ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<merge>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r10
  mov    %rsi,%rbp
  mov    %edx,%edi
  mov    %rcx,%rbx
  test   %edx,%edx
  jle    L94
  test   %r8d,%r8d
  jle    L94
  mov    %r10,%rsi
  mov    $0x1,%edx
  mov    $0x0,%eax
  mov    $0x0,%ecx
  jmp    L11
L45:
  add    $0x1,%eax
  mov    %edx,%r12d
  mov    %r9d,(%rsi)
L31:
  add    $0x1,%edx
  add    $0x4,%rsi
  cmp    %edi,%ecx
  jge    L63
  cmp    %r8d,%eax
  jge    L63
L11:
  movslq %ecx,%r9
  mov    0x0(%rbp,%r9,4),%r11d
  movslq %eax,%r9
  mov    (%rbx,%r9,4),%r9d
  cmp    %r9d,%r11d
  jge    L45
  add    $0x1,%ecx
  mov    %edx,%r12d
  mov    %r11d,(%rsi)
  jmp    L31
L94:
  mov    $0x0,%r12d
  mov    $0x0,%eax
  mov    $0x0,%ecx
L63:
  cmp    %ecx,%edi
  jle    L21
  movslq %ecx,%rdx
  movslq %r12d,%rsi
  sub    %rdx,%rsi
  lea    (%r10,%rsi,4),%r9
L108:
  mov    0x0(%rbp,%rdx,4),%esi
  mov    %esi,(%r9,%rdx,4)
  add    $0x1,%rdx
  cmp    %edx,%edi
  jg     L108
  add    %r12d,%edi
  sub    %ecx,%edi
  mov    %edi,%r12d
L21:
  cmp    %eax,%r8d
  jle    L9
  cltq
  movslq %r12d,%r12
  sub    %rax,%r12
  lea    (%r10,%r12,4),%rcx
L16:
  mov    (%rbx,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jg     L16
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void merge(int *A, int *L, int leftcount, int *R, int rightcount) {
  int i = 0, j = 0, k = 0;
  while (i < leftcount && j < rightcount) {
    if (L[i] < R[j])
      A[k++] = L[i++];
    else
      A[k++] = R[j++];
  }
  while (i < leftcount)
    A[k++] = L[i++];
  while (j < rightcount)
    A[k++] = R[j++];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rdi,%r10
  mov    %rcx,%r11
  mov    %edx,%edi
  push   %rbx
  mov    %rsi,%rbx
  test   %edx,%edx
  jle    L41
  test   %r8d,%r8d
  jle    L41
  mov    $0x1,%edx
  xor    %eax,%eax
  xor    %esi,%esi
  jmp    L39
  nopl   (%rax)
L12:
  add    $0x1,%esi
  mov    %ecx,-0x4(%r10,%rdx,4)
  add    $0x1,%rdx
  cmp    %edi,%esi
  jge    L54
L58:
  cmp    %r8d,%eax
  jge    L54
L39:
  movslq %esi,%rcx
  movslq %eax,%r9
  movslq %edx,%rbp
  mov    (%rbx,%rcx,4),%ecx
  mov    (%r11,%r9,4),%r9d
  cmp    %r9d,%ecx
  jl     L12
  mov    %r9d,%ecx
  add    $0x1,%eax
  mov    %ecx,-0x4(%r10,%rdx,4)
  add    $0x1,%rdx
  cmp    %edi,%esi
  jl     L58
L54:
  cmp    %esi,%edi
  jle    L63
  movslq %esi,%rdx
  movslq %ebp,%rcx
  sub    %rdx,%rcx
  lea    (%r10,%rcx,4),%r9
  xchg   %ax,%ax
L108:
  mov    (%rbx,%rdx,4),%ecx
  mov    %ecx,(%r9,%rdx,4)
  add    $0x1,%rdx
  cmp    %edx,%edi
  jg     L108
  add    %ebp,%edi
  sub    %esi,%edi
  movslq %edi,%rbp
L63:
  cmp    %eax,%r8d
  jle    L45
  cltq
  sub    %rax,%rbp
  lea    (%r10,%rbp,4),%rcx
  nopl   0x0(%rax)
L70:
  mov    (%r11,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jg     L70
L45:
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  xor    %ebp,%ebp
  xor    %eax,%eax
  xor    %esi,%esi
  jmp    L54
```

assistant: ```c
void merge(int *A, int *L, int leftcount, int *R, int rightcount) {
  int i = 0, j = 0, k = 0;
  while (i < leftcount && j < rightcount) {
    if (L[i] < R[j])
      A[k++] = L[i++];
    else
      A[k++] = R[j++];
  }
  while (i < leftcount)
    A[k++] = L[i++];
  while (j < rightcount)
    A[k++] = R[j++];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<updateCPneg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L5:
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  %xmm2,%xmm1
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  %xmm2,%xmm1
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  %xmm2,%xmm1
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x18(%rbp),%eax
  imul   -0x14(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L5
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void updateCPneg(int ncp, int kcp, double *cp, double *gnDir, double *dirOffset) {
  int i;
  for (i = 0; i < kcp * ncp; i++) {
    cp[4 * i] -= gnDir[i] * dirOffset[3 * i];
    cp[4 * i + 1] -= gnDir[i] * dirOffset[3 * i + 1];
    cp[4 * i + 2] -= gnDir[i] * dirOffset[3 * i + 2];
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<updateCPneg>:
  endbr64
  imul   %edi,%esi
  test   %esi,%esi
  jle    L0
  mov    %rdx,%rax
  lea    -0x1(%rsi),%esi
  shl    $0x5,%rsi
  lea    0x20(%rdx,%rsi,1),%rdx
L9:
  movsd  (%rcx),%xmm1
  mulsd  (%r8),%xmm1
  movsd  (%rax),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movsd  (%rcx),%xmm1
  mulsd  0x8(%r8),%xmm1
  movsd  0x8(%rax),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%rax)
  movsd  (%rcx),%xmm1
  mulsd  0x10(%r8),%xmm1
  movsd  0x10(%rax),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%rax)
  add    $0x20,%rax
  add    $0x8,%rcx
  add    $0x18,%r8
  cmp    %rdx,%rax
  jne    L9
L0:
  ret
```

