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

```asm
<mangle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L10
L15:
  mov    -0x24(%rbp),%eax
  sub    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0xc(%rbp),%eax
  shl    $0x3,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  cltq
  or     %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L10:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L15
  mov    -0x8(%rbp),%rax
  imul   $0x7c32b16d,%rax,%rax
  and    $0xffffffff,%eax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L22
L7:
  mov    -0xc(%rbp),%eax
  shl    $0x3,%eax
  mov    -0x8(%rbp),%rdx
  mov    %eax,%ecx
  shr    %cl,%rdx
  mov    %rdx,%rcx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
L22:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L7
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void mangle(const unsigned char *key, unsigned char *ret_key, int nbytes) {
  unsigned long long new_key = 0;
  int i;
  for (i = 0; i < nbytes; ++i) {
    new_key |= key[nbytes - i - 1] << (i * 8);
  }
  new_key = (new_key * 2083697005) & (0xffffffff);
  for (i = 0; i < nbytes; ++i) {
    ret_key[i] = (new_key >> (i * 8)) & 0xff;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mangle>:
  endbr64
  mov    %rsi,%r8
  test   %edx,%edx
  jle    L15
  movslq %edx,%rax
  lea    -0x1(%rdi,%rax,1),%rdi
  lea    0x0(,%rdx,8),%r9d
  mov    $0x0,%ecx
  mov    $0x0,%esi
L26:
  movzbl (%rdi),%eax
  shl    %cl,%eax
  cltq
  or     %rax,%rsi
  sub    $0x1,%rdi
  add    $0x8,%ecx
  cmp    %r9d,%ecx
  jne    L26
  imul   $0x7c32b16d,%esi,%esi
  mov    %edx,%edx
  mov    $0x0,%eax
L17:
  lea    0x0(,%rax,8),%ecx
  mov    %rsi,%rdi
  shr    %cl,%rdi
  mov    %dil,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L17
  ret
L15:
  ret
```

assistant: ```c
void mangle(const unsigned char *key, unsigned char *ret_key, int nbytes) {
  unsigned long long new_key = 0;
  int i;
  for (i = 0; i < nbytes; ++i) {
    new_key |= key[nbytes - i - 1] << (i * 8);
  }
  new_key = (new_key * 2083697005) & (0xffffffff);
  for (i = 0; i < nbytes; ++i) {
    ret_key[i] = (new_key >> (i * 8)) & 0xff;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mangle>:
  endbr64
  mov    %rsi,%r8
  test   %edx,%edx
  jle    L5
  movslq %edx,%r9
  xor    %ecx,%ecx
  shl    $0x3,%edx
  xor    %esi,%esi
  lea    -0x1(%rdi,%r9,1),%rdi
  nopw   0x0(%rax,%rax,1)
L4:
  movzbl (%rdi),%eax
  sub    $0x1,%rdi
  shl    %cl,%eax
  add    $0x8,%ecx
  cltq
  or     %rax,%rsi
  cmp    %ecx,%edx
  jne    L4
  imul   $0x7c32b16d,%esi,%esi
  xor    %eax,%eax
  nopl   (%rax)
L19:
  lea    0x0(,%rax,8),%ecx
  mov    %rsi,%rdx
  shr    %cl,%rdx
  mov    %dl,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%r9
  jne    L19
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  ret
```

assistant: ```c
void mangle(const unsigned char *key, unsigned char *ret_key, int nbytes) {
  unsigned long long new_key = 0;
  int i;
  for (i = 0; i < nbytes; ++i) {
    new_key |= key[nbytes - i - 1] << (i * 8);
  }
  new_key = (new_key * 2083697005) & (0xffffffff);
  for (i = 0; i < nbytes; ++i) {
    ret_key[i] = (new_key >> (i * 8)) & 0xff;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mangle>:
  endbr64
  mov    %rsi,%r8
  test   %edx,%edx
  jle    L13
  movslq %edx,%r9
  xor    %ecx,%ecx
  shl    $0x3,%edx
  xor    %esi,%esi
  lea    -0x1(%rdi,%r9,1),%rdi
  nopw   0x0(%rax,%rax,1)
L6:
