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
<int_matrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    -0x30(%rbp),%rax
  sub    -0x28(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  sub    -0x38(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  shl    $0x3,%rax
  mov    %rax,%rdi
  call   <xmalloc@plt>
  mov    %rax,-0x8(%rbp)
  addq   $0x8,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  shl    $0x3,%rax
  neg    %rax
  add    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  imul   -0x10(%rbp),%rax
  add    $0x1,%rax
  shl    $0x2,%rax
  mov    %rax,%rdi
  call   <xmalloc@plt>
  mov    -0x28(%rbp),%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,8),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  add    $0x4,%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x38(%rbp),%rax
  shl    $0x2,%rax
  neg    %rax
  mov    %rax,%rcx
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,8),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x20(%rbp)
  jmp    L1
L0:
  mov    -0x20(%rbp),%rax
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  addq   $0x1,-0x20(%rbp)
L1:
  mov    -0x20(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jle    L0
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int **int_matrix(long nrl, long nrh, long ncl, long nch) {
  long i, nrow, ncol;
  int **m;
  nrow = nrh - nrl + 1;
  ncol = nch - ncl + 1;
  m = (int **)xmalloc((size_t)((nrow + 1) * sizeof(int *)));
  m += 1;
  m -= nrl;
  m[nrl] = (int *)xmalloc((size_t)((nrow * ncol + 1) * sizeof(int)));
  m[nrl] += 1;
  m[nrl] -= ncl;
  for (i = nrl + 1; i <= nrh; i++)
    m[i] = m[i - 1] + ncol;
  return m;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<int_matrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %rsi,%r13
  mov    %rsi,%r12
  sub    %rdi,%r12
  mov    %rdx,0x8(%rsp)
  sub    %rdx,%rcx
  lea    0x1(%rcx),%rbp
  lea    0x10(,%r12,8),%rdi
  call   <xmalloc@plt>
  mov    %rax,%rbx
  mov    $0x1,%r14d
  mov    %r14,%rax
  mov    %r15,(%rsp)
  sub    %r15,%rax
  lea    (%rbx,%rax,8),%r15
  add    $0x1,%r12
  imul   %rbp,%r12
  lea    0x4(,%r12,4),%rdi
  call   <xmalloc@plt>
  sub    0x8(%rsp),%r14
  lea    (%rax,%r14,4),%rax
  mov    %rax,0x8(%rbx)
  cmp    (%rsp),%r13
  jle    L12
  shl    $0x2,%rbp
  add    $0x8,%rbx
  lea    (%r15,%r13,8),%rdx
L16:
  mov    %rbp,%rax
  add    (%rbx),%rax
  mov    %rax,0x8(%rbx)
  add    $0x8,%rbx
  cmp    %rdx,%rbx
  jne    L16
L12:
  mov    %r15,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **int_matrix(long nrl, long nrh, long ncl, long nch) {
  long i, nrow, ncol;
  int **m;
  nrow = nrh - nrl + 1;
  ncol = nch - ncl + 1;
  m = (int **)xmalloc((size_t)((nrow + 1) * sizeof(int *)));
  m += 1;
  m -= nrl;
  m[nrl] = (int *)xmalloc((size_t)((nrow * ncol + 1) * sizeof(int)));
  m[nrl] += 1;
  m[nrl] -= ncl;
  for (i = nrl + 1; i <= nrh; i++)
    m[i] = m[i - 1] + ncol;
  return m;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<int_matrix>:
  endbr64
  push   %r15
  sub    %rdx,%rcx
  mov    $0x1,%r15d
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  sub    %rdi,%r12
  lea    0x1(%rcx),%rbp
  push   %rbx
  lea    0x10(,%r12,8),%rdi
  add    $0x1,%r12
  imul   %rbp,%r12
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  call   <xmalloc@plt>
  lea    0x4(,%r12,4),%rdi
  mov    %rax,%rbx
  mov    %r15,%rax
  sub    %r13,%rax
  lea    (%rbx,%rax,8),%r8
  mov    %r8,(%rsp)
