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

```asm
<dcopy_codegen_9>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x28,-0x8(%rbp)
  movl   $0x28,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L38
L8:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L24
L0:
  mov    -0x4(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x10(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x10(%rbp)
L24:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L0
  addl   $0x1,-0xc(%rbp)
L38:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L8
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dcopy_codegen_9(double *A, double *C) {
  const int n = 4 + 0 * 40 + 9 * 4;
  const int lda = 4 + 0 * 40 + 9 * 4;
  int ii, jj;
  for (jj = 0; jj < n; jj++) {
    for (ii = jj; ii < n; ii++) {
      C[ii + lda * jj] = A[ii + lda * jj];
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<dcopy_codegen_9>:
  endbr64
  mov    $0x0,%edx
  jmp    L39
L47:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    $0x27,%eax
  jle    L47
L5:
  add    $0x1,%rdx
  add    $0x140,%rdi
  add    $0x140,%rsi
  cmp    $0x28,%rdx
  je     L48
L39:
  mov    %rdx,%rax
  cmp    $0x27,%edx
  jle    L47
  jmp    L5
L48:
  ret
```

assistant: ```c
void dcopy_codegen_9(double *A, double *C) {
  const int n = 4 + 0 * 40 + 9 * 4;
  const int lda = 4 + 0 * 40 + 9 * 4;
  int ii, jj;
  for (jj = 0; jj < n; jj++) {
    for (ii = jj; ii < n; ii++) {
      C[ii + lda * jj] = A[ii + lda * jj];
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<dcopy_codegen_9>:
  endbr64
  xor    %edx,%edx
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    %rdx,%rax
  nopl   0x0(%rax,%rax,1)
L2:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    $0x28,%eax
  jne    L2
  add    $0x1,%rdx
  add    $0x140,%rdi
  add    $0x140,%rsi
  cmp    $0x28,%rdx
  jne    L16
  ret
```

assistant: ```c
void dcopy_codegen_9(double *A, double *C) {
  const int n = 4 + 0 * 40 + 9 * 4;
  const int lda = 4 + 0 * 40 + 9 * 4;
  int ii, jj;
  for (jj = 0; jj < n; jj++) {
    for (ii = jj; ii < n; ii++) {
      C[ii + lda * jj] = A[ii + lda * jj];
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dcopy_codegen_9>:
  endbr64
  push   %r14
  mov    %rdi,%r9
  mov    %rsi,%r8
  lea    0x8(%rdi),%r11
  push   %r13
  mov    %rsi,%rcx
  xor    %r10d,%r10d
  push   %r12
  mov    $0x27,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x28,%edi
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  nop
L58:
  mov    %r12d,%edx
  mov    %r10d,%eax
  sub    %r10d,%edx
  cmp    $0x2,%edx
  jbe    L5
  cmp    %r11,%rcx
  je     L5
  mov    %edi,%r13d
  lea    -0x8(%r11),%r14
  xor    %edx,%edx
  shr    %r13d
  shl    $0x4,%r13
  nopl   0x0(%rax,%rax,1)
L38:
  movupd (%r14,%rdx,1),%xmm1
  movups %xmm1,(%rcx,%rdx,1)
  add    $0x10,%rdx
  cmp    %r13,%rdx
  jne    L38
  mov    %edi,%edx
  and    $0xfffffffe,%edx
  add    %edx,%eax
  cmp    %edx,%edi
  je     L7
  add    %esi,%eax
  cltq
  movsd  (%r9,%rax,8),%xmm0
  movsd  %xmm0,(%r8,%rax,8)
L7:
  add    $0x1,%r10
  add    $0x148,%r11
  add    $0x28,%esi
  add    $0x148,%rcx
  add    $0x140,%rbp
  add    $0x140,%rbx
  sub    $0x1,%edi
  jne    L58
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %r10,%rax
  nopl   0x0(%rax,%rax,1)
L11:
  movsd  0x0(%rbp,%rax,8),%xmm0
  movsd  %xmm0,(%rbx,%rax,8)
  add    $0x1,%rax
  cmp    $0x28,%eax
  jne    L11
  jmp    L7
```