assistant: ```c
void updateCPneg(int ncp, int kcp, double *cp, double *gnDir, double *dirOffset) {
  int i;
  for (i = 0; i < kcp * ncp; i++) {
    cp[4 * i] -= gnDir[i] * dirOffset[3 * i];
    cp[4 * i + 1] -= gnDir[i] * dirOffset[3 * i + 1];
    cp[4 * i + 2] -= gnDir[i] * dirOffset[3 * i + 2];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<updateCPneg>:
  endbr64
  imul   %edi,%esi
  test   %esi,%esi
  jle    L11
  lea    -0x1(%rsi),%eax
  shl    $0x5,%rax
  lea    0x20(%rdx,%rax,1),%rax
  nopw   0x0(%rax,%rax,1)
L4:
  movsd  (%rcx),%xmm1
  movsd  (%rdx),%xmm0
  add    $0x20,%rdx
  add    $0x8,%rcx
  mulsd  (%r8),%xmm1
  add    $0x18,%r8
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rdx)
  movsd  -0x8(%rcx),%xmm1
  mulsd  -0x10(%r8),%xmm1
  movsd  -0x18(%rdx),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rdx)
  movsd  -0x8(%rcx),%xmm1
  mulsd  -0x8(%r8),%xmm1
  movsd  -0x10(%rdx),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rdx)
  cmp    %rdx,%rax
  jne    L4
L11:
  ret
```

assistant: ```c
void updateCPneg(int ncp, int kcp, double *cp, double *gnDir, double *dirOffset) {
  int i;
  for (i = 0; i < kcp * ncp; i++) {
    cp[4 * i] -= gnDir[i] * dirOffset[3 * i];
    cp[4 * i + 1] -= gnDir[i] * dirOffset[3 * i + 1];
    cp[4 * i + 2] -= gnDir[i] * dirOffset[3 * i + 2];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fetch_section_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L16
L3:
  addq   $0x1,-0x8(%rbp)
L16:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L3
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t fetch_section_count(struct conf_sec **s) {
  size_t ret;
  for (ret = 0; s[ret]; ret++)
    ;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fetch_section_count>:
  endbr64
  cmpq   $0x0,(%rdi)
  je     L0
  mov    $0x0,%eax
L9:
  add    $0x1,%rax
  cmpq   $0x0,(%rdi,%rax,8)
  jne    L9
  ret
L0:
  mov    $0x0,%eax
  ret
```