  call   <xmalloc@plt>
  mov    0x8(%rsp),%rdx
  mov    (%rsp),%r8
  sub    %rdx,%r15
  cmp    %r13,%r14
  lea    (%rax,%r15,4),%rax
  mov    %rax,0x8(%rbx)
  jle    L13
  lea    0x0(,%rbp,4),%rcx
  lea    0x8(%rbx),%rax
  lea    (%r8,%r14,8),%rsi
  nopl   0x0(%rax)
L1:
  mov    (%rax),%rdx
  add    $0x8,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  cmp    %rax,%rsi
  jne    L1
L13:
  add    $0x18,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **int_matrix(long nrl, long nrh, long ncl, long nch) {
  long i, nrow, ncol;
  int **m;
  nrow = nrh - nrl + 1;
  ncol = nch - ncl + 1;
  m = (int **)xmalloc((size_t)((nrow + 1) * sizeof(int *)));
  m += 1;
  m -= nrl;
  m[nrl] = (int *)xmalloc((size_t)((nrow * ncol + 1) * sizeof(int)));
  m[nrl] += 1;
  m[nrl] -= ncl;
  for (i = nrl + 1; i <= nrh; i++)
    m[i] = m[i - 1] + ncol;
  return m;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<int_matrix>:
  endbr64
  push   %r15
  sub    %rdx,%rcx
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  sub    %rdi,%r12
  mov    %rdi,%rbp
  push   %rbx
  lea    0x10(,%r12,8),%rdi
  lea    0x1(%rcx),%rbx
  add    $0x1,%r12
  imul   %rbx,%r12
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  call   <xmalloc@plt>
  lea    0x4(,%r12,4),%rdi
  mov    %rax,%r14
  mov    $0x1,%eax
  sub    %rbp,%rax
  add    $0x1,%rbp
  lea    (%r14,%rax,8),%r15
  call   <xmalloc@plt>
  mov    0x8(%rsp),%rsi
  mov    $0x1,%edx
  sub    %rsi,%rdx
  lea    (%rax,%rdx,4),%rdx
  mov    %rdx,0x8(%r14)
  cmp    %rbp,%r13
  jl     L13
  lea    0x0(,%rbx,4),%rcx
  lea    (%r15,%rbp,8),%rax
  lea    0x8(%r15,%r13,8),%rsi
  nopw   0x0(%rax,%rax,1)
L9:
  add    %rcx,%rdx
  add    $0x8,%rax
  mov    %rdx,-0x8(%rax)
  cmp    %rsi,%rax
  jne    L9
L13:
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **int_matrix(long nrl, long nrh, long ncl, long nch) {
  long i, nrow, ncol;
  int **m;
  nrow = nrh - nrl + 1;
  ncol = nch - ncl + 1;
  m = (int **)xmalloc((size_t)((nrow + 1) * sizeof(int *)));
  m += 1;
  m -= nrl;
  m[nrl] = (int *)xmalloc((size_t)((nrow * ncol + 1) * sizeof(int)));
  m[nrl] += 1;
  m[nrl] -= ncl;
  for (i = nrl + 1; i <= nrh; i++)
    m[i] = m[i - 1] + ncol;
  return m;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<uDMAChannelRequest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0xffe0,%eax
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  and    $0x1f,%eax
  mov    $0x400ff014,%edx
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void uDMAChannelRequest(uint32_t ui32ChannelNum) {
  ASSERT((ui32ChannelNum & 0xffff) < 32);
  (*((volatile uint32_t *)(0x400FF014))) = 1 << (ui32ChannelNum & 0x1f);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<uDMAChannelRequest>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   $0xffe0,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  mov    $0x1,%eax
  mov    %ebx,%ecx
  shl    %cl,%eax
  mov    %eax,0x400ff014
  pop    %rbx
  ret
```

assistant: ```c
void uDMAChannelRequest(uint32_t ui32ChannelNum) {
  ASSERT((ui32ChannelNum & 0xffff) < 32);
  (*((volatile uint32_t *)(0x400FF014))) = 1 << (ui32ChannelNum & 0x1f);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<uDMAChannelRequest>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  test   $0xffe0,%ebx
  sete   %dil
  call   <ASSERT@plt>
  mov    %ebx,%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,0x400ff014
  pop    %rbx
  ret
```

