ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<_write_string_len>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <_write_int@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void _write_string_len(char **pptr, const char *string, int len) {
  _write_int(pptr, len);
  memcpy(*pptr, string, len);
  *pptr += len;
}
```
user: What is the c source code of the assembly code below?

```asm
<_write_string_len>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %edx,%ebx
  mov    %edx,%esi
  call   <_write_int@plt>
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    0x0(%rbp),%rdi
  call   <memcpy@plt>
  movslq %ebx,%rbx
  add    %rbx,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void _write_string_len(char **pptr, const char *string, int len) {
  _write_int(pptr, len);
  memcpy(*pptr, string, len);
  *pptr += len;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_write_string_len>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %edx,%rbx
  mov    %ebx,%esi
  call   <_write_int@plt>
  mov    0x0(%rbp),%rdi
  mov    %ebx,%edx
  mov    %r12,%rsi
  call   <memcpy@plt>
  add    %rbx,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void _write_string_len(char **pptr, const char *string, int len) {
  _write_int(pptr, len);
  memcpy(*pptr, string, len);
  *pptr += len;
}
```
user: What is the c source code of the assembly code below:

```asm
<_write_string_len>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %edx,%rbx
  mov    %ebx,%esi
  call   <_write_int@plt>
  mov    0x0(%rbp),%rdi
  mov    %ebx,%edx
  mov    %r12,%rsi
  call   <memcpy@plt>
  add    %rbx,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void _write_string_len(char **pptr, const char *string, int len) {
  _write_int(pptr, len);
  memcpy(*pptr, string, len);
  *pptr += len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<clear_dmatrix3d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    %r9d,-0x2c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L59
L23:
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L10
L49:
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L21
L55:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L21:
  mov    -0x4(%rbp),%eax
  cmp    0x10(%rbp),%eax
  jle    L55
  addl   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jle    L49
  addl   $0x1,-0xc(%rbp)
L59:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L23
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void clear_dmatrix3d(double ***m, int nsl, int nsh, int nrl, int nrh, int ncl, int nch) {
  int i, j, k;
  for (i = nsl; i <= nsh; i++) {
    for (j = nrl; j <= nrh; j++)
      for (k = ncl; k <= nch; k++)
        m[i][j][k] = 0.0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<clear_dmatrix3d>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x28(%rsp),%r10d
  cmp    %edx,%esi
  jg     L51
  mov    %esi,%r11d
  mov    %edx,%eax
  mov    %ecx,%ebx
  mov    %r8d,%ebp
  mov    %r9d,%r8d
  movslq %esi,%rdx
  lea    (%rdi,%rdx,8),%rsi
  sub    %r11d,%eax
  add    %rdx,%rax
  lea    0x8(%rdi,%rax,8),%r13
  movslq %ecx,%rax
  lea    0x0(,%rax,8),%r12
  mov    %ebp,%edx
  sub    %ecx,%edx
  lea    0x1(%rax,%rdx,1),%r9
  shl    $0x3,%r9
  movslq %r8d,%rax
  lea    0x0(,%rax,8),%r11
  mov    %r10d,%edx
  sub    %r8d,%edx
  lea    0x1(%rax,%rdx,1),%rdi
  shl    $0x3,%rdi
  jmp    L70
L19:
  mov    (%rsi),%rdx
  mov    (%rdx,%rcx,1),%rdx
  movq   $0x0,(%rdx,%rax,1)
  add    $0x8,%rax
  cmp    %rax,%rdi
  jne    L19
L0:
  add    $0x8,%rcx
  cmp    %r9,%rcx
  je     L80
L2:
  mov    %r11,%rax
  cmp    %r10d,%r8d
  jle    L19
  jmp    L0
L80:
  add    $0x8,%rsi
  cmp    %r13,%rsi
  je     L51
L70:
  mov    %r12,%rcx
  cmp    %ebp,%ebx
  jle    L2
  jmp    L80
L51:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void clear_dmatrix3d(double ***m, int nsl, int nsh, int nrl, int nrh, int ncl, int nch) {
  int i, j, k;
  for (i = nsl; i <= nsh; i++) {
    for (j = nrl; j <= nrh; j++)
      for (k = ncl; k <= nch; k++)