assistant: ```c
void dcopy_codegen_9(double *A, double *C) {
  const int n = 4 + 0 * 40 + 9 * 4;
  const int lda = 4 + 0 * 40 + 9 * 4;
  int ii, jj;
  for (jj = 0; jj < n; jj++) {
    for (ii = jj; ii < n; ii++) {
      C[ii + lda * jj] = A[ii + lda * jj];
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rsa_multip_cap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x5,-0x4(%rbp)
  cmpl   $0x3ff,-0x14(%rbp)
  jg     L55
  movl   $0x2,-0x4(%rbp)
  jmp    L57
L55:
  cmpl   $0xfff,-0x14(%rbp)
  jg     L58
  movl   $0x3,-0x4(%rbp)
  jmp    L57
L58:
  cmpl   $0x1fff,-0x14(%rbp)
  jg     L57
  movl   $0x4,-0x4(%rbp)
L57:
  cmpl   $0x5,-0x4(%rbp)
  jle    L31
  movl   $0x5,-0x4(%rbp)
L31:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int rsa_multip_cap(int bits) {
  int cap = 5;
  if (bits < 1024)
    cap = 2;
  else if (bits < 4096)
    cap = 3;
  else if (bits < 8192)
    cap = 4;
  if (cap > 5)
    cap = 5;
  return cap;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rsa_multip_cap>:
  endbr64
  mov    $0x2,%eax
  cmp    $0x3ff,%edi
  jle    L16
  mov    $0x3,%eax
  cmp    $0xfff,%edi
  jle    L16
  cmp    $0x1fff,%edi
  setg   %al
  movzbl %al,%eax
  add    $0x4,%eax
L16:
  ret
```

assistant: ```c
int rsa_multip_cap(int bits) {
  int cap = 5;
  if (bits < 1024)
    cap = 2;
  else if (bits < 4096)
    cap = 3;
  else if (bits < 8192)
    cap = 4;
  if (cap > 5)
    cap = 5;
  return cap;
}
```
user: What is the c source code of the assembly code below:

```asm
<rsa_multip_cap>:
  endbr64
  mov    $0x2,%eax
  cmp    $0x3ff,%edi
  jle    L13
  mov    $0x3,%eax
  cmp    $0xfff,%edi
  jg     L7
L13:
  ret
  nop
L7:
  xor    %eax,%eax
  cmp    $0x1fff,%edi
  setg   %al
  add    $0x4,%eax
  ret
```

assistant: ```c
int rsa_multip_cap(int bits) {
  int cap = 5;
  if (bits < 1024)
    cap = 2;
  else if (bits < 4096)
    cap = 3;
  else if (bits < 8192)
    cap = 4;
  if (cap > 5)
    cap = 5;
  return cap;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rsa_multip_cap>:
  endbr64
  mov    $0x2,%eax
  cmp    $0x3ff,%edi
  jle    L11
  mov    $0x3,%eax
  cmp    $0xfff,%edi
  jg     L1
L11:
  ret
  nop
L1:
  xor    %eax,%eax
  cmp    $0x1fff,%edi
  setg   %al
  add    $0x4,%eax
  ret
```

assistant: ```c
int rsa_multip_cap(int bits) {
  int cap = 5;
  if (bits < 1024)
    cap = 2;
  else if (bits < 4096)
    cap = 3;
  else if (bits < 8192)
    cap = 4;
  if (cap > 5)
    cap = 5;
  return cap;
}
```
user: What is the c source code of the assembly code below?

```asm
<check_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L33
  mov    $0x0,%eax
  jmp    L6
L33:
  movl   $0x2,-0x4(%rbp)
  jmp    L38
L36:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L45
  mov    $0x0,%eax
  jmp    L6
L45:
  addl   $0x1,-0x4(%rbp)
L38:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L36
  mov    $0x1,%eax
L6:
  pop    %rbp
  ret
```

assistant: ```c
int check_prime(int num) {
  int i = 1;
  if (num == 1)
    return (0);
  for (i = 2; i < num; i = i + 1) {
    if (num % i == 0)
      return (0);
  }
  return (1);
}
```
user: What is the c source code of the assembly code below?

