ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<auth_ssl_load>:
  endbr64
  push   %rbp
  test   %edi,%edi
  lea    D16(%rip),%rax
  lea    D1(%rip),%rdi
  cmove  %rax,%rdi
  call   <storage_load@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <openssl_set_certs@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1080 <safe_free_strs@plt>
```

tool_calls: name: parse_data
arguments: data_label: D16
type: string
name: parse_data
arguments: data_label: D1
type: string

tool: D16: connect

tool: D1: accept

assistant: ```c
void auth_ssl_load(int server) {
  char **str = storage_load(server ? "accept" : "connect");
  openssl_set_certs(str);
  safe_free_strs(str);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sf_f3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L0:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L18:
  cmpl   $0x3ff,-0x4(%rbp)
  jle    L0
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void __attribute__((no, noclone)) sf_f3(float *__restrict a, float *__restrict b, float *__restrict c, float *__restrict d) {
  for (int i = 0; i < 1024; ++i) {
    a[i] = d[i * 3];
    b[i] = d[i * 3 + 1];
    c[i] = d[i * 3 + 2];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sf_f3>:
  endbr64
  mov    $0x0,%eax
L6:
  movss  (%rcx),%xmm0
  movss  %xmm0,(%rdi,%rax,1)
  movss  0x4(%rcx),%xmm0
  movss  %xmm0,(%rsi,%rax,1)
  movss  0x8(%rcx),%xmm0
  movss  %xmm0,(%rdx,%rax,1)
  add    $0xc,%rcx
  add    $0x4,%rax
  cmp    $0x1000,%rax
  jne    L6
  ret
```

assistant: ```c
void __attribute__((no, noclone)) sf_f3(float *__restrict a, float *__restrict b, float *__restrict c, float *__restrict d) {
  for (int i = 0; i < 1024; ++i) {
    a[i] = d[i * 3];
    b[i] = d[i * 3 + 1];
    c[i] = d[i * 3 + 2];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<sf_f3>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L9:
  movss  (%rcx),%xmm0
  add    $0xc,%rcx
  movss  %xmm0,(%rdi,%rax,1)
  movss  -0x8(%rcx),%xmm0
  movss  %xmm0,(%rsi,%rax,1)
  movss  -0x4(%rcx),%xmm0
  movss  %xmm0,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    $0x1000,%rax
  jne    L9
  ret
```

assistant: ```c
void __attribute__((no, noclone)) sf_f3(float *__restrict a, float *__restrict b, float *__restrict c, float *__restrict d) {
  for (int i = 0; i < 1024; ++i) {
    a[i] = d[i * 3];
    b[i] = d[i * 3 + 1];
    c[i] = d[i * 3 + 2];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<sf_f3>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L5:
  movss  (%rcx),%xmm0
  add    $0xc,%rcx
  movss  %xmm0,(%rdi,%rax,1)
  movss  -0x8(%rcx),%xmm0
  movss  %xmm0,(%rsi,%rax,1)
  movss  -0x4(%rcx),%xmm0
  movss  %xmm0,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    $0x1000,%rax
  jne    L5
  ret
```

assistant: ```c
void __attribute__((no, noclone)) sf_f3(float *__restrict a, float *__restrict b, float *__restrict c, float *__restrict d) {
  for (int i = 0; i < 1024; ++i) {
    a[i] = d[i * 3];
    b[i] = d[i * 3 + 1];
    c[i] = d[i * 3 + 2];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x14(%rbp)
  jmp    L22
L1:
  mov    -0xc(%rbp),%eax
  add    %eax,-0x18(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x14(%rbp)
  addl   $0x1,-0x10(%rbp)
L22:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L1
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D34
type_guessed: string
name: read_data
arguments: data_tag: D6
type_guessed: string
name: read_data
arguments: data_tag: D0
type_guessed: string
name: read_data
arguments: data_tag: D19
type_guessed: string

tool: D34: Please enter the number of terms to calculate.

tool: D6: %d

tool: D0: %d 

tool: D19: 
The sum is %d.


