,%eax
  mov    %eax,%r11d
  mov    %eax,%ecx
  neg    %r11d
  lea    (%rdi,%rcx,2),%rcx
  lea    (%rdx,%r11,2),%edx
  cmp    %eax,%r10d
  je     L74
L81:
  sub    %eax,%r9d
  lea    0x1(%r9),%r10d
  cmp    $0x2,%r9d
  jbe    L19
  movd   %esi,%xmm1
  pshuflw $0x0,%xmm1,%xmm0
  movq   %xmm0,(%rdi,%rax,2)
  mov    %r10d,%eax
  and    $0xfffffffc,%eax
  mov    %eax,%esi
  lea    (%rcx,%rsi,2),%rcx
  mov    %eax,%esi
  neg    %esi
  lea    (%rdx,%rsi,2),%edx
  cmp    %eax,%r10d
  je     L123
L19:
  mov    %r8w,(%rcx)
  cmp    $0x2,%edx
  je     L123
  mov    %r8w,0x2(%rcx)
  cmp    $0x4,%edx
  je     L123
  mov    %r8w,0x4(%rcx)
  ret
  nopw   0x0(%rax,%rax,1)
L74:
  ret
  nopl   0x0(%rax)
L12:
  ret
  nopl   0x0(%rax)
L143:
  ret
L138:
  mov    %rdi,%rcx
  xor    %eax,%eax
  jmp    L81
```

assistant: ```c
void write_memory(unsigned long phys_addr, void *addr, int len, int iosize, unsigned long value) {
  switch (iosize) {
  case 1:
    while (len) {
      *(unsigned char *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  case 2:
    while (len) {
      *(unsigned short *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  case 4:
    while (len) {
      *(unsigned long *)addr = value;
      len -= iosize;
      addr += iosize;
    }
    break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int f() {
  int foo = 0;
  return foo;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<f>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int f() {
  int foo = 0;
  return foo;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<f>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int f() {
  int foo = 0;
  return foo;
}
```
user: Translate the below assembly code into C programming language:

```asm
<f>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int f() {
  int foo = 0;
  return foo;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<tapdisk_logfile_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    -0x20(%rbp),%rax
  mov    $0x80,%edx
  cmp    %rdx,%rax
  cmova  %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%ecx
  mov    -0x30(%rbp),%rdi
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  sub    $0x8,%rsp
  push   -0x38(%rbp)
  mov    %ecx,%r9d
  mov    %rdi,%r8
  mov    %rdx,%rcx
  lea    D0(%rip),%rdx
  mov    %rax,%rdi
  call   <snprintf@plt>
  add    $0x10,%rsp
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D0
type_guessed: string

tool: D0: %s/%s.%d%s

assistant: ```c
int tapdisk_logfile_name(char *path, size_t size, const char *dir, const char *ident, const char *suffix) {
  const size_t max = (((size) < (128UL)) ? (size) : (128UL));
  return snprintf(path, max, "%s/%s.%d%s", dir, ident, getpid(), suffix);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<tapdisk_logfile_name>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    %rcx,%r13
  mov    %r8,%r14
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%r9d
  mov    $0x80,%edx
  cmp    %rdx,%rbx
  mov    %rdx,%rsi
  cmovbe %rbx,%rsi
  sub    $0x8,%rsp
  push   %r14
  mov    %r13,%r8
  mov    %r12,%rcx
  lea    D0(%rip),%rdx
  mov    %rbp,%rdi
  call   <snprintf@plt>
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

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

tool: D0: %s/%s.%d%s

assistant: ```c
int tapdisk_logfile_name(char *path, size_t size, const char *dir, const char *ident, const char *suffix) {
  const size_t max = (((size) < (128UL)) ? (size) : (128UL));
  return snprintf(path, max, "%s/%s.%d%s", dir, ident, getpid(), suffix);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tapdisk_logfile_name>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %r8,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <getpid@plt>
  mov    $0x80,%edx
  mov    %r12,%rcx
  mov    %r13,%r8
  cmp    %rdx,%rbx
  mov    %rdx,%rsi
  mov    %rbp,%rdi
  mov    %eax,%r9d
  cmovbe %rbx,%rsi
  sub    $0x8,%rsp
  lea    D1(%rip),%rdx
  push   %r14
  call   <snprintf@plt>
  pop    %rdx