```asm
<check_prime>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x1,%edi
  je     L38
  cmp    $0x2,%edi
  jle    L17
  test   $0x1,%dil
  je     L36
  mov    $0x2,%ecx
L30:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  je     L34
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %edx,%edx
  jne    L30
L38:
  ret
L34:
  mov    $0x1,%eax
  ret
L17:
  mov    $0x1,%eax
  ret
L36:
  mov    $0x0,%eax
  jmp    L38
```

assistant: ```c
int check_prime(int num) {
  int i = 1;
  if (num == 1)
    return (0);
  for (i = 2; i < num; i = i + 1) {
    if (num % i == 0)
      return (0);
  }
  return (1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<check_prime>:
  endbr64
  xor    %edx,%edx
  cmp    $0x1,%edi
  je     L41
  cmp    $0x2,%edi
  jle    L34
  test   $0x1,%dil
  je     L41
  mov    $0x2,%ecx
  jmp    L5
  nopl   (%rax)
L27:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L41
L5:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L27
L34:
  mov    $0x1,%edx
L41:
  mov    %edx,%eax
  ret
```

assistant: ```c
int check_prime(int num) {
  int i = 1;
  if (num == 1)
    return (0);
  for (i = 2; i < num; i = i + 1) {
    if (num % i == 0)
      return (0);
  }
  return (1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<check_prime>:
  endbr64
  xor    %edx,%edx
  cmp    $0x1,%edi
  je     L18
  cmp    $0x2,%edi
  jle    L3
  test   $0x1,%dil
  je     L18
  mov    $0x2,%ecx
  jmp    L43
  nopl   (%rax)
L55:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L18
L43:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L55
L3:
  mov    $0x1,%edx
L18:
  mov    %edx,%eax
  ret
```

assistant: ```c
int check_prime(int num) {
  int i = 1;
  if (num == 1)
    return (0);
  for (i = 2; i < num; i = i + 1) {
    if (num % i == 0)
      return (0);
  }
  return (1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L25
  addl   $0x1,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L9
  mov    $0x0,%eax
  call   <abort@plt>
L9:
  addq   $0x1,-0x18(%rbp)
  jmp    L6
L25:
  nop
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int __attribute__((no, noclone)) foo(char *p) {
  int h = 0;
  do {
    if (*p == '\0')
      break;
    ++h;
    if (p == 0)
      abort();
    ++p;
  } while (1);
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L11
  mov    %rdi,%rax
  mov    $0x1,%edx
  sub    %edi,%edx
L25:
  lea    (%rdx,%rax,1),%ecx
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L25
L26:
  mov    %ecx,%eax
  ret
L11:
  mov    $0x0,%ecx
  jmp    L26
```

assistant: ```c
int __attribute__((no, noclone)) foo(char *p) {
  int h = 0;
  do {
    if (*p == '\0')
      break;
    ++h;
    if (p == 0)
      abort();
    ++p;
  } while (1);
  return h;
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L15
  mov    $0x1,%eax
  sub    %edi,%eax
L3:
  lea    (%rax,%rdi,1),%r8d
  add    $0x1,%rdi
  cmpb   $0x0,(%rdi)
  jne    L3
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L15:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int __attribute__((no, noclone)) foo(char *p) {
  int h = 0;
  do {
    if (*p == '\0')
      break;
    ++h;
    if (p == 0)
      abort();
    ++p;
  } while (1);
  return h;
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L12
  mov    $0x1,%eax
  sub    %edi,%eax
L5:
  lea    (%rax,%rdi,1),%r8d
  add    $0x1,%rdi
  cmpb   $0x0,(%rdi)
  jne    L5
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L12:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int __attribute__((no, noclone)) foo(char *p) {
  int h = 0;
  do {
    if (*p == '\0')
      break;
    ++h;
    if (p == 0)
      abort();
    ++p;
  } while (1);
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x20(%rbp)
  movl   $0x6,-0x1c(%rbp)
  movl   $0x7,-0x18(%rbp)
  movl   $0x8,-0x14(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

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

tool: D4: %u
%u
%i

assistant: ```c
int main() {
  int nums[] = {5, 6, 7, 8};
  int *pointer = &nums[0];
  printf("%u\n%u\n%i", pointer, nums[0], *pointer);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x5,(%rsp)
  movl   $0x6,0x4(%rsp)
  movl   $0x7,0x8(%rsp)
  movl   $0x8,0xc(%rsp)
  mov    %rsp,%rsi
  mov    $0x5,%ecx
  mov    $0x5,%edx
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

