ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <bar@plt>
  mov    $0x0,%eax
  call   <bar@plt>
  mov    $0x0,%eax
  call   <bar@plt>
  mov    $0x0,%eax
  call   <bar@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void foo(void) {
#pragma omp parallel
  bar();
#pragma omp p
  bar();

#pragma omp parallel

  bar();

#pragma omp p

  bar();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <bar@plt>
```

assistant: ```c
void foo(void) {
#pragma omp parallel
  bar();
#pragma omp p
  bar();

#pragma omp parallel

  bar();

#pragma omp p

  bar();
}
```
user: What is the c source code of the assembly code below?

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <bar@plt>
```

assistant: ```c
void foo(void) {
#pragma omp parallel
  bar();
#pragma omp p
  bar();

#pragma omp parallel

  bar();

#pragma omp p

  bar();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_ram>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x18(%rbp)
  lea    D19(%rip),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L43
  lea    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  lea    D7(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fscanf@plt>
  cmp    $0x1,%eax
  je     L25
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  jmp    L40
L25:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  je     L37
  mov    -0x18(%rbp),%rax
  add    $0x200,%rax
  shr    $0xa,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rcx
  lea    D39(%rip),%rdx
  mov    %rax,%rdi
  call   <snprintf@plt>
  jmp    L40
L43:
  nop
  jmp    L40
L37:
  nop
L40:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L67
  call   <__stack_chk_fail@plt>
L67:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D19
type_guessed: string
name: disassemble_data
arguments: tag: D17
type_guessed: string
name: disassemble_data
arguments: tag: D7
type_guessed: string
name: disassemble_data
arguments: tag: D39
type_guessed: string

tool: D19: r

tool: D17: /proc/meminfo

tool: D7: MemTotal: %llu kB

tool: D39: (%llu MB)

assistant: ```c
void get_ram(char *buffer, size_t size) {
  unsigned long long ram = 0;
  FILE *meminfo = fopen("/proc/meminfo", "r");
  if (meminfo == NULL)
    return;
  if (fscanf(meminfo, "MemTotal: %llu kB", &ram) != 1) {
    fclose(meminfo);
    return;
  }
  fclose(meminfo);
  if (ram == 0)
    return;
  snprintf(buffer, size, "(%llu MB)", (ram + 512) / 1024);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_ram>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  lea    D14(%rip),%rsi
  lea    D32(%rip),%rdi
  call   <fopen@plt>
  test   %rax,%rax
  je     L8
  mov    %rax,%rbx
  mov    %rsp,%rdx
  lea    D1(%rip),%rsi
  mov    %rax,%rdi
  call   <fscanf@plt>
  cmp    $0x1,%eax
  je     L18
  mov    %rbx,%rdi
  call   <fclose@plt>
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L8
  add    $0x200,%rcx
  shr    $0xa,%rcx
  lea    D29(%rip),%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <snprintf@plt>
  jmp    L8
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D14
datatype: string
name: disassemble_data
arguments: label: D32
datatype: string
name: disassemble_data
arguments: label: D1
datatype: string
name: disassemble_data
arguments: label: D29
datatype: string

tool: D14: r

tool: D32: /proc/meminfo

tool: D1: MemTotal: %llu kB

tool: D29: (%llu MB)

assistant: ```c
void get_ram(char *buffer, size_t size) {
  unsigned long long ram = 0;
  FILE *meminfo = fopen("/proc/meminfo", "r");
  if (meminfo == NULL)
    return;
  if (fscanf(meminfo, "MemTotal: %llu kB", &ram) != 1) {
    fclose(meminfo);
    return;
  }
  fclose(meminfo);
  if (ram == 0)
    return;
  snprintf(buffer, size, "(%llu MB)", (ram + 512) / 1024);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_ram>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  lea    D10(%rip),%rsi
  push   %r12
  mov    %rdi,%r12
  lea    D9(%rip),%rdi
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <fopen@plt>
  test   %rax,%rax
  je     L37
  mov    %rax,%rdi
  mov    %rsp,%rdx
  lea    D22(%rip),%rsi
  mov    %rax,%rbp
  call   <fscanf@plt>
  mov    %rbp,%rdi
  cmp    $0x1,%eax
  je     L0
  call   <fclose@plt>
L37:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x10,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L0:
  call   <fclose@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L37
  add    $0x200,%rcx
  lea    D36(%rip),%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  shr    $0xa,%rcx
  call   <snprintf@plt>
  jmp    L37
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
type_guessed: string
name: unpack_data
arguments: label: D9
type_guessed: string
name: unpack_data
arguments: label: D22
type_guessed: string
name: unpack_data
arguments: label: D36
type_guessed: string

tool: D10: r

tool: D9: /proc/meminfo

tool: D22: MemTotal: %llu kB

tool: D36: (%llu MB)

assistant: ```c
void get_ram(char *buffer, size_t size) {
  unsigned long long ram = 0;
  FILE *meminfo = fopen("/proc/meminfo", "r");
  if (meminfo == NULL)
    return;
  if (fscanf(meminfo, "MemTotal: %llu kB", &ram) != 1) {
    fclose(meminfo);
    return;
  }
  fclose(meminfo);
  if (ram == 0)
    return;
  snprintf(buffer, size, "(%llu MB)", (ram + 512) / 1024);
}
```
user: What is the c source code of the assembly code below:

```asm
<get_ram>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  lea    D2(%rip),%rsi
  push   %r12
  mov    %rdi,%r12
  lea    D20(%rip),%rdi
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <fopen@plt>
  test   %rax,%rax
  je     L34
  mov    %rax,%rdi
  mov    %rsp,%rdx
  lea    D33(%rip),%rsi
  mov    %rax,%rbp
  call   <fscanf@plt>
  mov    %rbp,%rdi
  cmp    $0x1,%eax
  je     L38
  call   <fclose@plt>
L34:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x10,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L38:
  call   <fclose@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L34
  add    $0x200,%rcx
  lea    D37(%rip),%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  shr    $0xa,%rcx
  call   <snprintf@plt>
  jmp    L34
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
datatype: string
name: unpack_data
arguments: data_label: D20
datatype: string
name: unpack_data
arguments: data_label: D33
datatype: string
name: unpack_data
arguments: data_label: D37
datatype: string

tool: D2: r

tool: D20: /proc/meminfo

tool: D33: MemTotal: %llu kB

tool: D37: (%llu MB)

assistant: ```c
void get_ram(char *buffer, size_t size) {
  unsigned long long ram = 0;
  FILE *meminfo = fopen("/proc/meminfo", "r");
  if (meminfo == NULL)
    return;
  if (fscanf(meminfo, "MemTotal: %llu kB", &ram) != 1) {
    fclose(meminfo);
    return;
  }
  fclose(meminfo);
  if (ram == 0)
    return;
  snprintf(buffer, size, "(%llu MB)", (ram + 512) / 1024);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<readwriteblock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <readblock@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  je     L3
  mov    -0x4(%rbp),%eax
  jmp    L9
L3:
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <r_write@plt>
L9:
  leave
  ret
```