assistant: ```c
int main() {
  int sum = 0, range = 0, i = 0, j = 0, term = 1;
  printf("Please enter the number of terms to calculate.\n");
  scanf("%d", &range);
  for (i = 1; j < range; i++, j++) {
    sum = sum + term;
    printf("%d ", term);
    term = term + i;
  }
  printf("\nThe sum is %d.\n", sum);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    D32(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L0
  mov    $0x1,%ebp
  mov    $0x1,%ebx
  mov    $0x0,%r12d
  lea    D30(%rip),%r13
L32:
  add    %ebp,%r12d
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    %ebx,%ebp
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,0x4(%rsp)
  jg     L32
L8:
  mov    %r12d,%esi
  lea    D7(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  mov    $0x0,%r12d
  jmp    L8
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D32
type_guessed: string
name: read_data
arguments: label: D14
type_guessed: string
name: read_data
arguments: label: D30
type_guessed: string
name: read_data
arguments: label: D7
type_guessed: string

tool: D32: Please enter the number of terms to calculate.

tool: D14: %d

tool: D30: %d 

tool: D7: 
The sum is %d.


assistant: ```c
int main() {
  int sum = 0, range = 0, i = 0, j = 0, term = 1;
  printf("Please enter the number of terms to calculate.\n");
  scanf("%d", &range);
  for (i = 1; j < range; i++, j++) {
    sum = sum + term;
    printf("%d ", term);
    term = term + i;
  }
  printf("\nThe sum is %d.\n", sum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  lea    D5(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L16
  mov    $0x1,%ebp
  mov    $0x1,%ebx
  xor    %r12d,%r12d
  lea    D4(%rip),%r13
  nop
L34:
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    %ebp,%r12d
  call   <printf@plt>
  add    %ebx,%ebp
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,0x4(%rsp)
  jg     L34
L6:
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  xor    %r12d,%r12d
  jmp    L6
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type_guessed: string
name: unpack_data
arguments: data_tag: D24
type_guessed: string
name: unpack_data
arguments: data_tag: D4
type_guessed: string
name: unpack_data
arguments: data_tag: D11
type_guessed: string

tool: D5: Please enter the number of terms to calculate.

tool: D24: %d

tool: D4: %d 

tool: D11: 
The sum is %d.


assistant: ```c
int main() {
  int sum = 0, range = 0, i = 0, j = 0, term = 1;
  printf("Please enter the number of terms to calculate.\n");
  scanf("%d", &range);
  for (i = 1; j < range; i++, j++) {
    sum = sum + term;
    printf("%d ", term);
    term = term + i;
  }
  printf("\nThe sum is %d.\n", sum);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D15(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L16
  mov    $0x1,%ebp
  mov    $0x1,%ebx
  xor    %r12d,%r12d
  lea    D18(%rip),%r13
  nop
L31:
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    %ebp,%r12d
  call   <printf@plt>
  add    %ebx,%ebp
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,0x4(%rsp)
  jg     L31
L2:
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  xor    %r12d,%r12d
  jmp    L2
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D15
type: string
name: parse_data
arguments: label: D21
type: string
name: parse_data
arguments: label: D18
type: string
name: parse_data
arguments: label: D6
type: string

tool: D15: Please enter the number of terms to calculate.

tool: D21: %d

tool: D18: %d 

tool: D6: 
The sum is %d.


assistant: ```c
int main() {
  int sum = 0, range = 0, i = 0, j = 0, term = 1;
  printf("Please enter the number of terms to calculate.\n");
  scanf("%d", &range);
  for (i = 1; j < range; i++, j++) {
    sum = sum + term;
    printf("%d ", term);
    term = term + i;
  }
  printf("\nThe sum is %d.\n", sum);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<vecexp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L19
L7:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  movq   %rax,%xmm0
  call   <exp@plt>
  movq   %xmm0,%rax
  mov    %rax,(%rbx)
  addl   $0x1,-0x14(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L7
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void vecexp(double *values, const int n) {
  int i;
  for (i = 0; i < n; ++i) {
    values[i] = exp(values[i]);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<vecexp>:
  endbr64
  test   %esi,%esi
  jle    L11
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rbp
L0:
  movsd  (%rbx),%xmm0
  call   <exp@plt>
  movsd  %xmm0,(%rbx)
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L11:
  ret
```

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

```asm
<vecexp>:
  endbr64
  test   %esi,%esi
  jle    L16
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x8(%rdi,%rax,8),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L11:
  movsd  (%rbx),%xmm0
  add    $0x8,%rbx
  call   <exp@plt>
  movsd  %xmm0,-0x8(%rbx)
  cmp    %rbp,%rbx
  jne    L11
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L16:
  ret
```

assistant: ```c
void vecexp(double *values, const int n) {
  int i;
  for (i = 0; i < n; ++i) {
    values[i] = exp(values[i]);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<vecexp>:
  endbr64
  test   %esi,%esi
  jle    L12
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x8(%rdi,%rax,8),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L9:
  movsd  (%rbx),%xmm0
  add    $0x8,%rbx
  call   <exp@plt>
  movsd  %xmm0,-0x8(%rbx)
  cmp    %rbp,%rbx
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L12:
  ret
```