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

tool: D1: %u
%u
%i

assistant: ```c
int main() {
  int nums[] = {5, 6, 7, 8};
  int *pointer = &nums[0];
  printf("%u\n%u\n%i", pointer, nums[0], *pointer);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x5,%ecx
  mov    $0x5,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x600000005,%rax
  mov    %rsp,%rsi
  lea    D6(%rip),%rdi
  mov    %rax,(%rsp)
  movabs $0x800000007,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D6
datatype: string

tool: D6: %u
%u
%i

assistant: ```c
int main() {
  int nums[] = {5, 6, 7, 8};
  int *pointer = &nums[0];
  printf("%u\n%u\n%i", pointer, nums[0], *pointer);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  movdqa D3(%rip),%xmm0
  mov    $0x5,%ecx
  mov    $0x5,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D0(%rip),%rdi
  movaps %xmm0,(%rsp)
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D3
type_infered: byte[16]
name: read_data
arguments: data_label: D0
type_infered: string

tool: {"D3": [0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: D0: %u
%u
%i

assistant: ```c
int main() {
  int nums[] = {5, 6, 7, 8};
  int *pointer = &nums[0];
  printf("%u\n%u\n%i", pointer, nums[0], *pointer);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ej9>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movsd  %xmm0,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L12
L27:
  movl   $0x0,-0x4(%rbp)
  jmp    L34
L8:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mulsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L34:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L8
  addl   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L27
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ej9(double **matriz, int M, int N, double s) {
  int i, j;
  for (i = 0; i < M; i++) {
    for (j = 0; j < N; j++) {
      matriz[i][j] = s * matriz[i][j];
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ej9>:
  endbr64
  test   %esi,%esi
  jle    L56
  mov    %rdi,%r8
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rdi
  lea    -0x1(%rdx),%eax
  lea    0x8(,%rax,8),%rsi
  jmp    L50
L12:
  mov    %rax,%rcx
  add    (%r8),%rcx
  movapd %xmm0,%xmm1
  mulsd  (%rcx),%xmm1
  movsd  %xmm1,(%rcx)
  add    $0x8,%rax
  cmp    %rsi,%rax
  jne    L12
L5:
  add    $0x8,%r8
  cmp    %rdi,%r8
  je     L56
L50:
  mov    $0x0,%eax
  test   %edx,%edx
  jg     L12
  jmp    L5
L56:
  ret
```

assistant: ```c
void ej9(double **matriz, int M, int N, double s) {
  int i, j;
  for (i = 0; i < M; i++) {
    for (j = 0; j < N; j++) {
      matriz[i][j] = s * matriz[i][j];
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ej9>:
  endbr64
  test   %esi,%esi
  jle    L10
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rsi
  lea    -0x1(%rdx),%eax
  lea    0x8(,%rax,8),%r8
  nopl   0x0(%rax,%rax,1)
L21:
  test   %edx,%edx
  jle    L24
  mov    (%rdi),%rax
  lea    (%r8,%rax,1),%rcx
  nopl   0x0(%rax,%rax,1)
L9:
  movsd  (%rax),%xmm1
  add    $0x8,%rax
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,-0x8(%rax)
  cmp    %rcx,%rax
  jne    L9
L24:
  add    $0x8,%rdi
  cmp    %rsi,%rdi
  jne    L21
L10:
  ret
```

assistant: ```c
void ej9(double **matriz, int M, int N, double s) {
  int i, j;
  for (i = 0; i < M; i++) {
    for (j = 0; j < N; j++) {
      matriz[i][j] = s * matriz[i][j];
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ej9>:
  endbr64
  mov    %edx,%r8d
  test   %esi,%esi
  jle    L25
  test   %edx,%edx
  jle    L25
  lea    -0x1(%rsi),%eax
  movapd %xmm0,%xmm2
  mov    %edx,%esi
  lea    0x8(%rdi,%rax,8),%r10
  mov    %edx,%eax
  unpcklpd %xmm2,%xmm2
  and    $0xfffffffe,%esi
  shr    %eax
  lea    -0x1(%rax),%r9d
  add    $0x1,%r9
  shl    $0x4,%r9
  nopl   0x0(%rax)
L55:
  mov    (%rdi),%rdx
  mov    %rdx,%rax
  lea    (%r9,%rdx,1),%rcx
  cmp    $0x1,%r8d
  je     L41
  nopl   0x0(%rax,%rax,1)
L53:
  movupd (%rax),%xmm1
  add    $0x10,%rax
  mulpd  %xmm2,%xmm1
  movups %xmm1,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L53
  movslq %esi,%rax
  cmp    %esi,%r8d
  je     L52
L18:
  lea    (%rdx,%rax,8),%rax
  movsd  (%rax),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,(%rax)
L52:
  add    $0x8,%rdi
  cmp    %r10,%rdi
  jne    L55
L25:
  ret
L41:
  xor    %eax,%eax
  jmp    L18
```

assistant: ```c
void ej9(double **matriz, int M, int N, double s) {
  int i, j;
  for (i = 0; i < M; i++) {
    for (j = 0; j < N; j++) {
      matriz[i][j] = s * matriz[i][j];
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<neio_bwrite>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  jmp    L15
L10:
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <neio_write@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L37
  mov    $0xffffffff,%eax
  jmp    L30
L37:
  cmpl   $0x0,-0x4(%rbp)
  jle    L15
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,-0x20(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x18(%rbp)
L15:
  cmpl   $0x0,-0x18(%rbp)
  jne    L10
  mov    $0x0,%eax
L30:
  leave
  ret
```

