ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bcd2_bin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x6,-0xc(%rbp)
  movq   $0xf000000,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movq   $0x0,(%rax)
  jmp    L0
L14:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rsi
  mov    -0x20(%rbp),%rax
  and    -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0xc(%rbp),%eax
  shl    $0x2,%eax
  mov    %eax,%ecx
  sar    %cl,%rdx
  mov    %rdx,%rax
  lea    (%rsi,%rax,1),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  sarq   $0x4,-0x8(%rbp)
  subl   $0x1,-0xc(%rbp)
L0:
  cmpq   $0x0,-0x8(%rbp)
  jne    L14
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int bcd2_bin(long *dest, long bcd) {
  int count;
  long mask;
  count = 7 - 1;
  mask = 0xF000000;
  *dest = 0;
  while (mask) {
    *dest *= 10;
    *dest += (bcd & mask) >> (4 * count);
    mask = mask >> 4;
    count -= 1;
  }
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bcd2_bin>:
  endbr64
  mov    $0x18,%ecx
  mov    $0x0,%edx
  mov    $0xf000000,%r8d
L0:
  mov    %rsi,%rax
  and    %r8,%rax
  sar    %cl,%rax
  lea    (%rdx,%rdx,4),%rdx
  lea    (%rax,%rdx,2),%rdx
  sar    $0x4,%r8
  sub    $0x4,%ecx
  cmp    $0xfffffffc,%ecx
  jne    L0
  mov    %rdx,(%rdi)
  mov    $0x0,%eax
  ret
```

assistant: ```c
int bcd2_bin(long *dest, long bcd) {
  int count;
  long mask;
  count = 7 - 1;
  mask = 0xF000000;
  *dest = 0;
  while (mask) {
    *dest *= 10;
    *dest += (bcd & mask) >> (4 * count);
    mask = mask >> 4;
    count -= 1;
  }
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<bcd2_bin>:
  endbr64
  mov    $0x18,%ecx
  xor    %edx,%edx
  mov    $0xf000000,%r8d
  nopl   0x0(%rax)
L5:
  mov    %rsi,%rax
  lea    (%rdx,%rdx,4),%rdx
  and    %r8,%rax
  sar    $0x4,%r8
  sar    %cl,%rax
  sub    $0x4,%ecx
  lea    (%rax,%rdx,2),%rdx
  cmp    $0xfffffffc,%ecx
  jne    L5
  mov    %rdx,(%rdi)
  xor    %eax,%eax
  ret
```

assistant: ```c
int bcd2_bin(long *dest, long bcd) {
  int count;
  long mask;
  count = 7 - 1;
  mask = 0xF000000;
  *dest = 0;
  while (mask) {
    *dest *= 10;
    *dest += (bcd & mask) >> (4 * count);
    mask = mask >> 4;
    count -= 1;
  }
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bcd2_bin>:
  endbr64
  mov    %rsi,%rdx
  mov    %rsi,%rax
  sar    $0x18,%rdx
  and    $0xf,%eax
  and    $0xf,%edx
  lea    (%rdx,%rdx,4),%rcx
  mov    %rsi,%rdx
  sar    $0x14,%rdx
  and    $0xf,%edx
  lea    (%rdx,%rcx,2),%rdx
  lea    (%rdx,%rdx,4),%rcx
  mov    %rsi,%rdx
  sar    $0x10,%rdx
  and    $0xf,%edx
  lea    (%rdx,%rcx,2),%rdx
  lea    (%rdx,%rdx,4),%rcx
  mov    %rsi,%rdx
  sar    $0xc,%rdx
  and    $0xf,%edx
  lea    (%rdx,%rcx,2),%rdx
  lea    (%rdx,%rdx,4),%rcx
  mov    %rsi,%rdx
  sar    $0x8,%rdx
  and    $0xf,%edx
  lea    (%rdx,%rcx,2),%rdx
  lea    (%rdx,%rdx,4),%rcx
  mov    %rsi,%rdx
  sar    $0x4,%rdx
  and    $0xf,%edx
  lea    (%rdx,%rcx,2),%rdx
  lea    (%rdx,%rdx,4),%rdx