assistant: ```c
void uDMAChannelRequest(uint32_t ui32ChannelNum) {
  ASSERT((ui32ChannelNum & 0xffff) < 32);
  (*((volatile uint32_t *)(0x400FF014))) = 1 << (ui32ChannelNum & 0x1f);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x38(%rbp)
  jmp    L48
L27:
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x38(%rbp),%eax
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L48:
  mov    -0x38(%rbp),%eax
  cmp    -0x44(%rbp),%eax
  jl     L27
  lea    D28(%rip),%rax
  mov    %rax,-0x30(%rbp)
  lea    D45(%rip),%rax
  mov    %rax,-0x28(%rbp)
  lea    D30(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D15(%rip),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x4,-0x34(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L49
L17:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,8),%rax
  mov    %rax,%rsi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L49:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L17
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D51
type: string
name: disassemble_data
arguments: data_label: D28
type: string
name: disassemble_data
arguments: data_label: D45
type: string
name: disassemble_data
arguments: data_label: D30
type: string
name: disassemble_data
arguments: data_label: D15
type: string
name: disassemble_data
arguments: data_label: D46
type: string

tool: D51: arg %d: %s


tool: D28: Oregon

tool: D45: Minnesota

tool: D30: Wisconsin

tool: D15: New York

tool: D46: state %d: %s


assistant: ```c
int main(int argc, const char *argv[]) {
  int i = 0;
  while (i < argc) {
    printf("arg %d: %s\n", i, argv[i]);
    i++;
  }
  char *states[] = {"Oregon", "Minnesota", "Wisconsin", "New York"};
  int num_states = 4;
  i = 0;
  while (i < num_states) {
    printf("state %d: %s\n", states[i]);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L0
  mov    %rsi,%r12
  mov    %edi,%ebp
  mov    $0x0,%ebx
  lea    D16(%rip),%r13
L29:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L29
L0:
  lea    D18(%rip),%rbp
  lea    D13(%rip),%rbx
L12:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L12
```

tool_calls: name: unpack_data
arguments: data_tag: D16
data_type: string
name: unpack_data
arguments: data_tag: D18
data_type: string
name: unpack_data
arguments: data_tag: D13
data_type: string

tool: D16: arg %d: %s


tool: D18: Oregon

tool: D13: state %d: %s


assistant: ```c
int main(int argc, const char *argv[]) {
  int i = 0;
  while (i < argc) {
    printf("arg %d: %s\n", i, argv[i]);
    i++;
  }
  char *states[] = {"Oregon", "Minnesota", "Wisconsin", "New York"};
  int num_states = 4;
  i = 0;
  while (i < num_states) {
    printf("state %d: %s\n", states[i]);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L29
  mov    %rsi,%r12
  movslq %edi,%rbp
  lea    D3(%rip),%r13
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L2:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L2
L29:
  lea    D6(%rip),%rbp
  lea    D14(%rip),%rbx
  nop
L25:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L25
```

tool_calls: name: read_data
arguments: data_tag: D3
type: string
name: read_data
arguments: data_tag: D6
type: string
name: read_data
arguments: data_tag: D14
type: string

tool: D3: arg %d: %s


tool: D6: Oregon

tool: D14: state %d: %s


assistant: ```c
int main(int argc, const char *argv[]) {
  int i = 0;
  while (i < argc) {
    printf("arg %d: %s\n", i, argv[i]);
    i++;
  }
  char *states[] = {"Oregon", "Minnesota", "Wisconsin", "New York"};
  int num_states = 4;
  i = 0;
  while (i < num_states) {
    printf("state %d: %s\n", states[i]);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L11
  mov    %rsi,%r12
  movslq %edi,%rbp
  lea    D15(%rip),%r13
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L3:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L3
L11:
  lea    D11(%rip),%rbp
  lea    D1(%rip),%rbx
  nop
L6:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L6
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
datatype: string
name: extract_symbolic_data
arguments: tag: D11
datatype: string
name: extract_symbolic_data
arguments: tag: D1
datatype: string