        m[i][j][k] = 0.0;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<clear_dmatrix3d>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %ecx,0x18(%rsp)
  mov    %r8d,0x1c(%rsp)
  cmp    %edx,%esi
  jg     L46
  mov    %edx,%eax
  movslq %esi,%rdx
  movslq %r9d,%r13
  mov    %r9d,%ebx
  sub    %esi,%eax
  lea    (%rdi,%rdx,8),%r15
  shl    $0x3,%r13
  add    %rdx,%rax
  mov    %r8d,%edx
  lea    0x8(%rdi,%rax,8),%rax
  sub    %ecx,%edx
  mov    %rax,0x10(%rsp)
  movslq %ecx,%rax
  lea    0x0(,%rax,8),%rsi
  lea    0x1(%rax,%rdx,1),%rbp
  mov    0x60(%rsp),%eax
  mov    %rsi,0x8(%rsp)
  shl    $0x3,%rbp
  sub    %r9d,%eax
  lea    0x8(,%rax,8),%r14
  nopw   0x0(%rax,%rax,1)
L3:
  mov    0x8(%rsp),%r12
  mov    0x1c(%rsp),%ecx
  cmp    %ecx,0x18(%rsp)
  jg     L30
  nopw   0x0(%rax,%rax,1)
L29:
  cmp    0x60(%rsp),%ebx
  jg     L31
  mov    (%r15),%rax
  mov    %r14,%rdx
  xor    %esi,%esi
  mov    (%rax,%r12,1),%rdi
  add    %r13,%rdi
  call   <memset@plt>
L31:
  add    $0x8,%r12
  cmp    %rbp,%r12
  jne    L29
L30:
  add    $0x8,%r15
  cmp    0x10(%rsp),%r15
  jne    L3
L46:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void clear_dmatrix3d(double ***m, int nsl, int nsh, int nrl, int nrh, int ncl, int nch) {
  int i, j, k;
  for (i = nsl; i <= nsh; i++) {
    for (j = nrl; j <= nrh; j++)
      for (k = ncl; k <= nch; k++)
        m[i][j][k] = 0.0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<clear_dmatrix3d>:
  endbr64
  push   %r15
  mov    %esi,%r10d
  mov    %edx,%eax
  mov    %r8d,%esi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%r8d
  cmp    %edx,%r10d
  jg     L28
  mov    %rdi,%r11
  mov    %ecx,%edi
  cmp    %esi,%ecx
  jg     L28
  movslq %r9d,%rdx
  cmp    %r8d,%edx
  jg     L28
  movslq %r10d,%r9
  sub    %r10d,%eax
  lea    0x0(,%rdx,8),%rbp
  sub    %edi,%esi
  add    %r9,%rax
  lea    (%r11,%r9,8),%r14
  lea    0x8(%r11,%rax,8),%r15
  movslq %ecx,%rax
  lea    0x0(,%rax,8),%rcx
  lea    0x1(%rax,%rsi,1),%r12
  mov    %r8d,%eax
  mov    %rcx,0x8(%rsp)
  sub    %edx,%eax
  shl    $0x3,%r12
  lea    0x8(,%rax,8),%r13
  nopl   0x0(%rax,%rax,1)
L8:
  mov    0x8(%rsp),%rbx
  nopl   (%rax)
L15:
  mov    (%r14),%rax
  mov    %r13,%rdx
  xor    %esi,%esi
  mov    (%rax,%rbx,1),%rdi
  add    $0x8,%rbx
  add    %rbp,%rdi
  call   <memset@plt>
  cmp    %rbx,%r12
  jne    L15
  add    $0x8,%r14
  cmp    %r15,%r14
  jne    L8
L28:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void clear_dmatrix3d(double ***m, int nsl, int nsh, int nrl, int nrh, int ncl, int nch) {
  int i, j, k;
  for (i = nsl; i <= nsh; i++) {
    for (j = nrl; j <= nrh; j++)
      for (k = ncl; k <= nch; k++)
        m[i][j][k] = 0.0;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<readfile>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L4
  mov    $0x3e8,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
L16:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L29
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  add    %eax,-0x14(%rbp)
  jmp    L16
L29:
  nop
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
L4:
  mov    -0x14(%rbp),%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
type_infered: string

tool: D0: r

assistant: ```c
int readfile(char *filename) {
  int sum;
  char *line;
  FILE *file = NULL;
  sum = 0;
  file = fopen(filename, "r");
  if (file != NULL) {
    line = (char *)malloc(1000);
    while (1) {
      if (fgets(line, 1000, file))
        sum = sum + atoi(line);
      else
        break;
    }
    fclose(file);
  }
  return (sum);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<readfile>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D0(%rip),%rsi
  call   <fopen@plt>
  mov    $0x0,%ebx
  test   %rax,%rax
  je     L10
  mov    %rax,%r12
  mov    $0x3e8,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  jmp    L4
L15:
  mov    %rbp,%rdi
  call   <atoi@plt>
  add    %eax,%ebx
L4:
  mov    %r12,%rdx
  mov    $0x3e8,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L15
  mov    %r12,%rdi
  call   <fclose@plt>
L10:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
type_infered: string