  lea    (%rax,%rdx,2),%rax
  mov    %rax,(%rdi)
  xor    %eax,%eax
  ret
```

assistant: ```c
int bcd2_bin(long *dest, long bcd) {
  int count;
  long mask;
  count = 7 - 1;
  mask = 0xF000000;
  *dest = 0;
  while (mask) {
    *dest *= 10;
    *dest += (bcd & mask) >> (4 * count);
    mask = mask >> 4;
    count -= 1;
  }
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<p3opt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %rdx,-0x58(%rbp)
  mov    %rcx,-0x60(%rbp)
  mov    -0x44(%rbp),%edx
  movslq %edx,%rax
  sub    $0x1,%rax
  mov    %rax,-0x28(%rbp)
  movslq %edx,%rax
  mov    %rax,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rax
  sub    $0x1,%rax
  mov    %rax,-0x20(%rbp)
  movslq %ecx,%rax
  mov    %rax,%r10
  mov    $0x0,%r11d
  mov    -0x44(%rbp),%eax
  movslq %eax,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x18(%rbp)
  movslq %eax,%rsi
  mov    %rsi,%r8
  mov    $0x0,%r9d
  movl   $0x0,-0x2c(%rbp)
  jmp    L56
L48:
  mov    -0x2c(%rbp),%esi
  mov    %esi,-0x34(%rbp)
  jmp    L14
L11:
  movl   $0x0,-0x30(%rbp)
  jmp    L34
L18:
  mov    -0x2c(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,8),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x34(%rbp),%esi
  movslq %esi,%rsi
  movsd  (%rdi,%rsi,8),%xmm1
  mov    -0x2c(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,8),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x30(%rbp),%esi
  movslq %esi,%rsi
  movsd  (%rdi,%rsi,8),%xmm0
  mulsd  %xmm0,%xmm1
  mov    -0x34(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,8),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x30(%rbp),%esi
  movslq %esi,%rsi
  movsd  (%rdi,%rsi,8),%xmm0
  mov    -0x34(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,8),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  addsd  %xmm1,%xmm0
  mov    -0x30(%rbp),%esi
  movslq %esi,%rsi
  movsd  %xmm0,(%rdi,%rsi,8)
  addl   $0x1,-0x30(%rbp)
L34:
  mov    -0x30(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L18
  addl   $0x1,-0x34(%rbp)
L14:
  mov    -0x34(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L11
  addl   $0x1,-0x2c(%rbp)
L56:
  mov    -0x2c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L48
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

assistant: ```c
void p3opt(int n, double a[][n], double b[][n], double c[][n]) {
  int i, j, k;
  for (k = 0; k < n; k++)
    for (i = k; i < n; i++)
      for (j = 0; j < n; j++)
        c[i][j] += a[k][i] * b[k][j];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<p3opt>:
  endbr64
  test   %edi,%edi
  jle    L2
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %rcx,%rbp
  movslq %edi,%r9
  shl    $0x3,%r9
  lea    0x8(%r9),%rax
  mov    %rax,-0x8(%rsp)
  mov    %rsi,%r14
  lea    -0x1(%rdi),%r15d
  mov    %edi,%esi
  mov    $0x0,%r11d
  jmp    L36
L54:
  mov    %r8,%rdx
  mov    $0x0,%eax
L49:
  movsd  (%rcx),%xmm0
  mulsd  (%rdi,%rax,8),%xmm0
  addsd  (%rdx,%rax,8),%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L49