assistant: ```c
size_t fetch_section_count(struct conf_sec **s) {
  size_t ret;
  for (ret = 0; s[ret]; ret++)
    ;
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<fetch_section_count>:
  endbr64
  xor    %eax,%eax
  cmpq   $0x0,(%rdi)
  je     L6
  nopl   0x0(%rax)
L4:
  add    $0x1,%rax
  cmpq   $0x0,(%rdi,%rax,8)
  jne    L4
  ret
  nopl   0x0(%rax)
L6:
  ret
```

assistant: ```c
size_t fetch_section_count(struct conf_sec **s) {
  size_t ret;
  for (ret = 0; s[ret]; ret++)
    ;
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fetch_section_count>:
  endbr64
  xor    %eax,%eax
  cmpq   $0x0,(%rdi)
  je     L3
  nopl   0x0(%rax)
L8:
  add    $0x1,%rax
  cmpq   $0x0,(%rdi,%rax,8)
  jne    L8
  ret
  nopl   0x0(%rax)
L3:
  ret
```

assistant: ```c
size_t fetch_section_count(struct conf_sec **s) {
  size_t ret;
  for (ret = 0; s[ret]; ret++)
    ;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    $0x0,%eax
  call   <getProduct@plt>
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x10(%rbp),%eax
  mov    -0xc(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D10
type_infered: string
name: parse_data
arguments: data_label: D0
type_infered: string

tool: D10: maxproduct = %d

tool: D0: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  printf("maxproduct = %d", getProduct());
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    $0x0,%eax
  call   <getProduct@plt>
  mov    %eax,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    (%rsp),%esi
  sub    0x4(%rsp),%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
type: string
name: retrieve_data
arguments: data_tag: D6
type: string

tool: D18: maxproduct = %d

tool: D6: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  printf("maxproduct = %d", getProduct());
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <gettimeofday@plt>
  xor    %eax,%eax
  call   <getProduct@plt>
  lea    D19(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%esi
  lea    D0(%rip),%rdi
  sub    0x4(%rsp),%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D19
type_guessed: string
name: parse_data
arguments: data_tag: D0
type_guessed: string

tool: D19: maxproduct = %d

tool: D0: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  printf("maxproduct = %d", getProduct());
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <gettimeofday@plt>
  xor    %eax,%eax
  call   <getProduct@plt>
  lea    D11(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%esi
  lea    D18(%rip),%rdi
  sub    0x4(%rsp),%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D11
datatype: string
name: parse_data
arguments: data_tag: D18
datatype: string