tool: D0: r

assistant: ```c
int readfile(char *filename) {
  int sum;
  char *line;
  FILE *file = NULL;
  sum = 0;
  file = fopen(filename, "r");
  if (file != NULL) {
    line = (char *)malloc(1000);
    while (1) {
      if (fgets(line, 1000, file))
        sum = sum + atoi(line);
      else
        break;
    }
    fclose(file);
  }
  return (sum);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<readfile>:
  endbr64
  push   %r12
  lea    D1(%rip),%rsi
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  call   <fopen@plt>
  test   %rax,%rax
  je     L26
  mov    $0x3e8,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rax,%rbx
  jmp    L1
  xchg   %ax,%ax
L28:
  mov    %rbx,%rdi
  call   <atoi@plt>
  add    %eax,%r12d
L1:
  mov    %rbp,%rdx
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L28
  mov    %rbp,%rdi
  call   <fclose@plt>
L26:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: string

tool: D1: r

assistant: ```c
int readfile(char *filename) {
  int sum;
  char *line;
  FILE *file = NULL;
  sum = 0;
  file = fopen(filename, "r");
  if (file != NULL) {
    line = (char *)malloc(1000);
    while (1) {
      if (fgets(line, 1000, file))
        sum = sum + atoi(line);
      else
        break;
    }
    fclose(file);
  }
  return (sum);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<readfile>:
  endbr64
  push   %r12
  lea    D8(%rip),%rsi
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  call   <fopen@plt>
  test   %rax,%rax
  je     L23
  mov    $0x3e8,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rax,%rbx
  jmp    L2
  xchg   %ax,%ax
L14:
  mov    %rbx,%rdi
  call   <atoi@plt>
  add    %eax,%r12d
L2:
  mov    %rbp,%rdx
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L14
  mov    %rbp,%rdi
  call   <fclose@plt>
L23:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
data_type: string

