ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<write_double_array_to_socket>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  je     L0
  cmpl   $0x0,-0x4(%rbp)
  jle    L10
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x8(%rbp)
  jmp    L10
L18:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  je     L12
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x8(%rbp)
L10:
  cmpl   $0x0,-0x8(%rbp)
  jg     L18
  jmp    L0
L12:
  nop
L0:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int write_double_array_to_socket(int sockfd, double *elements, int size) {
  int total = 0;
  int bytesremaining = size;
  int n;
  n = write(sockfd, elements, size);
  if (n != size) {
    if (n > 0) {
      total += n;
      bytesremaining -= n;
    }
    while (bytesremaining > 0) {
      n = write(sockfd, elements + total, bytesremaining);
      if (n == -1) {
        break;
      }
      total += n;
      bytesremaining -= n;
    }
  }
  return n;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<write_double_array_to_socket>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %edx,%ebx
  call   <write@plt>
  cmp    %eax,%ebx
  je     L42
  mov    $0x0,%r13d
  test   %eax,%eax
  jle    L38
  sub    %eax,%ebx
  mov    %eax,%r13d
L38:
  test   %ebx,%ebx
  jle    L42
L3:
  movslq %r13d,%rax
  lea    (%r12,%rax,8),%rsi
  mov    %ebx,%edx
  mov    %ebp,%edi
  call   <write@plt>
  cmp    $0xffffffff,%eax
  je     L42
  add    %eax,%r13d
  sub    %eax,%ebx
  test   %ebx,%ebx
  jg     L3
L42:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int write_double_array_to_socket(int sockfd, double *elements, int size) {
  int total = 0;
  int bytesremaining = size;
  int n;
  n = write(sockfd, elements, size);
  if (n != size) {
    if (n > 0) {
      total += n;
      bytesremaining -= n;
    }
    while (bytesremaining > 0) {
      n = write(sockfd, elements + total, bytesremaining);
      if (n == -1) {
        break;
      }
      total += n;
      bytesremaining -= n;
    }
  }
  return n;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<write_double_array_to_socket>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <write@plt>
  cmp    %eax,%ebx
  je     L26
  xor    %r13d,%r13d
  test   %eax,%eax
  jle    L34
  sub    %eax,%ebx
  mov    %eax,%r13d
  test   %ebx,%ebx
  jle    L26
  xchg   %ax,%ax
L10:
  movslq %r13d,%rax
  mov    %ebx,%edx
  mov    %ebp,%edi
  lea    (%r12,%rax,8),%rsi
  call   <write@plt>
  cmp    $0xffffffff,%eax
  je     L26
  add    %eax,%r13d
  sub    %eax,%ebx
L34:
  test   %ebx,%ebx
  jg     L10
L26:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int write_double_array_to_socket(int sockfd, double *elements, int size) {
  int total = 0;
  int bytesremaining = size;
  int n;
  n = write(sockfd, elements, size);
  if (n != size) {
    if (n > 0) {
      total += n;
      bytesremaining -= n;
    }
    while (bytesremaining > 0) {
      n = write(sockfd, elements + total, bytesremaining);
      if (n == -1) {
        break;
      }
      total += n;
      bytesremaining -= n;
    }
  }
  return n;
}
```
user: Translate the below assembly code into C programming language:

```asm
<write_double_array_to_socket>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <write@plt>
  cmp    %eax,%ebx
  je     L8
  xor    %r13d,%r13d
  test   %eax,%eax
  jle    L18
  sub    %eax,%ebx
  mov    %eax,%r13d
  test   %ebx,%ebx
  jle    L8
  xchg   %ax,%ax
L22:
  movslq %r13d,%rax
  mov    %ebx,%edx
  mov    %ebp,%edi
  lea    (%r12,%rax,8),%rsi
  call   <write@plt>
  cmp    $0xffffffff,%eax
  je     L8
  add    %eax,%r13d
  sub    %eax,%ebx
L18:
  test   %ebx,%ebx
  jg     L22
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int write_double_array_to_socket(int sockfd, double *elements, int size) {
  int total = 0;
  int bytesremaining = size;
  int n;
  n = write(sockfd, elements, size);
  if (n != size) {
    if (n > 0) {
      total += n;
      bytesremaining -= n;
    }
    while (bytesremaining > 0) {
      n = write(sockfd, elements + total, bytesremaining);
      if (n == -1) {
        break;
      }
      total += n;
      bytesremaining -= n;
    }
  }
  return n;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    $0x1,%eax
  pop    %rbp
  ret
```

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

tool: D8: char

assistant: ```c
int main(void) {
  char a;
  char *b = "char";
  a == b[0];
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int main(void) {
  char a;
  char *b = "char";
  a == b[0];
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int main(void) {
  char a;
  char *b = "char";
  a == b[0];
  return 1;
}
```
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    $0xa0,%rsp
  mov    %edi,-0x94(%rbp)
  mov    %rsi,-0xa0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  fldt   0xe66(%rip)        # 2010 <_fini+0xd70>
  fstpt  -0x50(%rbp)
  fldt   0xe6d(%rip)        # 2020 <_fini+0xd80>
  fstpt  -0x40(%rbp)
  fldt   0xe74(%rip)        # 2030 <_fini+0xd90>
  fstpt  -0x30(%rbp)
  fldt   0xe7b(%rip)        # 2040 <_fini+0xda0>
  fstpt  -0x20(%rbp)
  movl   $0x0,-0x84(%rbp)
  jmp    L0
L23:
  fildl  -0x84(%rbp)
  mov    -0x84(%rbp),%eax
  cltq
  shl    $0x4,%rax
  add    %rbp,%rax
  sub    $0x70,%rax
  fstpt  (%rax)
  addl   $0x1,-0x84(%rbp)
L0:
  cmpl   $0x1,-0x84(%rbp)
  jle    L23
  fldt   -0x20(%rbp)
  fldt   -0x70(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x80(%rbp)
  mov    -0xa0(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  push   -0x78(%rbp)
  push   -0x80(%rbp)
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <cov_log@plt>
  add    $0x10,%rsp
  mov    -0xa0(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rdx
  mov    -0xa0(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rcx
  mov    -0xa0(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <cov_check@plt>
  push   -0x78(%rbp)
  push   -0x80(%rbp)
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  add    $0x10,%rsp
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
data_type: string
name: retrieve_data
arguments: tag: D4
data_type: string

tool: D3: result

tool: D4: %1.15Le


assistant: ```c
int main(int argc, char *argv[]) {
  long double x[4] = {2.1, 2.2, 2.3, 2.4};
  long double y[2];
  int i;
  for (i = 0; i < 2; i++) {
    y[i] = i;
  }
  long double result = x[3] + y[0];
  cov_log(result, "result", argv[2]);
  cov_check(argv[2], argv[1], argv[3]);
  printf("%1.15Le\n", result);
  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
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    0x10(%rsi),%rsi
  fldl   0xea1(%rip)        # 2010 <_fini+0xe48>
  fstpt  (%rsp)
  mov    (%rsp),%rax
  mov    0x8(%rsp),%edx
  push   %rdx
  push   %rax
  lea    D12(%rip),%rdi
  call   <cov_log@plt>
  mov    0x18(%rbx),%rdx
  mov    0x8(%rbx),%rsi
  mov    0x10(%rbx),%rdi
  call   <cov_check@plt>
  fldl   0xe71(%rip)        # 2010 <_fini+0xe48>
  fstpt  0x10(%rsp)
  mov    0x10(%rsp),%rax
  mov    0x18(%rsp),%edx
  push   %rdx
  push   %rax
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D12
datatype: string
name: parse_data
arguments: label: D15
datatype: string