  add    $0x8,%rcx
  add    %r9,%r8
  cmp    %r10,%rcx
  jne    L54
L35:
  add    $0x1,%r11
  add    %r9,%r13
  add    -0x8(%rsp),%r14
  add    %r9,%rbp
  add    %r9,%rbx
  cmp    %rsi,%r11
  je     L7
L36:
  cmp    %r11d,%r12d
  jle    L35
  mov    %r13,%rdi
  mov    %r15d,%eax
  sub    %r11d,%eax
  add    %r11,%rax
  lea    0x8(%rbx,%rax,8),%r10
  mov    %rbp,%r8
  mov    %r14,%rcx
  jmp    L54
L7:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  ret
```

assistant: ```c
void p3opt(int n, double a[][n], double b[][n], double c[][n]) {
  int i, j, k;
  for (k = 0; k < n; k++)
    for (i = k; i < n; i++)
      for (j = 0; j < n; j++)
        c[i][j] += a[k][i] * b[k][j];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<p3opt>:
  endbr64
  movslq %edi,%rdi
  test   %edi,%edi
  jle    L33
  lea    0x0(,%rdi,8),%r9
  push   %r12
  mov    %rdi,%r10
  xor    %r11d,%r11d
  push   %rbp
  lea    0x8(%r9),%r12
  mov    %rsi,%rbp
  mov    %rdx,%rsi
  push   %rbx
  mov    %rcx,%rbx
  xchg   %ax,%ax
L4:
  mov    %rbx,%rdx
  mov    %rbp,%rcx
  mov    %r11d,%r8d
  nopl   0x0(%rax)
L16:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L14:
  movsd  (%rcx),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  addsd  (%rdx,%rax,8),%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L14
  add    $0x1,%r8d
  add    $0x8,%rcx
  add    %r9,%rdx
  cmp    %r8d,%r10d
  jg     L16
  add    $0x1,%r11d
  add    %r9,%rsi
  add    %r12,%rbp
  add    %r9,%rbx
  cmp    %r11d,%r10d
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L33:
  ret
```

assistant: ```c
void p3opt(int n, double a[][n], double b[][n], double c[][n]) {
  int i, j, k;
  for (k = 0; k < n; k++)
    for (i = k; i < n; i++)
      for (j = 0; j < n; j++)
        c[i][j] += a[k][i] * b[k][j];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<p3opt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  movslq %edi,%r12
  push   %rbp
  push   %rbx
  mov    %rcx,-0x10(%rsp)
  test   %r12d,%r12d
  jle    L79
  mov    %rsi,%rax
  mov    %r12d,%r15d
  mov    %rdx,%rcx
  mov    %r12,%rbx
  add    $0x8,%rax
  and    $0xfffffffe,%r15d
  lea    0x8(%rdx),%r14
  movq   $0x0,-0x20(%rsp)
  mov    %rax,-0x28(%rsp)
  lea    0x0(,%r12,8),%r13
  mov    %r12d,%eax
  movslq %r15d,%rdx
  lea    0x8(%r13),%rdi
  shr    %eax
  movl   $0x0,-0x2c(%rsp)
  mov    %rdi,-0x18(%rsp)
  lea    -0x1(%rax),%edi
  add    $0x1,%rdi
  shl    $0x4,%rdi
  nopl   0x0(%rax)
L50:
  mov    -0x20(%rsp),%rsi
  mov    -0x10(%rsp),%rax
  mov    -0x28(%rsp),%r8
  mov    -0x2c(%rsp),%r9d
  lea    (%rax,%rsi,8),%rax
  lea    -0x8(%r8),%rsi
  nopl   0x0(%rax)
L40:
  lea    -0x8(%r8),%r10
  lea    (%rax,%r13,1),%r11
  cmp    %r11,%r10
  setae  %r10b
  cmp    %r8,%rax
  setae  %bpl
  or     %bpl,%r10b
  je     L76
  cmp    %r14,%rax
  setne  %bpl
  cmp    $0x1,%ebx
  setne  %r10b
  test   %r10b,%bpl
  je     L76
  movsd  (%rsi),%xmm1
  xor    %r10d,%r10d