tool: D15: arg %d: %s


tool: D11: Oregon

tool: D1: state %d: %s


assistant: ```c
int main(int argc, const char *argv[]) {
  int i = 0;
  while (i < argc) {
    printf("arg %d: %s\n", i, argv[i]);
    i++;
  }
  char *states[] = {"Oregon", "Minnesota", "Wisconsin", "New York"};
  int num_states = 4;
  i = 0;
  while (i < num_states) {
    printf("state %d: %s\n", states[i]);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tr_partialcopy>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r11
  mov    %rsi,%rbx
  mov    %rcx,%r10
  mov    0x38(%rsp),%ecx
  mov    %r8,%r12
  sub    %rsi,%r12
  sar    $0x2,%r12
  sub    $0x1,%r12d
  sub    $0x4,%r10
  cmp    %r10,%rdx
  ja     L137
  mov    %rdx,%rbp
  mov    %rdx,%rax
  mov    $0xffffffff,%edx
  mov    $0xffffffff,%r14d
  jmp    L34
L125:
  add    $0x4,%rax
  cmp    %rax,%r10
  jb     L157
L34:
  mov    (%rax),%esi
  mov    %esi,%r15d
  sub    %ecx,%r15d
  js     L125
  movslq %r15d,%rdi
  lea    (%r11,%rdi,4),%rdi
  cmp    %r12d,(%rdi)
  jne    L125
  lea    0x4(%r10),%r13
  mov    %r15d,0x4(%r10)
  movslq %esi,%rsi
  mov    (%r11,%rsi,4),%r10d
  mov    %r13,%rsi
  sub    %rbx,%rsi
  sar    $0x2,%rsi
  cmp    %r14d,%r10d
  cmovne %esi,%edx
  mov    %edx,(%rdi)
  mov    %r10d,%r14d
  mov    %r13,%r10
  jmp    L125
L157:
  cmp    %rbp,%r10
  jb     L15
  mov    %r10,%rax
  mov    $0xffffffff,%esi
  jmp    L25
L124:
  sub    $0x4,%rax
  cmp    %rax,%rbp
  ja     L15
L25:
  movslq (%rax),%rdi
  lea    (%r11,%rdi,4),%r13
  mov    %esi,%r14d
  mov    0x0(%r13),%esi
  mov    %rax,%rdi
  sub    %rbx,%rdi
  sar    $0x2,%rdi
  cmp    %r14d,%esi
  cmovne %edi,%edx
  cmp    %esi,%edx
  je     L124
  mov    %edx,0x0(%r13)
  jmp    L124
L137:
  mov    $0xffffffff,%edx
L15:
  sub    $0x4,%r9
  add    $0x4,%r10
  cmp    %r10,%r8
  jbe    L155
  mov    $0xffffffff,%ebp
  jmp    L58
L107:
  sub    $0x4,%r9
  cmp    %r10,%r8
  jbe    L155
L58:
  mov    (%r9),%eax
  mov    %eax,%esi
  sub    %ecx,%esi
  js     L107
  movslq %esi,%rdi
  lea    (%r11,%rdi,4),%rdi
  cmp    %r12d,(%rdi)
  jne    L107
  lea    -0x4(%r8),%r13
  mov    %esi,-0x4(%r8)
  cltq
  mov    (%r11,%rax,4),%esi
  mov    %r13,%rax
  sub    %rbx,%rax
  sar    $0x2,%rax
  cmp    %ebp,%esi
  cmovne %eax,%edx
  mov    %edx,(%rdi)
  mov    %esi,%ebp
  mov    %r13,%r8
  jmp    L107
L155:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void tr_partialcopy(int *ISA, const int *SA, int *first, int *a, int *b, int *last, int depth) {
  int *c, *d, *e;
  int s, v;
  int rank, lastrank, newrank = -1;
  v = b - SA - 1;
  lastrank = -1;
  for (c = first, d = a - 1; c <= d; ++c) {
    if ((0 <= (s = *c - depth)) && (ISA[s] == v)) {
      *++d = s;
      rank = ISA[s + depth];
      if (lastrank != rank) {
        lastrank = rank;
        newrank = d - SA;
      }
      ISA[s] = newrank;
    }
  }
  lastrank = -1;
  for (e = d; first <= e; --e) {
    rank = ISA[*e];
    if (lastrank != rank) {
      lastrank = rank;
      newrank = e - SA;
    }
    if (newrank != rank) {
      ISA[*e] = newrank;
    }
  }
  lastrank = -1;
  for (c = last - 1, e = d + 1, d = b; e < d; --c) {
    if ((0 <= (s = *c - depth)) && (ISA[s] == v)) {
      *--d = s;
      rank = ISA[s + depth];
      if (lastrank != rank) {
        lastrank = rank;
        newrank = d - SA;
      }
      ISA[s] = newrank;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<tr_partialcopy>:
  endbr64
  push   %r15
  mov    %rdi,%r10
  mov    %rsi,%r11
  push   %r14
  push   %r13
  push   %r12
  mov    %r8,%r12
  push   %rbp
  sub    %rsi,%r12
  mov    %rcx,%rbp
  sar    $0x2,%r12
  lea    -0x4(%rbp),%rax
  push   %rbx
  mov    0x38(%rsp),%ecx
  sub    $0x1,%r12d
  cmp    %rax,%rdx
  ja     L58
  mov    %rdx,%rbx
  mov    %rdx,%rsi
  mov    $0xffffffff,%r14d
  mov    $0xffffffff,%edx
  nopw   0x0(%rax,%rax,1)
L67:
  movslq (%rsi),%rdi
  mov    %edi,%r15d
  sub    %ecx,%r15d
  js     L78
  movslq %r15d,%rbp
  lea    (%r10,%rbp,4),%rbp
  cmp    %r12d,0x0(%rbp)
  jne    L78
  lea    0x4(%rax),%r13
  mov    %r15d,0x4(%rax)
  mov    (%r10,%rdi,4),%edi
  mov    %r13,%rax
  sub    %r11,%rax
  sar    $0x2,%rax
  cmp    %r14d,%edi