assistant: ```c
int neio_bwrite(int fd, const char *buf, unsigned int size) {
  while (size) {
    int wd = neio_write(fd, buf, size);
    do {
      if ((wd))
        return -1;
    } while (0);
    if (wd > 0) {
      buf += wd;
      size -= wd;
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<neio_bwrite>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %edx,%ebx
L8:
  test   %ebx,%ebx
  je     L29
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <neio_write@plt>
  test   %eax,%eax
  je     L8
  mov    $0xffffffff,%eax
  jmp    L27
L29:
  mov    $0x0,%eax
L27:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int neio_bwrite(int fd, const char *buf, unsigned int size) {
  while (size) {
    int wd = neio_write(fd, buf, size);
    do {
      if ((wd))
        return -1;
    } while (0);
    if (wd > 0) {
      buf += wd;
      size -= wd;
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<neio_bwrite>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <neio_write@plt>
  test   %eax,%eax
  jne    L21
L17:
  test   %ebx,%ebx
  jne    L4
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L21:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int neio_bwrite(int fd, const char *buf, unsigned int size) {
  while (size) {
    int wd = neio_write(fd, buf, size);
    do {
      if ((wd))
        return -1;
    } while (0);
    if (wd > 0) {
      buf += wd;
      size -= wd;
    }
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<neio_bwrite>:
  endbr64
  test   %edx,%edx
  je     L16
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  nopl   0x0(%rax)
L8:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <neio_write@plt>
  test   %eax,%eax
  je     L8
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  ret
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
int neio_bwrite(int fd, const char *buf, unsigned int size) {
  while (size) {
    int wd = neio_write(fd, buf, size);
    do {
      if ((wd))
        return -1;
    } while (0);
    if (wd > 0) {
      buf += wd;
      size -= wd;
    }
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Calc_CheckSum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movw   $0x0,-0x2(%rbp)
  jmp    L6
L7:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %ax,-0x2(%rbp)
L6:
  mov    -0x20(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  test   %rax,%rax
  jne    L7
  movzwl -0x2(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned short Calc_CheckSum(unsigned char *mem, unsigned long size) {
  unsigned short u = 0;
  while (size--)
    u += *mem++;
  return (unsigned short)(u & 0xffff);
}
```
user: What is the c source code of the assembly code below?

```asm
<Calc_CheckSum>:
  endbr64
  test   %rsi,%rsi
  je     L19
  add    %rdi,%rsi
  mov    $0x0,%eax
L3:
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%edx
  add    %edx,%eax
  cmp    %rdi,%rsi
  jne    L3
  ret
L19:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned short Calc_CheckSum(unsigned char *mem, unsigned long size) {
  unsigned short u = 0;
  while (size--)
    u += *mem++;
  return (unsigned short)(u & 0xffff);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Calc_CheckSum>:
  endbr64
  test   %rsi,%rsi
  je     L16
  add    %rdi,%rsi
  xor    %eax,%eax
  xchg   %ax,%ax
L2:
  movzbl (%rdi),%edx
  add    $0x1,%rdi
  add    %edx,%eax
  cmp    %rdi,%rsi
  jne    L2
  ret
  nop
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned short Calc_CheckSum(unsigned char *mem, unsigned long size) {
  unsigned short u = 0;
  while (size--)
    u += *mem++;
  return (unsigned short)(u & 0xffff);
}
```