  unpcklpd %xmm1,%xmm1
  nopl   0x0(%rax)
L69:
  movupd (%rcx,%r10,1),%xmm0
  movupd (%rax,%r10,1),%xmm2
  mulpd  %xmm1,%xmm0
  addpd  %xmm2,%xmm0
  movups %xmm0,(%rax,%r10,1)
  add    $0x10,%r10
  cmp    %rdi,%r10
  jne    L69
  cmp    %r15d,%ebx
  je     L2
  movsd  (%rsi),%xmm0
  mulsd  (%rcx,%rdx,8),%xmm0
  addsd  (%rax,%rdx,8),%xmm0
  movsd  %xmm0,(%rax,%rdx,8)
L2:
  add    $0x1,%r9d
  mov    %r11,%rax
  add    $0x8,%r8
  add    $0x8,%rsi
  cmp    %r9d,%ebx
  jg     L40
  addl   $0x1,-0x2c(%rsp)
  mov    -0x2c(%rsp),%eax
  add    %r13,%rcx
  add    %r13,%r14
  mov    -0x18(%rsp),%r9
  add    %r12,-0x20(%rsp)
  add    %r9,-0x28(%rsp)
  cmp    %eax,%ebx
  jne    L50
L79:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L76:
  xor    %r10d,%r10d
  nopl   0x0(%rax,%rax,1)
L31:
  movsd  (%rcx,%r10,8),%xmm0
  mulsd  (%rsi),%xmm0
  addsd  (%rax,%r10,8),%xmm0
  movsd  %xmm0,(%rax,%r10,8)
  add    $0x1,%r10
  cmp    %r10,%r12
  jne    L31
  jmp    L2
```

assistant: ```c
void p3opt(int n, double a[][n], double b[][n], double c[][n]) {
  int i, j, k;
  for (k = 0; k < n; k++)
    for (i = k; i < n; i++)
      for (j = 0; j < n; j++)
        c[i][j] += a[k][i] * b[k][j];
}
```
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    $0x10,%rsp
  mov    $0x9,%edi
  call   <CountTree@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
data_type: string

tool: D2: %d is the number of trees


assistant: ```c
int main() {
  int n = CountTree(9);
  printf("%d is the number of trees\n", n);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x9,%edi
  call   <CountTree@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
data_type: string

tool: D4: %d is the number of trees


assistant: ```c
int main() {
  int n = CountTree(9);
  printf("%d is the number of trees\n", n);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x9,%edi
  call   <CountTree@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D7: %d is the number of trees


assistant: ```c
int main() {
  int n = CountTree(9);
  printf("%d is the number of trees\n", n);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x9,%edi
  call   <CountTree@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D1: %d is the number of trees


assistant: ```c
int main() {
  int n = CountTree(9);
  printf("%d is the number of trees\n", n);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strrchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L39
L1:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x9(%rbp)
  jne    L23
  mov    -0x8(%rbp),%rax
  jmp    L32
L23:
  subq   $0x1,-0x8(%rbp)
L39:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jae    L1
  mov    $0x0,%eax
L32:
  leave
  ret
```

assistant: ```c
char *strrchr(const char *s, int c) {
  char cb = c;
  char *ptr = (char *)s + strlen(s) - 1;
  while (ptr >= s) {
    if (*ptr == cb) {
      return ptr;
    }
    --ptr;
  }
  return NULL;
}
```
user: What is the c source code of the assembly code below?