tool: D11: maxproduct = %d

tool: D18: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  printf("maxproduct = %d", getProduct());
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<PB_Cnumroc>:
  endbr64
  mov    %esi,%eax
  mov    %ecx,%r10d
  mov    0x8(%rsp),%r11d
  cmp    $0xffffffff,%r9d
  je     L48
  cmp    $0x1,%r11d
  je     L48
  mov    %edx,%ecx
  sub    %esi,%ecx
  test   %ecx,%ecx
  jle    L65
L124:
  cmp    %r8d,%r9d
  je     L7
  cmp    %edi,%ecx
  jge    L156
  sub    %ecx,%edi
  mov    %edi,%eax
  cltd
  idiv   %r10d
  lea    0x1(%rax),%esi
  sub    %r9d,%r8d
  lea    (%r8,%r11,1),%edx
  cmovs  %edx,%r8d
  cmp    %esi,%r11d
  jle    L154
  mov    %r10d,%ecx
  cmp    %esi,%r8d
  jl     L106
  mov    $0x0,%ecx
  jne    L106
  imul   %r10d,%eax
  sub    %eax,%edi
  mov    %edi,%ecx
  jmp    L106
L65:
  sub    %edx,%eax
  cltd
  idiv   %r10d
  lea    0x1(%rax),%esi
  lea    (%r9,%rsi,1),%eax
  cltd
  idiv   %r11d
  mov    %edx,%r9d
  imul   %r10d,%esi
  add    %esi,%ecx
  jmp    L124
L7:
  cmp    %edi,%ecx
  jge    L66
  mov    %edi,%eax
  sub    %ecx,%eax
  cltd
  idiv   %r10d
  lea    0x1(%rax),%esi
  cmp    %esi,%r11d
  jg     L106
  mov    %esi,%eax
  cltd
  idiv   %r11d
  imul   %eax,%r11d
  cmp    %esi,%r11d
  je     L21
  imul   %r10d,%eax
  add    %eax,%ecx
  jmp    L106
L21:
  sub    %esi,%eax
  imul   %r10d,%eax
  lea    (%rax,%rdi,1),%ecx
  jmp    L106
L154:
  mov    %esi,%eax
  cltd
  idiv   %r11d
  sub    %edx,%r8d
  js     L140
  test   %r8d,%r8d
  jle    L150
  mov    %r10d,%ecx
  imul   %eax,%ecx
  jmp    L106
L140:
  lea    0x1(%rax),%ecx
  imul   %r10d,%ecx
  jmp    L106
L150:
  sub    %esi,%eax
  mov    %eax,%ecx
  add    $0x1,%ecx
  imul   %r10d,%ecx
  add    %edi,%ecx
  jmp    L106
L48:
  mov    %edi,%ecx
L106:
  mov    %ecx,%eax
  ret
L66:
  mov    %edi,%ecx
  jmp    L106
L156:
  mov    $0x0,%ecx
  jmp    L106
```

assistant: ```c
int PB_Cnumroc(int N, int I, int INB, int NB, int PROC, int SRCPROC, int NPROCS) {
  int ilocblk, mydist, nblocks;
  if ((SRCPROC == -1) || (NPROCS == 1))
    return (N);
  if ((INB -= I) <= 0) {
    nblocks = (-INB) / NB + 1;
    SRCPROC += nblocks;
    SRCPROC -= (SRCPROC / NPROCS) * NPROCS;
    INB += nblocks * NB;
  }
  if (PROC == SRCPROC) {
    if (N <= INB)
      return (N);
    nblocks = (N - INB) / NB + 1;
    if (nblocks < NPROCS)
      return (INB);
    ilocblk = nblocks / NPROCS;