tool: D12: result

tool: D15: %1.15Le


assistant: ```c
int main(int argc, char *argv[]) {
  long double x[4] = {2.1, 2.2, 2.3, 2.4};
  long double y[2];
  int i;
  for (i = 0; i < 2; i++) {
    y[i] = i;
  }
  long double result = x[3] + y[0];
  cov_log(result, "result", argv[2]);
  cov_check(argv[2], argv[1], argv[3]);
  printf("%1.15Le\n", result);
  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
  push   %rbx
  fldl   0xf65(%rip)        # 2010 <_fini+0xe54>
  mov    %rsi,%rbx
  lea    D3(%rip),%rdi
  mov    0x10(%rsi),%rsi
  sub    $0x10,%rsp
  fstpt  (%rsp)
  call   <cov_log@plt>
  mov    0x18(%rbx),%rdx
  mov    0x8(%rbx),%rsi
  mov    0x10(%rbx),%rdi
  call   <cov_check@plt>
  fldt   0xf44(%rip)        # 2020 <_fini+0xe64>
  sub    $0x10,%rsp
  lea    D10(%rip),%rdi
  fstpt  (%rsp)
  call   <printf@plt>
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
data_type: string
name: retrieve_data
arguments: tag: D10
data_type: string

tool: D3: result

tool: D10: %1.15Le


assistant: ```c
int main(int argc, char *argv[]) {
  long double x[4] = {2.1, 2.2, 2.3, 2.4};
  long double y[2];
  int i;
  for (i = 0; i < 2; i++) {
    y[i] = i;
  }
  long double result = x[3] + y[0];
  cov_log(result, "result", argv[2]);
  cov_check(argv[2], argv[1], argv[3]);
  printf("%1.15Le\n", result);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  fldl   0xf65(%rip)        # 2010 <_fini+0xe54>
  mov    %rsi,%rbx
  lea    D11(%rip),%rdi
  mov    0x10(%rsi),%rsi
  sub    $0x10,%rsp
  fstpt  (%rsp)
  call   <cov_log@plt>
  mov    0x18(%rbx),%rdx
  mov    0x8(%rbx),%rsi
  mov    0x10(%rbx),%rdi
  call   <cov_check@plt>
  fldt   0xf44(%rip)        # 2020 <_fini+0xe64>
  sub    $0x10,%rsp
  lea    D19(%rip),%rdi
  fstpt  (%rsp)
  call   <printf@plt>
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
data_type: string
name: retrieve_data
arguments: data_tag: D19
data_type: string

tool: D11: result

tool: D19: %1.15Le


assistant: ```c
int main(int argc, char *argv[]) {
  long double x[4] = {2.1, 2.2, 2.3, 2.4};
  long double y[2];
  int i;
  for (i = 0; i < 2; i++) {
    y[i] = i;
  }
  long double result = x[3] + y[0];
  cov_log(result, "result", argv[2]);
  cov_check(argv[2], argv[1], argv[3]);
  printf("%1.15Le\n", result);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<snormal_rndc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L16
L3:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <snorm@plt>
  movq   %xmm0,%rax
  mov    %rax,(%rbx)
  addl   $0x1,-0x14(%rbp)
L16:
  mov    -0x14(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L3
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void snormal_rndc(int n, double *randvec) {
  int i;
  for (i = 0; i < n; i++)
    randvec[i] = snorm();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<snormal_rndc>:
  endbr64
  test   %edi,%edi
  jle    L8
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  lea    -0x1(%rdi),%eax
  lea    0x8(%rsi,%rax,8),%rbp
L11:
  mov    $0x0,%eax
  call   <snorm@plt>
  movsd  %xmm0,(%rbx)
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L11
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  ret
```

assistant: ```c
void snormal_rndc(int n, double *randvec) {
  int i;
  for (i = 0; i < n; i++)
    randvec[i] = snorm();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<snormal_rndc>:
  endbr64
  test   %edi,%edi
  jle    L15
  push   %rbp
  lea    -0x1(%rdi),%eax
  push   %rbx
  lea    0x8(%rsi,%rax,8),%rbp
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L6:
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <snorm@plt>
  movsd  %xmm0,-0x8(%rbx)
  cmp    %rbp,%rbx
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L15:
  ret
```