  movzbl (%rdi),%eax
  sub    $0x1,%rdi
  shl    %cl,%eax
  add    $0x8,%ecx
  cltq
  or     %rax,%rsi
  cmp    %ecx,%edx
  jne    L6
  imul   $0x7c32b16d,%esi,%esi
  xor    %eax,%eax
  nopl   (%rax)
L7:
  lea    0x0(,%rax,8),%ecx
  mov    %rsi,%rdx
  shr    %cl,%rdx
  mov    %dl,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %r9,%rax
  jne    L7
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  ret
```

assistant: ```c
void mangle(const unsigned char *key, unsigned char *ret_key, int nbytes) {
  unsigned long long new_key = 0;
  int i;
  for (i = 0; i < nbytes; ++i) {
    new_key |= key[nbytes - i - 1] << (i * 8);
  }
  new_key = (new_key * 2083697005) & (0xffffffff);
  for (i = 0; i < nbytes; ++i) {
    ret_key[i] = (new_key >> (i * 8)) & 0xff;
  }
}
```
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_chars_bin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fwrite@plt>
  cmp    %eax,-0x14(%rbp)
  je     L12
  mov    $0xffffffff,%eax
  jmp    L14
L12:
  mov    $0x0,%eax
L14:
  leave
  ret
```

assistant: ```c
int write_chars_bin(FILE *out, const char *s, int count) {
  if (fwrite(s, 1, count, out) != count)
    return -1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<write_chars_bin>:
  endbr64
  push   %rbx
  mov    %rdi,%rcx
  mov    %rsi,%rdi
  mov    %edx,%ebx
  mov    $0x1,%esi
  call   <fwrite@plt>
  cmp    %ebx,%eax
  setne  %al
  movzbl %al,%eax
  neg    %eax
  pop    %rbx
  ret
```

assistant: ```c
int write_chars_bin(FILE *out, const char *s, int count) {
  if (fwrite(s, 1, count, out) != count)
    return -1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<write_chars_bin>:
  endbr64
  push   %rbx
  mov    %rdi,%rcx
  mov    %rsi,%rdi
  mov    $0x1,%esi
  mov    %edx,%ebx
  call   <fwrite@plt>
  cmp    %ebx,%eax
  pop    %rbx
  setne  %al
  movzbl %al,%eax
  neg    %eax
  ret
```

assistant: ```c
int write_chars_bin(FILE *out, const char *s, int count) {
  if (fwrite(s, 1, count, out) != count)
    return -1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<write_chars_bin>:
  endbr64
  push   %rbx
  mov    %rdi,%rcx
  mov    %rsi,%rdi
  mov    $0x1,%esi
  mov    %edx,%ebx
  call   <fwrite@plt>
  cmp    %ebx,%eax
  pop    %rbx
  setne  %al
  movzbl %al,%eax
  neg    %eax
  ret
```

assistant: ```c
int write_chars_bin(FILE *out, const char *s, int count) {
  if (fwrite(s, 1, count, out) != count)
    return -1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<fiat_p224_divstep_precomp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x800000,%edi
  call   <UINT32_C@plt>
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rbx
  mov    $0x800000,%edi
  call   <UINT32_C@plt>
  mov    %eax,(%rbx)
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rbx
  mov    $0xfe000000,%edi
  call   <UINT32_C@plt>
  mov    %eax,(%rbx)
  mov    -0x18(%rbp),%rax
  lea    0xc(%rax),%rbx
  mov    $0xffffff,%edi
  call   <UINT32_C@plt>
  mov    %eax,(%rbx)
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  movl   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x14(%rax),%rbx
  mov    $0xff800000,%edi
  call   <UINT32_C@plt>
  mov    %eax,(%rbx)
  mov    -0x18(%rbp),%rax
  lea    0x18(%rax),%rbx
  mov    $0x17fffff,%edi
  call   <UINT32_C@plt>
  mov    %eax,(%rbx)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void fiat_p224_divstep_precomp(uint32_t out1[7]) {
  out1[0] = UINT32_C(0x800000);
  out1[1] = UINT32_C(0x800000);
  out1[2] = UINT32_C(0xfe000000);