    return ((nblocks - ilocblk * NPROCS) ? INB + ilocblk * NB : N + (ilocblk - nblocks) * NB);
  } else {
    if (N <= INB)
      return (0);
    nblocks = (N - INB) / NB + 1;
    if ((mydist = PROC - SRCPROC) < 0)
      mydist += NPROCS;
    if (nblocks < NPROCS)
      return ((mydist < nblocks) ? NB : ((mydist > nblocks) ? 0 : N - INB + NB * (1 - nblocks)));
    ilocblk = nblocks / NPROCS;
    mydist -= nblocks - ilocblk * NPROCS;
    return ((mydist < 0) ? (ilocblk + 1) * NB : ((mydist > 0) ? ilocblk * NB : N - INB + NB * (ilocblk - nblocks + 1)));
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<PB_Cnumroc>:
  endbr64
  mov    %esi,%eax
  mov    0x8(%rsp),%r11d
  mov    %ecx,%esi
  cmp    $0xffffffff,%r9d
  je     L41
  cmp    $0x1,%r11d
  je     L41
  mov    %edx,%r10d
  sub    %eax,%r10d
  test   %r10d,%r10d
  jle    L154
  cmp    %r8d,%r9d
  je     L25
L97:
  cmp    %edi,%r10d
  jge    L196
  sub    %r10d,%edi
  mov    %edi,%eax
  cltd
  idiv   %esi
  sub    %r9d,%r8d
  lea    (%r8,%r11,1),%edx
  cmovs  %edx,%r8d
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r11d
  jle    L72
  mov    %esi,%r10d
  cmp    %ecx,%r8d
  jl     L111
  mov    $0x0,%r10d
  jne    L111
  imul   %esi,%eax
  sub    %eax,%edi
  mov    %edi,%r10d
  jmp    L111
  nop
L154:
  sub    %edx,%eax
  cltd
  idiv   %ecx
  lea    0x1(%rax),%ecx
  lea    (%r9,%rcx,1),%eax
  cltd
  imul   %esi,%ecx
  idiv   %r11d
  add    %ecx,%r10d
  mov    %edx,%r9d
  cmp    %r8d,%r9d
  jne    L97
L25:
  cmp    %edi,%r10d
  jge    L41
  mov    %edi,%eax
  sub    %r10d,%eax
  cltd
  idiv   %esi
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r11d
  jg     L111
  mov    %ecx,%eax
  cltd
  idiv   %r11d
  imul   %eax,%r11d
  cmp    %ecx,%r11d
  je     L123
  imul   %esi,%eax
  add    %eax,%r10d
  jmp    L111
  nopw   0x0(%rax,%rax,1)
L41:
  mov    %edi,%r10d
L111:
  mov    %r10d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L72:
  mov    %ecx,%eax
  cltd
  idiv   %r11d
  cmp    %edx,%r8d
  js     L10
  je     L89
  imul   %eax,%esi
  mov    %esi,%r10d
  jmp    L111
  nopl   (%rax)
L196:
  xor    %r10d,%r10d
  mov    %r10d,%eax
  ret
  nop
L10:
  lea    0x1(%rax),%r10d
  imul   %esi,%r10d
  jmp    L111
  nopw   0x0(%rax,%rax,1)
L89:
  sub    %ecx,%eax
  mov    %eax,%r10d
  add    $0x1,%r10d
  imul   %esi,%r10d
  add    %edi,%r10d
  jmp    L111
  nopw   0x0(%rax,%rax,1)
L123:
  sub    %r11d,%eax
  imul   %esi,%eax
  lea    (%rax,%rdi,1),%r10d
  jmp    L111
```