  pop    %rcx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: tag: D1
data_type: string

tool: D1: %s/%s.%d%s

assistant: ```c
int tapdisk_logfile_name(char *path, size_t size, const char *dir, const char *ident, const char *suffix) {
  const size_t max = (((size) < (128UL)) ? (size) : (128UL));
  return snprintf(path, max, "%s/%s.%d%s", dir, ident, getpid(), suffix);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<tapdisk_logfile_name>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %r8,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <getpid@plt>
  mov    $0x80,%edx
  mov    %r12,%rcx
  mov    %r13,%r8
  cmp    %rdx,%rbx
  mov    %rdx,%rsi
  mov    %rbp,%rdi
  mov    %eax,%r9d
  cmovbe %rbx,%rsi
  sub    $0x8,%rsp
  lea    D6(%rip),%rdx
  push   %r14
  call   <snprintf@plt>
  pop    %rdx
  pop    %rcx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

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

tool: D6: %s/%s.%d%s

assistant: ```c
int tapdisk_logfile_name(char *path, size_t size, const char *dir, const char *ident, const char *suffix) {
  const size_t max = (((size) < (128UL)) ? (size) : (128UL));
  return snprintf(path, max, "%s/%s.%d%s", dir, ident, getpid(), suffix);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vec2_mul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  movss  %xmm0,-0xc(%rbp)
  mov    %rsi,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  -0xc(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mulss  -0xc(%rbp),%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vec2_mul(const float a[2], float k, float out[2]) {
  out[0] = a[0] * k;
  out[1] = a[1] * k;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<vec2_mul>:
  endbr64
  movaps %xmm0,%xmm1
  mulss  (%rdi),%xmm1
  movss  %xmm1,(%rsi)
  mulss  0x4(%rdi),%xmm0
  movss  %xmm0,0x4(%rsi)
  ret
```

assistant: ```c
void vec2_mul(const float a[2], float k, float out[2]) {
  out[0] = a[0] * k;
  out[1] = a[1] * k;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vec2_mul>:
  endbr64
  movss  (%rdi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rsi)
  mulss  0x4(%rdi),%xmm0
  movss  %xmm0,0x4(%rsi)
  ret
```

assistant: ```c
void vec2_mul(const float a[2], float k, float out[2]) {
  out[0] = a[0] * k;
  out[1] = a[1] * k;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<vec2_mul>:
  endbr64
  movss  (%rdi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rsi)
  mulss  0x4(%rdi),%xmm0
  movss  %xmm0,0x4(%rsi)
  ret
```

assistant: ```c
void vec2_mul(const float a[2], float k, float out[2]) {
  out[0] = a[0] * k;
  out[1] = a[1] * k;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<optk_get_basename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L32
  mov    $0x0,%eax
  jmp    L8
L32:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L46:
  mov    -0x10(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L26
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L46
L26:
  nop
  cmpq   $0x0,-0x20(%rbp)
  je     L36
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
L36:
  mov    -0x10(%rbp),%rax
L8:
  leave
  ret
```

assistant: ```c
char *optk_get_basename(const char *path, size_t *basename_len) {
  const char *basename, *ptr;
  if (path == NULL)
    return NULL;
  basename = ptr = path;
  for (;;) {
    ptr = strpbrk(basename, "/");
    if (ptr == NULL)
      break;
    basename = ptr + 1;
  }
  if (basename_len != NULL)
    *basename_len = strlen(basename);
  return (char *)basename;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<optk_get_basename>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  test   %rdi,%rdi
  jne    L27
  jmp    L24
L7:
  lea    0x1(%rax),%rbx
L27:
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L7
  test   %rbp,%rbp
  je     L24
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rax,0x0(%rbp)
L24:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *optk_get_basename(const char *path, size_t *basename_len) {
  const char *basename, *ptr;
  if (path == NULL)
    return NULL;