  out1[3] = UINT32_C(0xffffff);
  out1[4] = 0x0;
  out1[5] = UINT32_C(0xff800000);
  out1[6] = UINT32_C(0x17fffff);
}
```
user: Translate the below assembly code into C programming language:

```asm
<fiat_p224_divstep_precomp>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x800000,%edi
  call   <UINT32_C@plt>
  mov    %eax,(%rbx)
  mov    $0x800000,%edi
  call   <UINT32_C@plt>
  mov    %eax,0x4(%rbx)
  mov    $0xfe000000,%edi
  call   <UINT32_C@plt>
  mov    %eax,0x8(%rbx)
  mov    $0xffffff,%edi
  call   <UINT32_C@plt>
  mov    %eax,0xc(%rbx)
  movl   $0x0,0x10(%rbx)
  mov    $0xff800000,%edi
  call   <UINT32_C@plt>
  mov    %eax,0x14(%rbx)
  mov    $0x17fffff,%edi
  call   <UINT32_C@plt>
  mov    %eax,0x18(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void fiat_p224_divstep_precomp(uint32_t out1[7]) {
  out1[0] = UINT32_C(0x800000);
  out1[1] = UINT32_C(0x800000);
  out1[2] = UINT32_C(0xfe000000);
  out1[3] = UINT32_C(0xffffff);
  out1[4] = 0x0;
  out1[5] = UINT32_C(0xff800000);
  out1[6] = UINT32_C(0x17fffff);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<fiat_p224_divstep_precomp>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x800000,%edi
  call   <UINT32_C@plt>
  mov    $0x800000,%edi
  mov    %eax,(%rbx)
  call   <UINT32_C@plt>
  mov    $0xfe000000,%edi
  mov    %eax,0x4(%rbx)
  call   <UINT32_C@plt>
  mov    $0xffffff,%edi
  mov    %eax,0x8(%rbx)
  call   <UINT32_C@plt>
  movl   $0x0,0x10(%rbx)
  mov    $0xff800000,%edi
  mov    %eax,0xc(%rbx)
  call   <UINT32_C@plt>
  mov    $0x17fffff,%edi
  mov    %eax,0x14(%rbx)
  call   <UINT32_C@plt>
  mov    %eax,0x18(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void fiat_p224_divstep_precomp(uint32_t out1[7]) {
  out1[0] = UINT32_C(0x800000);
  out1[1] = UINT32_C(0x800000);
  out1[2] = UINT32_C(0xfe000000);
  out1[3] = UINT32_C(0xffffff);
  out1[4] = 0x0;
  out1[5] = UINT32_C(0xff800000);
  out1[6] = UINT32_C(0x17fffff);
}
```
user: What is the c source code of the assembly code below?

```asm
<fiat_p224_divstep_precomp>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x800000,%edi
  call   <UINT32_C@plt>
  mov    $0x800000,%edi
  mov    %eax,(%rbx)
  call   <UINT32_C@plt>
  mov    $0xfe000000,%edi
  mov    %eax,0x4(%rbx)
  call   <UINT32_C@plt>
  mov    $0xffffff,%edi
  mov    %eax,0x8(%rbx)
  call   <UINT32_C@plt>
  movl   $0x0,0x10(%rbx)
  mov    $0xff800000,%edi
  mov    %eax,0xc(%rbx)
  call   <UINT32_C@plt>
  mov    $0x17fffff,%edi
  mov    %eax,0x14(%rbx)
  call   <UINT32_C@plt>
  mov    %eax,0x18(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void fiat_p224_divstep_precomp(uint32_t out1[7]) {
  out1[0] = UINT32_C(0x800000);
  out1[1] = UINT32_C(0x800000);
  out1[2] = UINT32_C(0xfe000000);
  out1[3] = UINT32_C(0xffffff);
  out1[4] = 0x0;
  out1[5] = UINT32_C(0xff800000);
  out1[6] = UINT32_C(0x17fffff);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movl   $0x0,-0x24(%rbp)
  jmp    L12
L6:
  mov    $0xc3500,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    $0x10c8e0,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x124f80,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0xdbba0,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x24(%rbp)
L12:
  cmpl   $0x5db,-0x24(%rbp)
  jle    L6
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int i, *x1, *x2, *x3, *x4;
  for (i = 0; i < 1500; i++) {
    x1 = malloc(800 * 1000);
    x2 = malloc(1100 * 1000);