assistant: ```c
int PB_Cnumroc(int N, int I, int INB, int NB, int PROC, int SRCPROC, int NPROCS) {
  int ilocblk, mydist, nblocks;
  if ((SRCPROC == -1) || (NPROCS == 1))
    return (N);
  if ((INB -= I) <= 0) {
    nblocks = (-INB) / NB + 1;
    SRCPROC += nblocks;
    SRCPROC -= (SRCPROC / NPROCS) * NPROCS;
    INB += nblocks * NB;
  }
  if (PROC == SRCPROC) {
    if (N <= INB)
      return (N);
    nblocks = (N - INB) / NB + 1;
    if (nblocks < NPROCS)
      return (INB);
    ilocblk = nblocks / NPROCS;
    return ((nblocks - ilocblk * NPROCS) ? INB + ilocblk * NB : N + (ilocblk - nblocks) * NB);
  } else {
    if (N <= INB)
      return (0);
    nblocks = (N - INB) / NB + 1;
    if ((mydist = PROC - SRCPROC) < 0)
      mydist += NPROCS;
    if (nblocks < NPROCS)
      return ((mydist < nblocks) ? NB : ((mydist > nblocks) ? 0 : N - INB + NB * (1 - nblocks)));
    ilocblk = nblocks / NPROCS;
    mydist -= nblocks - ilocblk * NPROCS;
    return ((mydist < 0) ? (ilocblk + 1) * NB : ((mydist > 0) ? ilocblk * NB : N - INB + NB * (ilocblk - nblocks + 1)));
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<PB_Cnumroc>:
  endbr64
  mov    %esi,%eax
  mov    0x8(%rsp),%r11d
  mov    %ecx,%esi
  cmp    $0xffffffff,%r9d
  je     L134
  cmp    $0x1,%r11d
  je     L134
  mov    %edx,%r10d
  sub    %eax,%r10d
  test   %r10d,%r10d
  jle    L124
  cmp    %r8d,%r9d
  je     L72
L155:
  cmp    %edi,%r10d
  jge    L165
  sub    %r10d,%edi
  mov    %edi,%eax
  cltd
  idiv   %esi
  sub    %r9d,%r8d
  lea    (%r8,%r11,1),%edx
  cmovs  %edx,%r8d
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r11d
  jle    L139
  mov    %esi,%r10d
  cmp    %ecx,%r8d
  jl     L144
  mov    $0x0,%r10d
  jne    L144
  imul   %esi,%eax
  sub    %eax,%edi
  mov    %edi,%r10d
  jmp    L144
  nop
L124:
  sub    %edx,%eax
  cltd
  idiv   %ecx
  lea    0x1(%rax),%ecx
  lea    (%r9,%rcx,1),%eax
  cltd
  imul   %esi,%ecx
  idiv   %r11d
  add    %ecx,%r10d
  mov    %edx,%r9d
  cmp    %r8d,%r9d
  jne    L155
L72:
  cmp    %edi,%r10d
  jge    L134
  mov    %edi,%eax
  sub    %r10d,%eax
  cltd
  idiv   %esi
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r11d
  jg     L144
  mov    %ecx,%eax
  cltd
  idiv   %r11d
  imul   %eax,%r11d
  cmp    %ecx,%r11d
  je     L126
  imul   %esi,%eax
  add    %eax,%r10d
  jmp    L144
  nopw   0x0(%rax,%rax,1)
L134:
  mov    %edi,%r10d
L144:
  mov    %r10d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L139:
  mov    %ecx,%eax
  cltd
  idiv   %r11d
  cmp    %edx,%r8d
  js     L136
  je     L79
  imul   %eax,%esi
  mov    %esi,%r10d
  jmp    L144
  nopl   (%rax)
L165:
  xor    %r10d,%r10d
  mov    %r10d,%eax
  ret
  nop
L136:
  lea    0x1(%rax),%r10d
  imul   %esi,%r10d
  jmp    L144
  nopw   0x0(%rax,%rax,1)
L79:
  sub    %ecx,%eax
  mov    %eax,%r10d
  add    $0x1,%r10d
  imul   %esi,%r10d
  add    %edi,%r10d
  jmp    L144
  nopw   0x0(%rax,%rax,1)
L126:
  sub    %r11d,%eax
  imul   %esi,%eax
  lea    (%rax,%rdi,1),%r10d
  jmp    L144
```

assistant: ```c
int PB_Cnumroc(int N, int I, int INB, int NB, int PROC, int SRCPROC, int NPROCS) {
  int ilocblk, mydist, nblocks;
  if ((SRCPROC == -1) || (NPROCS == 1))
    return (N);
  if ((INB -= I) <= 0) {
    nblocks = (-INB) / NB + 1;
    SRCPROC += nblocks;
    SRCPROC -= (SRCPROC / NPROCS) * NPROCS;
    INB += nblocks * NB;
  }
  if (PROC == SRCPROC) {
    if (N <= INB)
      return (N);
    nblocks = (N - INB) / NB + 1;
    if (nblocks < NPROCS)
      return (INB);
    ilocblk = nblocks / NPROCS;
    return ((nblocks - ilocblk * NPROCS) ? INB + ilocblk * NB : N + (ilocblk - nblocks) * NB);
  } else {
    if (N <= INB)
      return (0);
    nblocks = (N - INB) / NB + 1;
    if ((mydist = PROC - SRCPROC) < 0)
      mydist += NPROCS;
    if (nblocks < NPROCS)
      return ((mydist < nblocks) ? NB : ((mydist > nblocks) ? 0 : N - INB + NB * (1 - nblocks)));
    ilocblk = nblocks / NPROCS;
    mydist -= nblocks - ilocblk * NPROCS;
    return ((mydist < 0) ? (ilocblk + 1) * NB : ((mydist > 0) ? ilocblk * NB : N - INB + NB * (ilocblk - nblocks + 1)));
  }
}
```