assistant: ```c
void snormal_rndc(int n, double *randvec) {
  int i;
  for (i = 0; i < n; i++)
    randvec[i] = snorm();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<snormal_rndc>:
  endbr64
  test   %edi,%edi
  jle    L13
  push   %rbp
  lea    -0x1(%rdi),%eax
  push   %rbx
  lea    0x8(%rsi,%rax,8),%rbp
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L7:
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <snorm@plt>
  movsd  %xmm0,-0x8(%rbx)
  cmp    %rbp,%rbx
  jne    L7
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L13:
  ret
```

assistant: ```c
void snormal_rndc(int n, double *randvec) {
  int i;
  for (i = 0; i < n; i++)
    randvec[i] = snorm();
}
```
user: What is the c source code of the assembly code below:

```asm
<sum_solve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8d,-0x2c(%rbp)
  cmpl   $0x0,-0x2c(%rbp)
  jns    L55
  mov    $0x0,%eax
  jmp    L10
L55:
  cmpl   $0x0,-0x2c(%rbp)
  jne    L71
  mov    $0x1,%eax
  jmp    L10
L71:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  je     L26
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  jmp    L10
L26:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  movl   $0x0,-0x8(%rbp)
  jmp    L62
L48:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L34
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x2c(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,%edi
  mov    -0x20(%rbp),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <sum_solve>
  mov    %eax,-0x4(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  add    %ecx,%edx
  mov    %edx,(%rax)
L34:
  addl   $0x1,-0x8(%rbp)
L62:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L48
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
L10:
  leave
  ret
```