    free(x1);
    x3 = malloc(1200 * 1000);
    free(x2);
    free(x3);
    x4 = malloc(900 * 1000);
    free(x4);
  }
  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   %r12
  push   %rbp
  push   %rbx
  mov    $0x5dc,%r12d
L7:
  mov    $0xc3500,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x10c8e0,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x124f80,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0xdbba0,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  sub    $0x1,%r12d
  jne    L7
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(void) {
  int i, *x1, *x2, *x3, *x4;
  for (i = 0; i < 1500; i++) {
    x1 = malloc(800 * 1000);
    x2 = malloc(1100 * 1000);
    free(x1);
    x3 = malloc(1200 * 1000);
    free(x2);
    free(x3);
    x4 = malloc(900 * 1000);
    free(x4);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x5dc,%ebx
  nopl   (%rax)
L0:
  mov    $0xc3500,%edi
  call   <malloc@plt>
  mov    $0x10c8e0,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
  mov    $0x124f80,%edi
  call   <malloc@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0xdbba0,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  sub    $0x1,%ebx
  jne    L0
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(void) {
  int i, *x1, *x2, *x3, *x4;
  for (i = 0; i < 1500; i++) {
    x1 = malloc(800 * 1000);
    x2 = malloc(1100 * 1000);
    free(x1);
    x3 = malloc(1200 * 1000);
    free(x2);
    free(x3);
    x4 = malloc(900 * 1000);
    free(x4);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x5dc,%ebx
  nopl   (%rax)
L3:
  mov    $0xc3500,%edi
  call   <malloc@plt>
  mov    $0x10c8e0,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
  mov    $0x124f80,%edi
  call   <malloc@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0xdbba0,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  sub    $0x1,%ebx
  jne    L3
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(void) {
  int i, *x1, *x2, *x3, *x4;
  for (i = 0; i < 1500; i++) {
    x1 = malloc(800 * 1000);
    x2 = malloc(1100 * 1000);
    free(x1);
    x3 = malloc(1200 * 1000);
    free(x2);
    free(x3);
    x4 = malloc(900 * 1000);
    free(x4);
  }
  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
<dkth_smallest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L8
L51:
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L75
L82:
  addl   $0x1,-0x20(%rbp)
L75:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  -0x10(%rbp),%xmm0
  comisd %xmm1,%xmm0
  ja     L82
  jmp    L47
L80:
  subl   $0x1,-0x1c(%rbp)
L47:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  comisd -0x10(%rbp),%xmm0
  ja     L80
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jg     L92
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x20(%rbp)
  subl   $0x1,-0x1c(%rbp)
L92:
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L75
  mov    -0x1c(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jge    L73
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L73:
  mov    -0x30(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L8
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L8:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L51
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double dkth_smallest(double a[], int n, int k) {
  int i, j, l, m;
  double t, x;
  l = 0;
  m = n - 1;
  while (l < m) {
    x = a[k];
    i = l;
    j = m;
    do {
      while (a[i] < x)