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

```asm
<BN_X931_generate_Xpq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %rcx,-0x40(%rbp)
  cmpl   $0x3ff,-0x34(%rbp)
  jle    L78
  mov    -0x34(%rbp),%eax
  movzbl %al,%eax
  test   %eax,%eax
  je     L16
L78:
  mov    $0x0,%eax
  jmp    L38
L16:
  sarl   -0x34(%rbp)
  mov    -0x34(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    $0x0,%ecx
  mov    $0x1,%edx
  mov    %rax,%rdi
  call   <BN_rand@plt>
  test   %eax,%eax
  jne    L112
  mov    $0x0,%eax
  jmp    L38
L112:
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_start@plt>
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L100
  mov    $0x0,%eax
  jmp    L38
L100:
  movl   $0x0,-0x1c(%rbp)
  jmp    L6
L24:
  mov    -0x34(%rbp),%esi
  mov    -0x30(%rbp),%rax
  mov    $0x0,%ecx
  mov    $0x1,%edx
  mov    %rax,%rdi
  call   <BN_rand@plt>
  test   %eax,%eax
  jne    L118
  mov    $0x0,%eax
  jmp    L38
L118:
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <BN_sub@plt>
  mov    -0x34(%rbp),%eax
  lea    -0x63(%rax),%ebx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_num_bits@plt>
  cmp    %eax,%ebx
  jle    L40
  addl   $0x1,-0x1c(%rbp)
L6:
  cmpl   $0x3e7,-0x1c(%rbp)
  jle    L24
  jmp    L5
L40:
  nop
L5:
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_end@plt>
  cmpl   $0x3e7,-0x1c(%rbp)
  jg     L35
  mov    $0x1,%eax
  jmp    L38
L35:
  mov    $0x0,%eax
L38:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int BN_X931_generate_Xpq(BIGNUM *Xp, BIGNUM *Xq, int nbits, BN_CTX *ctx) {
  BIGNUM *t;
  int i;
  if ((nbits < 1024) || (nbits & 0xff))
    return 0;
  nbits >>= 1;
  if (!BN_rand(Xp, nbits, 1, 0))
    return 0;
  BN_CTX_start(ctx);
  if ((t = BN_CTX_get(ctx)) == NULL)
    return 0;
  for (i = 0; i < 1000; i++) {
    if (!BN_rand(Xq, nbits, 1, 0))
      return 0;
    BN_sub(t, Xp, Xq);
    if (BN_num_bits(t) > (nbits - 100))
      break;
  }
  BN_CTX_end(ctx);
  if (i < 1000)
    return 1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<BN_X931_generate_Xpq>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rcx,0x8(%rsp)
  mov    $0x0,%eax
  cmp    $0x3ff,%edx
  jle    L61
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %edx,%ebx
  mov    %rcx,%r15
  mov    %edx,%r12d
  and    $0xff,%r12d
  je     L13
L61:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  sar    %ebx
  mov    $0x0,%ecx
  mov    $0x1,%edx
  mov    %ebx,%esi
  call   <BN_rand@plt>
  test   %eax,%eax
  je     L61
  mov    %r15,%rdi
  call   <BN_CTX_start@plt>
  mov    %r15,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L33
  lea    -0x63(%rbx),%r15d
L78:
  mov    $0x0,%ecx
  mov    $0x1,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <BN_rand@plt>
  test   %eax,%eax
  je     L61
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r14,%rdi
  call   <BN_sub@plt>
  mov    %r14,%rdi
  call   <BN_num_bits@plt>
  cmp    %eax,%r15d
  jle    L60
  add    $0x1,%r12d
  cmp    $0x3e8,%r12d
  jne    L78
L60:
  mov    0x8(%rsp),%rdi
  call   <BN_CTX_end@plt>
  cmp    $0x3e7,%r12d
  setle  %al
  movzbl %al,%eax
  jmp    L61
L33:
  mov    %r12d,%eax
  jmp    L61
```