```asm
<strrchr>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  ja     L18
L5:
  cmp    %bpl,(%rax)
  je     L22
  sub    $0x1,%rax
  cmp    %rax,%rbx
  jbe    L5
  mov    $0x0,%eax
L22:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  mov    $0x0,%eax
  jmp    L22
```

assistant: ```c
char *strrchr(const char *s, int c) {
  char cb = c;
  char *ptr = (char *)s + strlen(s) - 1;
  while (ptr >= s) {
    if (*ptr == cb) {
      return ptr;
    }
    --ptr;
  }
  return NULL;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strrchr>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  jbe    L38
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L3:
  sub    $0x1,%rax
  cmp    %rax,%rbx
  ja     L13
L38:
  cmp    %bpl,(%rax)
  jne    L3
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L13:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strrchr(const char *s, int c) {
  char cb = c;
  char *ptr = (char *)s + strlen(s) - 1;
  while (ptr >= s) {
    if (*ptr == cb) {
      return ptr;
    }
    --ptr;
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strrchr>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  jbe    L27
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L13:
  sub    $0x1,%rax
  cmp    %rax,%rbx
  ja     L18
L27:
  cmp    %bpl,(%rax)
  jne    L13
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L18:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strrchr(const char *s, int c) {
  char cb = c;
  char *ptr = (char *)s + strlen(s) - 1;
  while (ptr >= s) {
    if (*ptr == cb) {
      return ptr;
    }
    --ptr;
  }
  return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<capitalize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L24:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L72
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jg     L72
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  add    $0x20,%eax
  jmp    L45
L72:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
L45:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    D63(%rip),%rcx
  mov    %al,(%rdx,%rcx,1)
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L24
  mov    -0x4(%rbp),%eax
  cltq
  lea    D63(%rip),%rdx
  movb   $0x0,(%rax,%rdx,1)
  movzbl D63(%rip),%eax
  cmp    $0x60,%al
  jle    L29
  movzbl D63(%rip),%eax
  cmp    $0x7a,%al
  jg     L29
  movzbl D63(%rip),%eax
  sub    $0x20,%eax
  jmp    L59
L29:
  movzbl D63(%rip),%eax
L59:
  mov    %al,D46(%rip)
  lea    D46(%rip),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *capitalize(const char *str) {
  static char strcap[1024];
  int i;
  for (i = 0; str[i] != '\0'; i++)
    strcap[i] = ((str[i]) >= 'A' && (str[i]) <= 'Z' ? (str[i]) + 'a' - 'A' : (str[i]));
  strcap[i] = '\0';
  strcap[0] = ((strcap[0]) >= 'a' && (strcap[0]) <= 'z' ? (strcap[0]) + 'A' - 'a' : (strcap[0]));
  return strcap;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<capitalize>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L11
  mov    $0x1,%edx
  lea    D39(%rip),%r8
L19:
  lea    -0x41(%rax),%esi
  lea    0x20(%rax),%ecx
  cmp    $0x1a,%sil
  cmovb  %ecx,%eax
  mov    %al,(%rdx,%r8,1)
  mov    %edx,%ecx
  add    $0x1,%rdx
  movzbl -0x1(%rdi,%rdx,1),%eax
  test   %al,%al
  jne    L19
L15:
  movslq %ecx,%rcx
  lea    D38(%rip),%rax
  movb   $0x0,(%rax,%rcx,1)
  movzbl D38(%rip),%eax
  lea    -0x61(%rax),%ecx
  lea    -0x20(%rax),%edx
  cmp    $0x1a,%cl
  cmovb  %edx,%eax
  mov    %al,D4(%rip)
  lea    D4(%rip),%rax
  ret
L11:
  mov    $0x0,%ecx
  jmp    L15
```

assistant: ```c
char *capitalize(const char *str) {
  static char strcap[1024];
  int i;
  for (i = 0; str[i] != '\0'; i++)