        i++;
      while (x < a[j])
        j--;
      if (i <= j) {
        t = a[i];
        a[i] = a[j];
        a[j] = t;
        i++;
        j--;
      }
    } while (i <= j);
    if (j < k)
      l = i;
    if (k < i)
      m = j;
  }
  return a[k];
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<dkth_smallest>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r11
  mov    %edx,%r12d
  lea    -0x1(%rsi),%ebp
  test   %ebp,%ebp
  jle    L20
  movslq %edx,%rax
  lea    (%rdi,%rax,8),%r13
  mov    $0x0,%ebx
  jmp    L65
L6:
  movsd  %xmm0,(%r10)
  movsd  %xmm2,(%rcx)
  mov    %r8d,%r9d
  sub    $0x1,%edx
  cmp    %edx,%r8d
  jg     L100
L57:
  add    $0x8,%rsi
  add    $0x1,%r8d
  add    $0x1,%edi
L87:
  mov    %edi,%r9d
  mov    %rsi,%r10
  movsd  (%rsi),%xmm2
  comisd %xmm2,%xmm1
  ja     L57
  movslq %edx,%rax
  lea    (%r11,%rax,8),%rcx
  movsd  (%rcx),%xmm0
  lea    -0x8(%r11,%rax,8),%rax
  comisd %xmm1,%xmm0
  jbe    L39
L64:
  sub    $0x1,%edx
  mov    %rax,%rcx
  sub    $0x8,%rax
  movsd  0x8(%rax),%xmm0
  comisd %xmm1,%xmm0
  ja     L64
L39:
  cmp    %edx,%r9d
  jle    L6
L100:
  cmp    %edx,%r12d
  jg     L91
  cmp    %r9d,%r12d
  cmovl  %edx,%ebp
L27:
  cmp    %ebx,%ebp
  jle    L20
L65:
  movsd  0x0(%r13),%xmm1
  movslq %ebx,%rax
  lea    (%r11,%rax,8),%rsi
  mov    %ebx,%edi
  lea    0x1(%rbx),%r8d
  mov    %ebp,%edx
  jmp    L87
L91:
  cmp    %r9d,%r12d
  jl     L20
  mov    %r9d,%ebx
  jmp    L27
L20:
  movslq %r12d,%r12
  movsd  (%r11,%r12,8),%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
double dkth_smallest(double a[], int n, int k) {
  int i, j, l, m;
  double t, x;
  l = 0;
  m = n - 1;
  while (l < m) {
    x = a[k];
    i = l;
    j = m;
    do {
      while (a[i] < x)
        i++;
      while (x < a[j])
        j--;
      if (i <= j) {
        t = a[i];
        a[i] = a[j];
        a[j] = t;
        i++;
        j--;
      }
    } while (i <= j);
    if (j < k)
      l = i;
    if (k < i)
      m = j;
  }
  return a[k];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dkth_smallest>:
  endbr64
  push   %r12
  movslq %edx,%rax
  push   %rbp
  lea    (%rdi,%rax,8),%r12
  push   %rbx
  lea    -0x1(%rsi),%ebx
  test   %ebx,%ebx
  jle    L54
  mov    %rdi,%r9
  mov    %rax,%rbp
  xor    %r11d,%r11d
  nop
L77:
  movslq %r11d,%rax
  movsd  (%r12),%xmm1
  mov    %r11d,%esi
  lea    0x1(%r11),%r8d
  lea    (%r9,%rax,8),%rdi
  mov    %ebx,%edx
  movslq %ebx,%rax
  nopl   0x0(%rax)
L14:
  movsd  (%rdi),%xmm2
  lea    (%r9,%rax,8),%rcx
  mov    %esi,%r10d
  movsd  (%rcx),%xmm0
  comisd %xmm2,%xmm1
  ja     L18
  comisd %xmm1,%xmm0
  lea    -0x8(%r9,%rax,8),%rax
  jbe    L75
L72:
  mov    %rax,%rcx
  movsd  (%rax),%xmm0
  sub    $0x8,%rax
  sub    $0x1,%edx
  comisd %xmm1,%xmm0
  ja     L72
L75:
  cmp    %edx,%esi
  jle    L22
L12:
  cmp    %edx,%ebp
  cmovg  %r10d,%r11d
  cmp    %r10d,%ebp
  cmovl  %edx,%ebx
  cmp    %r11d,%ebx
  jg     L77
L54:
  movsd  (%r12),%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L22:
  sub    $0x1,%edx
  movsd  %xmm0,(%rdi)
  mov    %r8d,%r10d
  movsd  %xmm2,(%rcx)
  cmp    %edx,%r8d
  jg     L12
  movslq %edx,%rax
L18:
  add    $0x8,%rdi
  add    $0x1,%r8d
  add    $0x1,%esi
  jmp    L14
```

assistant: ```c
double dkth_smallest(double a[], int n, int k) {
  int i, j, l, m;
  double t, x;
  l = 0;
  m = n - 1;
  while (l < m) {
    x = a[k];
    i = l;
    j = m;
    do {
      while (a[i] < x)
        i++;