assistant: ```c
int sum_solve(int dp[], int size, int set[], int num, int n) {
  if (n < 0)
    return 0;
  if (n == 0)
    return 1;
  if (dp[n] != -1)
    return dp[n];
  dp[n] = 0;
  for (int i = 0; i < num; i++) {
    if (n >= set[i]) {
      int ret = sum_solve(dp, size, set, num, n - set[i]);
      dp[n] += ret;
    }
  }
  return dp[n];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sum_solve>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,0x4(%rsp)
  mov    %rdx,0x8(%rsp)
  mov    $0x0,%eax
  test   %r8d,%r8d
  js     L10
  mov    %rdi,%rbp
  mov    %ecx,%r12d
  mov    %r8d,%ebx
  mov    $0x1,%eax
  je     L10
  movslq %r8d,%rax
  lea    (%rdi,%rax,4),%r14
  mov    (%r14),%eax
  cmp    $0xffffffff,%eax
  jne    L10
  movl   $0x0,(%r14)
  test   %ecx,%ecx
  jle    L78
  mov    %rdx,%r13
  lea    -0x1(%rcx),%eax
  lea    0x4(%rdx,%rax,4),%r15
  jmp    L72
L75:
  add    $0x4,%r13
  cmp    %r15,%r13
  je     L78
L72:
  mov    0x0(%r13),%eax
  cmp    %ebx,%eax
  jg     L75
  mov    %ebx,%r8d
  sub    %eax,%r8d
  mov    %r12d,%ecx
  mov    0x8(%rsp),%rdx
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <sum_solve>
  add    %eax,(%r14)
  jmp    L75
L78:
  mov    (%r14),%eax
L10:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int sum_solve(int dp[], int size, int set[], int num, int n) {
  if (n < 0)
    return 0;
  if (n == 0)
    return 1;
  if (dp[n] != -1)
    return dp[n];
  dp[n] = 0;
  for (int i = 0; i < num; i++) {
    if (n >= set[i]) {
      int ret = sum_solve(dp, size, set, num, n - set[i]);
      dp[n] += ret;
    }
  }
  return dp[n];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum_solve>:
  endbr64
  xor    %eax,%eax
  test   %r8d,%r8d
  js     L39
  mov    $0x1,%eax
  je     L39
  push   %r15
  movslq %r8d,%rax
  push   %r14
  lea    (%rdi,%rax,4),%r9
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %r8d,%ebx
  sub    $0x18,%rsp
  mov    (%r9),%eax
  cmp    $0xffffffff,%eax
  jne    L62
  movl   $0x0,(%r9)
  mov    %ecx,%r13d
  xor    %eax,%eax
  test   %ecx,%ecx
  jle    L62
  lea    -0x1(%rcx),%eax
  mov    %esi,%r14d
  mov    %rdx,%r15
  mov    %rdx,%rbp
  lea    0x4(%rdx,%rax,4),%r10
L14:
  mov    0x0(%rbp),%eax
  cmp    %ebx,%eax
  jle    L50
L28:
  add    $0x4,%rbp
  cmp    %r10,%rbp
  jne    L14
  mov    (%r9),%eax
L62:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L39:
  ret
  nopl   0x0(%rax)
L50:
  mov    %ebx,%r8d
  mov    %r13d,%ecx
  mov    %r15,%rdx
  mov    %r14d,%esi
  sub    %eax,%r8d
  mov    %r12,%rdi
  mov    %r10,0x8(%rsp)
  mov    %r9,(%rsp)
  call   <sum_solve>
  mov    (%rsp),%r9
  mov    0x8(%rsp),%r10
  add    %eax,(%r9)
  jmp    L28
```

assistant: ```c
int sum_solve(int dp[], int size, int set[], int num, int n) {
  if (n < 0)
    return 0;
  if (n == 0)
    return 1;
  if (dp[n] != -1)
    return dp[n];
  dp[n] = 0;
  for (int i = 0; i < num; i++) {
    if (n >= set[i]) {
      int ret = sum_solve(dp, size, set, num, n - set[i]);
      dp[n] += ret;
    }
  }
  return dp[n];
}
```
user: Translate the below assembly code into C programming language:

```asm
<sum_solve>:
  endbr64
  xor    %eax,%eax
  test   %r8d,%r8d
  js     L30
  mov    $0x1,%eax
  je     L30
  push   %r15
  movslq %r8d,%rax
  push   %r14
  lea    (%rdi,%rax,4),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %r8d,%ebx
  sub    $0x28,%rsp
  mov    (%r14),%eax
  cmp    $0xffffffff,%eax
  jne    L40
  movl   $0x0,(%r14)
  xor    %eax,%eax
  test   %ecx,%ecx
  jle    L40
  lea    -0x1(%rcx),%eax
  mov    %rdx,%r12
  mov    %r14,%r9
  lea    0x4(%rdx,%rax,4),%r15
L38:
  mov    (%r12),%eax
  cmp    %ebx,%eax
  jle    L49
L32:
  add    $0x4,%r12
  cmp    %r15,%r12
  jne    L38
  mov    (%r9),%eax
L40:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L30:
  ret
  nopl   0x0(%rax)
L49:
  mov    %ebx,%r13d
  mov    $0x1,%r8d
  sub    %eax,%r13d
  je     L73
  movslq %r13d,%rax
  lea    (%rdi,%rax,4),%r14
  mov    (%r14),%r8d
  cmp    $0xffffffff,%r8d
  jne    L73
  movl   $0x0,(%r14)
  mov    %rdx,%rbp
  mov    %r9,0x8(%rsp)
  mov    %r12,0x10(%rsp)
  mov    %esi,%r12d
  mov    %ebx,0x1c(%rsp)
  mov    %rdi,%rbx
L24:
  mov    0x0(%rbp),%eax
  cmp    %eax,%r13d
  jge    L115
L37:
  add    $0x4,%rbp
  cmp    %r15,%rbp
  jne    L24
  mov    %rbx,%rdi
  mov    %r12d,%esi
  mov    0x8(%rsp),%r9
  mov    0x10(%rsp),%r12
  mov    0x1c(%rsp),%ebx
  mov    (%r14),%r8d
L73:
  add    %r8d,(%r9)
  jmp    L32
  nopl   0x0(%rax)
L115:
  mov    %r13d,%r8d
  mov    %r12d,%esi
  mov    %rbx,%rdi
  mov    %ecx,0x18(%rsp)
  sub    %eax,%r8d
  mov    %rdx,(%rsp)
  call   <sum_solve>
  mov    0x18(%rsp),%ecx
  add    %eax,(%r14)
  mov    (%rsp),%rdx
  jmp    L37
```