assistant: ```c
int BN_X931_generate_Xpq(BIGNUM *Xp, BIGNUM *Xq, int nbits, BN_CTX *ctx) {
  BIGNUM *t;
  int i;
  if ((nbits < 1024) || (nbits & 0xff))
    return 0;
  nbits >>= 1;
  if (!BN_rand(Xp, nbits, 1, 0))
    return 0;
  BN_CTX_start(ctx);
  if ((t = BN_CTX_get(ctx)) == NULL)
    return 0;
  for (i = 0; i < 1000; i++) {
    if (!BN_rand(Xq, nbits, 1, 0))
      return 0;
    BN_sub(t, Xp, Xq);
    if (BN_num_bits(t) > (nbits - 100))
      break;
  }
  BN_CTX_end(ctx);
  if (i < 1000)
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<BN_X931_generate_Xpq>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rcx,0x8(%rsp)
  cmp    $0x3ff,%edx
  jle    L64
  mov    %edx,%r15d
  mov    %edx,%ebx
  and    $0xff,%r15d
  je     L109
L64:
  xor    %r15d,%r15d
L8:
  add    $0x18,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L109:
  sar    %ebx
  mov    %rsi,%rbp
  mov    %rcx,%r14
  mov    $0x1,%edx
  xor    %ecx,%ecx
  mov    %ebx,%esi
  mov    %rdi,%r12
  call   <BN_rand@plt>
  test   %eax,%eax
  je     L64
  mov    %r14,%rdi
  call   <BN_CTX_start@plt>
  mov    %r14,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L64
  lea    -0x63(%rbx),%eax
  mov    $0x3e8,%r13d
  mov    %eax,0x4(%rsp)
  jmp    L23
  nopl   0x0(%rax)
L12:
  mov    %r14,%rdi
  mov    %rbp,%rdx
  mov    %r12,%rsi
  call   <BN_sub@plt>
  mov    %r14,%rdi
  call   <BN_num_bits@plt>
  cmp    %eax,0x4(%rsp)
  jle    L24
  sub    $0x1,%r13d
  je     L5
L23:
  xor    %ecx,%ecx
  mov    $0x1,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <BN_rand@plt>
  test   %eax,%eax
  jne    L12
  jmp    L64
  nopl   0x0(%rax)
L24:
  mov    0x8(%rsp),%rdi
  mov    $0x1,%r15d
  call   <BN_CTX_end@plt>
  jmp    L8
L5:
  mov    0x8(%rsp),%rdi
  call   <BN_CTX_end@plt>
  jmp    L8
```

assistant: ```c
int BN_X931_generate_Xpq(BIGNUM *Xp, BIGNUM *Xq, int nbits, BN_CTX *ctx) {
  BIGNUM *t;
  int i;
  if ((nbits < 1024) || (nbits & 0xff))
    return 0;
  nbits >>= 1;
  if (!BN_rand(Xp, nbits, 1, 0))
    return 0;
  BN_CTX_start(ctx);
  if ((t = BN_CTX_get(ctx)) == NULL)
    return 0;
  for (i = 0; i < 1000; i++) {
    if (!BN_rand(Xq, nbits, 1, 0))
      return 0;
    BN_sub(t, Xp, Xq);
    if (BN_num_bits(t) > (nbits - 100))
      break;
  }
  BN_CTX_end(ctx);
  if (i < 1000)
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<BN_X931_generate_Xpq>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rcx,0x8(%rsp)
  cmp    $0x3ff,%edx
  jle    L38
  mov    %edx,%r15d
  mov    %edx,%ebx
  and    $0xff,%r15d
  je     L0
L38:
  xor    %r15d,%r15d
L94:
  add    $0x18,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  sar    %ebx
  mov    %rsi,%rbp
  mov    %rcx,%r14
  mov    $0x1,%edx
  xor    %ecx,%ecx
  mov    %ebx,%esi
  mov    %rdi,%r12
  call   <BN_rand@plt>
  test   %eax,%eax
  je     L38
  mov    %r14,%rdi
  call   <BN_CTX_start@plt>
  mov    %r14,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L38
  lea    -0x63(%rbx),%eax
  mov    $0x3e8,%r13d
  mov    %eax,0x4(%rsp)
  jmp    L50
  nopl   0x0(%rax)
L43:
  mov    %r14,%rdi
  mov    %rbp,%rdx
  mov    %r12,%rsi
  call   <BN_sub@plt>
  mov    %r14,%rdi
  call   <BN_num_bits@plt>
  cmp    %eax,0x4(%rsp)
  jle    L68
  sub    $0x1,%r13d
  je     L67
L50:
  xor    %ecx,%ecx
  mov    $0x1,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <BN_rand@plt>
  test   %eax,%eax
  jne    L43
  jmp    L38
  nopl   0x0(%rax)
L68:
  mov    0x8(%rsp),%rdi
  mov    $0x1,%r15d
  call   <BN_CTX_end@plt>
  jmp    L94
L67:
  mov    0x8(%rsp),%rdi
  call   <BN_CTX_end@plt>
  jmp    L94
```