  mov    %edi,%r14d
  cmovne %eax,%edx
  mov    %r13,%rax
  mov    %edx,0x0(%rbp)
L78:
  add    $0x4,%rsi
  cmp    %rsi,%rax
  jae    L67
  lea    0x4(%rax),%rbp
  cmp    %rbx,%rax
  jb     L49
  mov    $0xffffffff,%esi
  nopl   0x0(%rax)
L115:
  movslq (%rax),%rdi
  mov    %esi,%r14d
  lea    (%r10,%rdi,4),%r13
  mov    %rax,%rdi
  mov    0x0(%r13),%esi
  sub    %r11,%rdi
  sar    $0x2,%rdi
  cmp    %r14d,%esi
  cmovne %edi,%edx
  cmp    %esi,%edx
  je     L87
  mov    %edx,0x0(%r13)
L87:
  sub    $0x4,%rax
  cmp    %rax,%rbx
  jbe    L115
L49:
  sub    $0x4,%r9
  mov    $0xffffffff,%ebx
  cmp    %rbp,%r8
  jbe    L54
  nopl   (%rax)
L6:
  movslq (%r9),%rax
  mov    %eax,%esi
  sub    %ecx,%esi
  js     L61
  movslq %esi,%rdi
  lea    (%r10,%rdi,4),%rdi
  cmp    %r12d,(%rdi)
  jne    L61
  lea    -0x4(%r8),%r13
  mov    %esi,-0x4(%r8)
  mov    (%r10,%rax,4),%esi
  mov    %r13,%rax
  mov    %r13,%r8
  sub    %r11,%rax
  sar    $0x2,%rax
  cmp    %ebx,%esi
  mov    %esi,%ebx
  cmovne %eax,%edx
  mov    %edx,(%rdi)
L61:
  sub    $0x4,%r9
  cmp    %rbp,%r8
  ja     L6
L54:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L58:
  mov    $0xffffffff,%edx
  jmp    L49
```

assistant: ```c
void tr_partialcopy(int *ISA, const int *SA, int *first, int *a, int *b, int *last, int depth) {
  int *c, *d, *e;
  int s, v;
  int rank, lastrank, newrank = -1;
  v = b - SA - 1;
  lastrank = -1;
  for (c = first, d = a - 1; c <= d; ++c) {
    if ((0 <= (s = *c - depth)) && (ISA[s] == v)) {
      *++d = s;
      rank = ISA[s + depth];
      if (lastrank != rank) {
        lastrank = rank;
        newrank = d - SA;
      }
      ISA[s] = newrank;
    }
  }
  lastrank = -1;
  for (e = d; first <= e; --e) {
    rank = ISA[*e];
    if (lastrank != rank) {
      lastrank = rank;
      newrank = e - SA;
    }
    if (newrank != rank) {
      ISA[*e] = newrank;
    }
  }
  lastrank = -1;
  for (c = last - 1, e = d + 1, d = b; e < d; --c) {
    if ((0 <= (s = *c - depth)) && (ISA[s] == v)) {
      *--d = s;
      rank = ISA[s + depth];
      if (lastrank != rank) {
        lastrank = rank;
        newrank = d - SA;
      }
      ISA[s] = newrank;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tr_partialcopy>:
  endbr64
  push   %r15
  mov    %rdi,%r10
  mov    %rsi,%r11
  push   %r14
  push   %r13
  push   %r12
  mov    %r8,%r12
  push   %rbp
  sub    %rsi,%r12
  mov    %rcx,%rbp
  sar    $0x2,%r12
  lea    -0x4(%rbp),%rax
  push   %rbx
  mov    0x38(%rsp),%ecx
  sub    $0x1,%r12d
  cmp    %rax,%rdx
  ja     L80
  mov    %rdx,%rbx
  mov    %rdx,%rsi
  mov    $0xffffffff,%r14d
  mov    $0xffffffff,%edx
  nopw   0x0(%rax,%rax,1)
L21:
  movslq (%rsi),%rdi
  mov    %edi,%r15d
  sub    %ecx,%r15d
  js     L90
  movslq %r15d,%rbp
  lea    (%r10,%rbp,4),%rbp
  cmp    %r12d,0x0(%rbp)
  jne    L90
  lea    0x4(%rax),%r13
  mov    %r15d,0x4(%rax)
  mov    (%r10,%rdi,4),%edi
  mov    %r13,%rax
  sub    %r11,%rax
  sar    $0x2,%rax
  cmp    %r14d,%edi
  mov    %edi,%r14d
  cmovne %eax,%edx
  mov    %r13,%rax
  mov    %edx,0x0(%rbp)
L90:
  add    $0x4,%rsi
  cmp    %rsi,%rax
  jae    L21
  lea    0x4(%rax),%rbp
  cmp    %rbx,%rax
  jb     L8
  mov    $0xffffffff,%esi
  nopl   0x0(%rax)
L11:
  movslq (%rax),%rdi
  mov    %esi,%r14d
  lea    (%r10,%rdi,4),%r13
  mov    %rax,%rdi
  mov    0x0(%r13),%esi
  sub    %r11,%rdi
  sar    $0x2,%rdi
  cmp    %r14d,%esi
  cmovne %edi,%edx
  cmp    %esi,%edx
  je     L7
  mov    %edx,0x0(%r13)
L7:
  sub    $0x4,%rax
  cmp    %rax,%rbx
  jbe    L11
L8:
  sub    $0x4,%r9
  mov    $0xffffffff,%ebx
  cmp    %rbp,%r8
  jbe    L9
  nopl   (%rax)
L105:
  movslq (%r9),%rax
  mov    %eax,%esi
  sub    %ecx,%esi
  js     L71
  movslq %esi,%rdi
  lea    (%r10,%rdi,4),%rdi
  cmp    %r12d,(%rdi)
  jne    L71
  lea    -0x4(%r8),%r13
  mov    %esi,-0x4(%r8)
  mov    (%r10,%rax,4),%esi
  mov    %r13,%rax
  mov    %r13,%r8
  sub    %r11,%rax
  sar    $0x2,%rax
  cmp    %ebx,%esi
  mov    %esi,%ebx
  cmovne %eax,%edx
  mov    %edx,(%rdi)
L71:
  sub    $0x4,%r9
  cmp    %rbp,%r8
  ja     L105
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L80:
  mov    $0xffffffff,%edx
  jmp    L8
```