tool: D8: r

assistant: ```c
int readfile(char *filename) {
  int sum;
  char *line;
  FILE *file = NULL;
  sum = 0;
  file = fopen(filename, "r");
  if (file != NULL) {
    line = (char *)malloc(1000);
    while (1) {
      if (fgets(line, 1000, file))
        sum = sum + atoi(line);
      else
        break;
    }
    fclose(file);
  }
  return (sum);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<base64_encode>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%rbx
  mov    %rcx,%r12
  test   %rsi,%rsi
  je     L28
  mov    %rdi,%r13
  mov    %rsi,%rbp
  test   %rcx,%rcx
  je     L28
  lea    D6(%rip),%r14
  jmp    L21
L125:
  movsbl 0x1(%r13),%edi
  call   <to_uchar@plt>
  sar    $0x4,%eax
  add    %r15d,%eax
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x1(%rbx)
  cmp    $0x2,%r12
  je     L150
  movsbl 0x1(%r13),%edi
  call   <to_uchar@plt>
  lea    0x0(,%rax,4),%r15d
  mov    $0x0,%eax
  sub    $0x2,%rbp
  jne    L146
L24:
  add    %eax,%r15d
  and    $0x3f,%r15d
  movzbl (%r14,%r15,1),%eax
  jmp    L54
L146:
  movsbl 0x2(%r13),%edi
  call   <to_uchar@plt>
  sar    $0x6,%eax
  jmp    L24
L1:
  movsbl 0x2(%r13),%edi
  call   <to_uchar@plt>
  add    $0x4,%rbx
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,-0x1(%rbx)
  sub    $0x4,%r12
  je     L150
  sub    $0x1,%rbp
  je     L126
  add    $0x3,%r13
L21:
  movsbl 0x0(%r13),%edi
  call   <to_uchar@plt>
  sar    $0x2,%eax
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,(%rbx)
  cmp    $0x1,%r12
  je     L150
  movsbl 0x0(%r13),%edi
  call   <to_uchar@plt>
  shl    $0x4,%eax
  mov    %eax,%r15d
  cmp    $0x1,%rbp
  jne    L125
  and    $0x30,%r15d
  movzbl (%r14,%r15,1),%eax
  mov    %al,0x1(%rbx)
  cmp    $0x2,%r12
  je     L150
  mov    $0x3d,%eax
  mov    $0x0,%ebp
L54:
  mov    %al,0x2(%rbx)
  cmp    $0x3,%r12
  je     L150
  test   %rbp,%rbp
  jne    L1
  add    $0x4,%rbx
  movb   $0x3d,-0x1(%rbx)
  cmp    $0x4,%r12
  je     L150
L126:
  movb   $0x0,(%rbx)
L150:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L28:
  test   %r12,%r12
  jne    L126
  jmp    L150
```

assistant: ```c
void base64_encode(const char *in, size_t inlen, char *out, size_t outlen) {
  static const char b64str[64] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
  while (inlen && outlen) {
    *out++ = b64str[(to_uchar(in[0]) >> 2) & 0x3f];
    if (!--outlen)
      break;
    *out++ = b64str[((to_uchar(in[0]) << 4) + (--inlen ? to_uchar(in[1]) >> 4 : 0)) & 0x3f];
    if (!--outlen)
      break;
    *out++ = (inlen ? b64str[((to_uchar(in[1]) << 2) + (--inlen ? to_uchar(in[2]) >> 6 : 0)) & 0x3f] : '=');
    if (!--outlen)
      break;
    *out++ = inlen ? b64str[to_uchar(in[2]) & 0x3f] : '=';
    if (!--outlen)
      break;
    if (inlen)
      inlen--;
    if (inlen)
      in += 3;
  }
  if (outlen)
    *out = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<base64_encode>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L137
  test   %rcx,%rcx
  je     L35
  mov    %rdi,%r12
  mov    %rsi,%r13
  lea    D3(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L106:
  movsbl (%r12),%edi
  call   <to_uchar@plt>
  sar    $0x2,%eax
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,(%rbx)
  cmp    $0x1,%rbp
  je     L35
  movsbl (%r12),%edi
  call   <to_uchar@plt>
  shl    $0x4,%eax
  mov    %eax,%r15d
  cmp    $0x1,%r13
  jne    L104
  and    $0x30,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x1(%rbx)
  cmp    $0x2,%rbp
  je     L35
  movb   $0x3d,0x2(%rbx)
  cmp    $0x3,%rbp
  je     L35
L118:
  movb   $0x3d,0x3(%rbx)
  add    $0x4,%rbx
  cmp    $0x4,%rbp
  je     L35
L92:
  movb   $0x0,(%rbx)
L35:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L104:
  movsbl 0x1(%r12),%edi
  call   <to_uchar@plt>
  sar    $0x4,%eax
  add    %r15d,%eax
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x1(%rbx)
  cmp    $0x2,%rbp
  je     L35
  movsbl 0x1(%r12),%edi
  call   <to_uchar@plt>
  lea    0x0(,%rax,4),%r15d
  cmp    $0x2,%r13
  jne    L135
  mov    %r15d,%eax
  and    $0x3c,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x2(%rbx)
  cmp    $0x3,%rbp
  jne    L118
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L135:
  movsbl 0x2(%r12),%edi
  call   <to_uchar@plt>
  sar    $0x6,%eax
  add    %r15d,%eax
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x2(%rbx)
  cmp    $0x3,%rbp
  je     L35
  movsbl 0x2(%r12),%edi
  add    $0x4,%rbx
  call   <to_uchar@plt>
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,-0x1(%rbx)
  sub    $0x4,%rbp
  je     L35
  sub    $0x3,%r13
  je     L92
  add    $0x3,%r12
  jmp    L106
L137:
  test   %rcx,%rcx
  jne    L92
  jmp    L35
```