  basename = ptr = path;
  for (;;) {
    ptr = strpbrk(basename, "/");
    if (ptr == NULL)
      break;
    basename = ptr + 1;
  }
  if (basename_len != NULL)
    *basename_len = strlen(basename);
  return (char *)basename;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<optk_get_basename>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L27
  mov    %rsi,%rbx
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L14:
  lea    0x1(%rax),%r12
L39:
  mov    $0x2f,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L14
  test   %rbx,%rbx
  je     L27
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rax,(%rbx)
L27:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *optk_get_basename(const char *path, size_t *basename_len) {
  const char *basename, *ptr;
  if (path == NULL)
    return NULL;
  basename = ptr = path;
  for (;;) {
    ptr = strpbrk(basename, "/");
    if (ptr == NULL)
      break;
    basename = ptr + 1;
  }
  if (basename_len != NULL)
    *basename_len = strlen(basename);
  return (char *)basename;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<optk_get_basename>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L2
  mov    %rsi,%rbx
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L8:
  lea    0x1(%rax),%r12
L10:
  mov    $0x2f,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L8
  test   %rbx,%rbx
  je     L2
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rax,(%rbx)
L2:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *optk_get_basename(const char *path, size_t *basename_len) {
  const char *basename, *ptr;
  if (path == NULL)
    return NULL;
  basename = ptr = path;
  for (;;) {
    ptr = strpbrk(basename, "/");
    if (ptr == NULL)
      break;
    basename = ptr + 1;
  }
  if (basename_len != NULL)
    *basename_len = strlen(basename);
  return (char *)basename;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x84(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L37
L2:
  movl   $0x0,-0x7c(%rbp)
  lea    -0x70(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0xd,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  lea    -0x80(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x78(%rbp)
  jmp    L32
L10:
  mov    $0x0,%eax
  call   <getchar@plt>
  movsbl %al,%eax
  sub    $0x61,%eax
  mov    %eax,-0x74(%rbp)
  mov    -0x74(%rbp),%eax
  cltq
  mov    -0x70(%rbp,%rax,4),%eax
  sub    -0x78(%rbp),%eax
  cmp    %eax,-0x7c(%rbp)
  jge    L13
  mov    -0x74(%rbp),%eax
  cltq
  mov    -0x70(%rbp,%rax,4),%eax
  sub    -0x78(%rbp),%eax
  mov    %eax,-0x7c(%rbp)
L13:
  mov    -0x80(%rbp),%edx
  mov    -0x78(%rbp),%eax
  add    %eax,%edx
  mov    -0x74(%rbp),%eax
  cltq
  mov    %edx,-0x70(%rbp,%rax,4)
  addl   $0x1,-0x78(%rbp)
L32:
  mov    -0x80(%rbp),%eax
  cmp    %eax,-0x78(%rbp)
  jl     L10
  mov    -0x7c(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L37:
  mov    -0x84(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x84(%rbp)
  test   %eax,%eax
  jne    L2
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D8
type_guessed: string
name: read_data
arguments: label: D6
type_guessed: string
name: read_data
arguments: label: D1
type_guessed: string

tool: D8: %d

tool: D6: %d 

tool: D1: %d


assistant: ```c
int main() {
  int t, n;
  scanf("%d", &t);
  while (t--) {
    int k = 0, hope[26] = {0};
    scanf("%d ", &n);
    for (int i = 0; i < n; ++i) {
      int c = getchar() - 'a';
      if (hope[c] - i > k)
        k = hope[c] - i;
      hope[c] = n + i;
    }
    printf("%d\n", k);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  je     L25
  mov    $0x0,%r15d
  lea    0xc(%rsp),%r14
  lea    D20(%rip),%r13
  lea    D4(%rip),%r12
L40:
  lea    0x10(%rsp),%rdi
  mov    $0xd,%ecx
  mov    %r15,%rax
  rep stos %rax,%es:(%rdi)
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L56
  mov    $0x0,%ebx
  mov    $0x0,%ebp
L17:
  mov    $0x0,%eax
  call   <getchar@plt>
  movsbl %al,%eax
  sub    $0x61,%eax
  cltq
  mov    0x10(%rsp,%rax,4),%edx