    strcap[i] = ((str[i]) >= 'A' && (str[i]) <= 'Z' ? (str[i]) + 'a' - 'A' : (str[i]));
  strcap[i] = '\0';
  strcap[0] = ((strcap[0]) >= 'a' && (strcap[0]) <= 'z' ? (strcap[0]) + 'A' - 'a' : (strcap[0]));
  return strcap;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<capitalize>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L15
  mov    $0x1,%edx
  lea    D25(%rip),%r8
  nopw   0x0(%rax,%rax,1)
L28:
  lea    -0x41(%rax),%esi
  lea    0x20(%rax),%ecx
  cmp    $0x1a,%sil
  cmovb  %ecx,%eax
  movslq %edx,%rcx
  mov    %al,(%r8,%rdx,1)
  add    $0x1,%rdx
  movzbl -0x1(%rdi,%rdx,1),%eax
  test   %al,%al
  jne    L28
L21:
  lea    D20(%rip),%r8
  movb   $0x0,(%r8,%rcx,1)
  movzbl D20(%rip),%eax
  lea    -0x61(%rax),%ecx
  lea    -0x20(%rax),%edx
  cmp    $0x1a,%cl
  cmovb  %edx,%eax
  mov    %al,D9(%rip)
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  xor    %ecx,%ecx
  jmp    L21
```

assistant: ```c
char *capitalize(const char *str) {
  static char strcap[1024];
  int i;
  for (i = 0; str[i] != '\0'; i++)
    strcap[i] = ((str[i]) >= 'A' && (str[i]) <= 'Z' ? (str[i]) + 'a' - 'A' : (str[i]));
  strcap[i] = '\0';
  strcap[0] = ((strcap[0]) >= 'a' && (strcap[0]) <= 'z' ? (strcap[0]) + 'A' - 'a' : (strcap[0]));
  return strcap;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<capitalize>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L24
  mov    $0x1,%edx
  lea    D14(%rip),%rsi
  nopw   0x0(%rax,%rax,1)
L23:
  lea    -0x41(%rax),%ecx
  cmp    $0x19,%cl
  ja     L11
  add    $0x20,%eax
L11:
  mov    %al,(%rsi,%rdx,1)
  movslq %edx,%rcx
  add    $0x1,%rdx
  movzbl -0x1(%rdi,%rdx,1),%eax
  test   %al,%al
  jne    L23
L21:
  lea    D27(%rip),%r8
  movb   $0x0,(%r8,%rcx,1)
  movzbl D27(%rip),%eax
  lea    -0x61(%rax),%ecx
  lea    -0x20(%rax),%edx
  cmp    $0x1a,%cl
  cmovb  %edx,%eax
  mov    %al,D29(%rip)
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L24:
  xor    %ecx,%ecx
  jmp    L21
```

assistant: ```c
char *capitalize(const char *str) {
  static char strcap[1024];
  int i;
  for (i = 0; str[i] != '\0'; i++)
    strcap[i] = ((str[i]) >= 'A' && (str[i]) <= 'Z' ? (str[i]) + 'a' - 'A' : (str[i]));
  strcap[i] = '\0';
  strcap[0] = ((strcap[0]) >= 'a' && (strcap[0]) <= 'z' ? (strcap[0]) + 'A' - 'a' : (strcap[0]));
  return strcap;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rgb_shiftmask_get>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  sub    $0x8,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rgb_shiftmask_get(int *shift, unsigned *mask, unsigned len, unsigned pos) {
  *shift = pos + len - 8;
  *mask = ((1 << len) - 1) << pos;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rgb_shiftmask_get>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rdi
  mov    %ecx,%esi
  lea    -0x8(%rdx,%rcx,1),%ecx
  mov    %ecx,(%rax)
  mov    $0x1,%eax
  mov    %edx,%ecx
  shl    %cl,%eax
  sub    $0x1,%eax
  mov    %esi,%ecx
  shl    %cl,%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void rgb_shiftmask_get(int *shift, unsigned *mask, unsigned len, unsigned pos) {
  *shift = pos + len - 8;
  *mask = ((1 << len) - 1) << pos;
}
```
user: Translate the below assembly code into C programming language:

```asm
<rgb_shiftmask_get>:
  endbr64
  lea    -0x8(%rdx,%rcx,1),%eax
  mov    %rsi,%r8
  mov    %ecx,%esi
  mov    %edx,%ecx
  mov    %eax,(%rdi)
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %esi,%ecx
  sub    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,(%r8)
  ret
```

assistant: ```c
void rgb_shiftmask_get(int *shift, unsigned *mask, unsigned len, unsigned pos) {
  *shift = pos + len - 8;
  *mask = ((1 << len) - 1) << pos;
}
```