assistant: ```c
void base64_encode(const char *in, size_t inlen, char *out, size_t outlen) {
  static const char b64str[64] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
  while (inlen && outlen) {
    *out++ = b64str[(to_uchar(in[0]) >> 2) & 0x3f];
    if (!--outlen)
      break;
    *out++ = b64str[((to_uchar(in[0]) << 4) + (--inlen ? to_uchar(in[1]) >> 4 : 0)) & 0x3f];
    if (!--outlen)
      break;
    *out++ = (inlen ? b64str[((to_uchar(in[1]) << 2) + (--inlen ? to_uchar(in[2]) >> 6 : 0)) & 0x3f] : '=');
    if (!--outlen)
      break;
    *out++ = inlen ? b64str[to_uchar(in[2]) & 0x3f] : '=';
    if (!--outlen)
      break;
    if (inlen)
      inlen--;
    if (inlen)
      in += 3;
  }
  if (outlen)
    *out = '\0';
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<base64_encode>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L121
  test   %rcx,%rcx
  je     L78
  mov    %rdi,%r12
  mov    %rsi,%r13
  lea    D7(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L67:
  movsbl (%r12),%edi
  call   <to_uchar@plt>
  sar    $0x2,%eax
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,(%rbx)
  cmp    $0x1,%rbp
  je     L78
  movsbl (%r12),%edi
  call   <to_uchar@plt>
  shl    $0x4,%eax
  mov    %eax,%r15d
  cmp    $0x1,%r13
  jne    L122
  and    $0x30,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x1(%rbx)
  cmp    $0x2,%rbp
  je     L78
  movb   $0x3d,0x2(%rbx)
  cmp    $0x3,%rbp
  je     L78
L46:
  movb   $0x3d,0x3(%rbx)
  add    $0x4,%rbx
  cmp    $0x4,%rbp
  je     L78
L150:
  movb   $0x0,(%rbx)
L78:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L122:
  movsbl 0x1(%r12),%edi
  call   <to_uchar@plt>
  sar    $0x4,%eax
  add    %r15d,%eax
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x1(%rbx)
  cmp    $0x2,%rbp
  je     L78
  movsbl 0x1(%r12),%edi
  call   <to_uchar@plt>
  lea    0x0(,%rax,4),%r15d
  cmp    $0x2,%r13
  jne    L113
  mov    %r15d,%eax
  and    $0x3c,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x2(%rbx)
  cmp    $0x3,%rbp
  jne    L46
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L113:
  movsbl 0x2(%r12),%edi
  call   <to_uchar@plt>
  sar    $0x6,%eax
  add    %r15d,%eax
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,0x2(%rbx)
  cmp    $0x3,%rbp
  je     L78
  movsbl 0x2(%r12),%edi
  add    $0x4,%rbx
  call   <to_uchar@plt>
  and    $0x3f,%eax
  movzbl (%r14,%rax,1),%eax
  mov    %al,-0x1(%rbx)
  sub    $0x4,%rbp
  je     L78
  sub    $0x3,%r13
  je     L150
  add    $0x3,%r12
  jmp    L67
L121:
  test   %rcx,%rcx
  jne    L150
  jmp    L78
```

assistant: ```c
void base64_encode(const char *in, size_t inlen, char *out, size_t outlen) {
  static const char b64str[64] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
  while (inlen && outlen) {
    *out++ = b64str[(to_uchar(in[0]) >> 2) & 0x3f];
    if (!--outlen)
      break;
    *out++ = b64str[((to_uchar(in[0]) << 4) + (--inlen ? to_uchar(in[1]) >> 4 : 0)) & 0x3f];
    if (!--outlen)
      break;
    *out++ = (inlen ? b64str[((to_uchar(in[1]) << 2) + (--inlen ? to_uchar(in[2]) >> 6 : 0)) & 0x3f] : '=');
    if (!--outlen)
      break;
    *out++ = inlen ? b64str[to_uchar(in[2]) & 0x3f] : '=';
    if (!--outlen)
      break;
    if (inlen)
      inlen--;
    if (inlen)
      in += 3;
  }
  if (outlen)
    *out = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ucaseConvLuToAlphanum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  movq   $0x24,-0x28(%rbp)
  movq   $0x1,-0x20(%rbp)
L34:
  mov    -0x38(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jae    L35
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <base36enc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L6
  mov    -0x18(%rbp),%rax
  jmp    L37
L6:
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <griddoor_malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L66
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <griddoor_free@plt>
  mov    $0x0,%eax
  jmp    L37
L66:
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <griddoor_free@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x30,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L37
L35:
  mov    -0x28(%rbp),%rax
  sub    %rax,-0x38(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,-0x28(%rbp)
  addq   $0x1,-0x20(%rbp)
  jmp    L34
L37:
  leave
  ret
```

assistant: ```c
char *ucaseConvLuToAlphanum(long unsigned int l) {
  long unsigned int sum = 36;
  for (long unsigned int len = 1;;) {
    if (l < sum) {
      char *str = base36enc(l);
      long unsigned int diff = len - strlen(str);
      if (!diff)
        return str;
      char *valtoret = griddoor_malloc(len + 1);
      if (!valtoret) {
        griddoor_free(str);
        return NULL;
      }
      strcpy(valtoret + diff, str);
      griddoor_free(str);
      return memset(valtoret, '0', diff);
    } else {
      l -= sum;
      sum *= 36;
      len++;
    }
  }
}
```