assistant: ```c
int sum_solve(int dp[], int size, int set[], int num, int n) {
  if (n < 0)
    return 0;
  if (n == 0)
    return 1;
  if (dp[n] != -1)
    return dp[n];
  dp[n] = 0;
  for (int i = 0; i < num; i++) {
    if (n >= set[i]) {
      int ret = sum_solve(dp, size, set, num, n - set[i]);
      dp[n] += ret;
    }
  }
  return dp[n];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<calc_opencl>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  lea    D13(%rip),%rdi
  call   <puts@plt>
  lea    0x20(%rsp),%rcx
  lea    0x18(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  call   <init_cl@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L14
L5:
  lea    0x1f(%rbx),%eax
  cltd
  idiv   %ebx
  mov    %eax,%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  lea    D7(%rip),%rdi
  call   <puts@plt>
  mov    0x20(%rsp),%rdi
  call   <clReleaseCommandQueue@plt>
  mov    0x18(%rsp),%rdi
  call   <clReleaseContext@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x30,%rsp
  pop    %rbx
  ret
L14:
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L5
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D13
type_guessed: string
name: unpack_data
arguments: data_tag: D5
type_guessed: string
name: unpack_data
arguments: data_tag: D38
type_guessed: string
name: unpack_data
arguments: data_tag: D7
type_guessed: string

tool: D13:  [Initialize Device]

tool: D5:    * Wrok Items: %d


tool: D38:  [Calculation]

tool: D7:  [Profile]

assistant: ```c
void calc_opencl(int datasize, double *data1, double *data2, double *result, const char *cl_filename) {
  cl_int num_rand = 4096 * 256;
  int count_all, i;
  int num_generator = 32;
  double pi;
  cl_platform_id platform_id = NULL;
  cl_device_id device_id = NULL;
  cl_context context = NULL;
  cl_command_queue command_queue = NULL;
  cl_uint num_compute_unit;
  cl_uint num_work_item;
  cl_program program = NULL;
  cl_kernel kernel_mt = NULL, kernel_pi = NULL;
  cl_mem rand, count;
  size_t global_item_size[3], local_item_size[3];
  cl_mem dev_mts;
  cl_event ev_mt_end, ev_pi_end, ev_copy_end;
  cl_ulong prof_start, prof_mt_end, prof_pi_end, prof_copy_end;
  cl_int ret;
  printf(" [Initialize Device]\n");
  num_compute_unit = init_cl(&platform_id, &device_id, &context, &command_queue);
  if (num_compute_unit < 1) {
    exit(-1);
  }
  num_work_item = (num_generator + (num_compute_unit - 1)) / num_compute_unit;
  printf("   * Wrok Items: %d\n", num_work_item);
  global_item_size[0] = num_work_item * num_compute_unit;
  global_item_size[1] = 1;
  global_item_size[2] = 1;
  local_item_size[0] = num_work_item;
  local_item_size[1] = 1;
  local_item_size[2] = 1;
  printf(" [Calculation]\n");
  printf(" [Profile]\n");
  clReleaseCommandQueue(command_queue);
  clReleaseContext(context);
}
```