      while (x < a[j])
        j--;
      if (i <= j) {
        t = a[i];
        a[i] = a[j];
        a[j] = t;
        i++;
        j--;
      }
    } while (i <= j);
    if (j < k)
      l = i;
    if (k < i)
      m = j;
  }
  return a[k];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dkth_smallest>:
  endbr64
  push   %r12
  movslq %edx,%rax
  mov    %rdi,%r9
  xor    %r11d,%r11d
  push   %rbp
  lea    (%rdi,%rax,8),%r12
  mov    %rax,%rbp
  push   %rbx
  lea    -0x1(%rsi),%ebx
  test   %ebx,%ebx
  jle    L17
  nop
L73:
  movslq %r11d,%rax
  movsd  (%r12),%xmm1
  mov    %r11d,%esi
  lea    0x1(%r11),%r8d
  lea    (%r9,%rax,8),%rdi
  mov    %ebx,%edx
  movslq %ebx,%rax
  nopl   0x0(%rax)
L78:
  movsd  (%rdi),%xmm2
  lea    (%r9,%rax,8),%rcx
  mov    %esi,%r10d
  movsd  (%rcx),%xmm0
  comisd %xmm2,%xmm1
  ja     L4
  comisd %xmm1,%xmm0
  lea    -0x8(%r9,%rax,8),%rax
  jbe    L0
L58:
  mov    %rax,%rcx
  movsd  (%rax),%xmm0
  sub    $0x8,%rax
  sub    $0x1,%edx
  comisd %xmm1,%xmm0
  ja     L58
L0:
  cmp    %edx,%esi
  jle    L62
L82:
  cmp    %edx,%ebp
  cmovg  %r10d,%r11d
  cmp    %r10d,%ebp
  jl     L51
  cmp    %ebx,%r11d
  jl     L73
L17:
  movsd  (%r12),%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L62:
  sub    $0x1,%edx
  movsd  %xmm0,(%rdi)
  mov    %r8d,%r10d
  movsd  %xmm2,(%rcx)
  cmp    %edx,%r8d
  jg     L82
  movslq %edx,%rax
L4:
  add    $0x8,%rdi
  add    $0x1,%r8d
  add    $0x1,%esi
  jmp    L78
L51:
  cmp    %edx,%r11d
  jge    L17
  mov    %edx,%ebx
  jmp    L73
```

assistant: ```c
double dkth_smallest(double a[], int n, int k) {
  int i, j, l, m;
  double t, x;
  l = 0;
  m = n - 1;
  while (l < m) {
    x = a[k];
    i = l;
    j = m;
    do {
      while (a[i] < x)
        i++;
      while (x < a[j])
        j--;
      if (i <= j) {
        t = a[i];
        a[i] = a[j];
        a[j] = t;
        i++;
        j--;
      }
    } while (i <= j);
    if (j < k)
      l = i;
    if (k < i)
      m = j;
  }
  return a[k];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ln_get_mercury_sdiam>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  D2(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <ln_get_mercury_earth_dist@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x10(%rbp),%xmm0
  divsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
data_type: f64

tool: D2: 3.36

assistant: ```c
double ln_get_mercury_sdiam(double JD) {
  double So = 3.36;
  double dist;
  dist = ln_get_mercury_earth_dist(JD);
  return So / dist;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ln_get_mercury_sdiam>:
  endbr64
  sub    $0x8,%rsp
  call   <ln_get_mercury_earth_dist@plt>
  movapd %xmm0,%xmm1
  movsd  D2(%rip),%xmm0
  divsd  %xmm1,%xmm0
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
datatype: f64

tool: D2: 3.36

assistant: ```c
double ln_get_mercury_sdiam(double JD) {
  double So = 3.36;
  double dist;
  dist = ln_get_mercury_earth_dist(JD);
  return So / dist;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ln_get_mercury_sdiam>:
  endbr64
  sub    $0x8,%rsp
  call   <ln_get_mercury_earth_dist@plt>
  movapd %xmm0,%xmm1
  movsd  D2(%rip),%xmm0
  add    $0x8,%rsp
  divsd  %xmm1,%xmm0
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
type_infered: f64

tool: D2: 3.36

assistant: ```c
double ln_get_mercury_sdiam(double JD) {
  double So = 3.36;
  double dist;
  dist = ln_get_mercury_earth_dist(JD);
  return So / dist;
}
```