assistant: ```c
void tr_partialcopy(int *ISA, const int *SA, int *first, int *a, int *b, int *last, int depth) {
  int *c, *d, *e;
  int s, v;
  int rank, lastrank, newrank = -1;
  v = b - SA - 1;
  lastrank = -1;
  for (c = first, d = a - 1; c <= d; ++c) {
    if ((0 <= (s = *c - depth)) && (ISA[s] == v)) {
      *++d = s;
      rank = ISA[s + depth];
      if (lastrank != rank) {
        lastrank = rank;
        newrank = d - SA;
      }
      ISA[s] = newrank;
    }
  }
  lastrank = -1;
  for (e = d; first <= e; --e) {
    rank = ISA[*e];
    if (lastrank != rank) {
      lastrank = rank;
      newrank = e - SA;
    }
    if (newrank != rank) {
      ISA[*e] = newrank;
    }
  }
  lastrank = -1;
  for (c = last - 1, e = d + 1, d = b; e < d; --c) {
    if ((0 <= (s = *c - depth)) && (ISA[s] == v)) {
      *--d = s;
      rank = ISA[s + depth];
      if (lastrank != rank) {
        lastrank = rank;
        newrank = d - SA;
      }
      ISA[s] = newrank;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <chdir@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type_infered: string
name: disassemble_data
arguments: data_tag: D13
type_infered: string

tool: D9: /

tool: D13: ret is %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int ret = chdir("/");
  printf("ret"
         " is %d\n",
         (ret));
  return 0;
}
```