assistant: ```c
int BN_X931_generate_Xpq(BIGNUM *Xp, BIGNUM *Xq, int nbits, BN_CTX *ctx) {
  BIGNUM *t;
  int i;
  if ((nbits < 1024) || (nbits & 0xff))
    return 0;
  nbits >>= 1;
  if (!BN_rand(Xp, nbits, 1, 0))
    return 0;
  BN_CTX_start(ctx);
  if ((t = BN_CTX_get(ctx)) == NULL)
    return 0;
  for (i = 0; i < 1000; i++) {
    if (!BN_rand(Xq, nbits, 1, 0))
      return 0;
    BN_sub(t, Xp, Xq);
    if (BN_num_bits(t) > (nbits - 100))
      break;
  }
  BN_CTX_end(ctx);
  if (i < 1000)
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<corner_dtrma_3_lib4>:
  endbr64
  test   %edi,%edi
  je     L17
  cmp    $0x3,%edi
  je     L17
  cmp    $0x1,%edi
  je     L10
  movsd  (%rsi),%xmm0
  movsd  %xmm0,(%rdx)
  movsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  movsd  0x28(%rsi),%xmm0
  movsd  %xmm0,0x28(%rdx)
  lea    -0x2(,%rcx,4),%eax
  cltq
  lea    (%rdx,%rax,8),%rax
  movsd  0x10(%rsi),%xmm0
  movsd  %xmm0,(%rax)
  movsd  0x30(%rsi),%xmm0
  movsd  %xmm0,0x20(%rax)
  movsd  0x50(%rsi),%xmm0
  movsd  %xmm0,0x40(%rax)
  ret
L17:
  movsd  (%rsi),%xmm0
  movsd  %xmm0,(%rdx)
  movsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  movsd  0x10(%rsi),%xmm0
  movsd  %xmm0,0x10(%rdx)
  movsd  0x28(%rsi),%xmm0
  movsd  %xmm0,0x28(%rdx)
  movsd  0x30(%rsi),%xmm0
  movsd  %xmm0,0x30(%rdx)
  movsd  0x50(%rsi),%xmm0
  movsd  %xmm0,0x50(%rdx)
  ret
L10:
  movsd  (%rsi),%xmm0
  movsd  %xmm0,(%rdx)
  lea    -0x3(,%rcx,4),%eax
  cltq
  lea    (%rdx,%rax,8),%rax
  movsd  0x8(%rsi),%xmm0
  movsd  %xmm0,(%rax)
  movsd  0x10(%rsi),%xmm0
  movsd  %xmm0,0x8(%rax)
  movsd  0x28(%rsi),%xmm0
  movsd  %xmm0,0x20(%rax)
  movsd  0x30(%rsi),%xmm0
  movsd  %xmm0,0x28(%rax)
  movsd  0x50(%rsi),%xmm0
  movsd  %xmm0,0x48(%rax)
  ret
```

assistant: ```c
void corner_dtrma_3_lib4(int kna, double *A, double *C, int sdc) {
  const int bs = 4;
  if (kna == 0 || kna == 3) {
    C[0 + bs * 0] = A[0 + bs * 0];
    C[1 + bs * 0] = A[1 + bs * 0];
    C[2 + bs * 0] = A[2 + bs * 0];
    C[1 + bs * 1] = A[1 + bs * 1];
    C[2 + bs * 1] = A[2 + bs * 1];
    C[2 + bs * 2] = A[2 + bs * 2];
    return;
  } else if (kna == 1) {
    C[0 + bs * 0] = A[0 + bs * 0];
    C += 1 + bs * (sdc - 1);
    C[0 + bs * 0] = A[1 + bs * 0];
    C[1 + bs * 0] = A[2 + bs * 0];
    C[0 + bs * 1] = A[1 + bs * 1];
    C[1 + bs * 1] = A[2 + bs * 1];
    C[1 + bs * 2] = A[2 + bs * 2];
    return;
  } else {
    C[0 + bs * 0] = A[0 + bs * 0];
    C[1 + bs * 0] = A[1 + bs * 0];
    C[1 + bs * 1] = A[1 + bs * 1];
    C += 2 + bs * (sdc - 1);
    C[0 + bs * 0] = A[2 + bs * 0];
    C[0 + bs * 1] = A[2 + bs * 1];
    C[0 + bs * 2] = A[2 + bs * 2];
    return;
  }
}
```