  sub    %ebx,%edx
  cmp    %edx,%ebp
  cmovl  %edx,%ebp
  mov    0xc(%rsp),%edx
  lea    (%rdx,%rbx,1),%ecx
  mov    %ecx,0x10(%rsp,%rax,4)
  add    $0x1,%ebx
  cmp    %ebx,%edx
  jg     L17
L51:
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  jne    L40
L25:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  mov    $0x0,%eax
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L56:
  mov    $0x0,%ebp
  jmp    L51
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D24
type: string
name: disassemble_data
arguments: tag: D20
type: string
name: disassemble_data
arguments: tag: D4
type: string

tool: D24: %d

tool: D20: %d 

tool: D4: %d


assistant: ```c
int main() {
  int t, n;
  scanf("%d", &t);
  while (t--) {
    int k = 0, hope[26] = {0};
    scanf("%d ", &n);
    for (int i = 0; i < n; ++i) {
      int c = getchar() - 'a';
      if (hope[c] - i > k)
        k = hope[c] - i;
      hope[c] = n + i;
    }
    printf("%d\n", k);
  }
  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
<main>:
  endbr64
  push   %r15
  lea    D18(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  je     L15
  lea    0x10(%rsp),%rbx
  lea    0xc(%rsp),%r14
  lea    D23(%rip),%r13
  lea    D5(%rip),%r12
  nopl   0x0(%rax)
L28:
  xor    %eax,%eax
  mov    $0xd,%ecx
  mov    %rbx,%rdi
  mov    %r14,%rsi
  rep stos %rax,%es:(%rdi)
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L37
  xor    %r15d,%r15d
  xor    %ebp,%ebp
  nopl   (%rax)
L34:
  xor    %eax,%eax
  call   <getchar@plt>
  movsbl %al,%eax
  sub    $0x61,%eax
  cltq
  mov    0x10(%rsp,%rax,4),%edx
  sub    %r15d,%edx
  cmp    %edx,%ebp
  cmovl  %edx,%ebp
  mov    0xc(%rsp),%edx
  lea    (%rdx,%r15,1),%ecx
  add    $0x1,%r15d
  mov    %ecx,0x10(%rsp,%rax,4)
  cmp    %r15d,%edx
  jg     L34
L6:
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  jne    L28
L15:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L37:
  xor    %ebp,%ebp
  jmp    L6
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D18
data_type: string
name: parse_data
arguments: tag: D23
data_type: string
name: parse_data
arguments: tag: D5
data_type: string

tool: D18: %d

tool: D23: %d 

tool: D5: %d


assistant: ```c
int main() {
  int t, n;
  scanf("%d", &t);
  while (t--) {
    int k = 0, hope[26] = {0};
    scanf("%d ", &n);
    for (int i = 0; i < n; ++i) {
      int c = getchar() - 'a';
      if (hope[c] - i > k)
        k = hope[c] - i;
      hope[c] = n + i;
    }
    printf("%d\n", k);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  lea    D15(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  je     L29
  lea    0x10(%rsp),%rbx
  lea    0xc(%rsp),%r14
  lea    D22(%rip),%r13
  lea    D12(%rip),%r12
  nopl   0x0(%rax)
L31:
  xor    %eax,%eax
  mov    $0xd,%ecx
  mov    %rbx,%rdi
  mov    %r14,%rsi
  rep stos %rax,%es:(%rdi)
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L45
  xor    %r15d,%r15d
  xor    %ebp,%ebp
  nopl   (%rax)
L56:
  xor    %eax,%eax
  call   <getchar@plt>
  movsbl %al,%eax
  sub    $0x61,%eax
  cltq
  mov    0x10(%rsp,%rax,4),%edx
  sub    %r15d,%edx
  cmp    %edx,%ebp
  cmovl  %edx,%ebp
  mov    0xc(%rsp),%edx
  lea    (%rdx,%r15,1),%ecx
  add    $0x1,%r15d
  mov    %ecx,0x10(%rsp,%rax,4)
  cmp    %r15d,%edx
  jg     L56
L6:
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  jne    L31
L29:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L45:
  xor    %ebp,%ebp
  jmp    L6
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D15
data_type: string
name: disassemble_data
arguments: data_tag: D22
data_type: string
name: disassemble_data
arguments: data_tag: D12
data_type: string

tool: D15: %d

tool: D22: %d 

tool: D12: %d


assistant: ```c
int main() {
  int t, n;
  scanf("%d", &t);
  while (t--) {
    int k = 0, hope[26] = {0};
    scanf("%d ", &n);
    for (int i = 0; i < n; ++i) {
      int c = getchar() - 'a';
      if (hope[c] - i > k)
        k = hope[c] - i;
      hope[c] = n + i;
    }
    printf("%d\n", k);
  }
  return 0;
}
```