assistant: ```c
int readwriteblock(int fromfd, int tofd, char *buf, int size) {
  int bytesread;
  bytesread = readblock(fromfd, buf, size);
  if (bytesread != size)
    return bytesread;
  return r_write(tofd, buf, size);
}
```
user: Translate the below assembly code into C programming language:

```asm
<readwriteblock>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r12d
  mov    %rdx,%rbp
  mov    %ecx,%ebx
  mov    %ecx,%edx
  mov    %rbp,%rsi
  call   <readblock@plt>
  cmp    %eax,%ebx
  je     L12
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <r_write@plt>
  jmp    L8
```

assistant: ```c
int readwriteblock(int fromfd, int tofd, char *buf, int size) {
  int bytesread;
  bytesread = readblock(fromfd, buf, size);
  if (bytesread != size)
    return bytesread;
  return r_write(tofd, buf, size);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readwriteblock>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdx,%rbp
  mov    %ecx,%edx
  mov    %rbp,%rsi
  call   <readblock@plt>
  cmp    %eax,%r12d
  je     L6
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <r_write@plt>
```

assistant: ```c
int readwriteblock(int fromfd, int tofd, char *buf, int size) {
  int bytesread;
  bytesread = readblock(fromfd, buf, size);
  if (bytesread != size)
    return bytesread;
  return r_write(tofd, buf, size);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<readwriteblock>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdx,%rbp
  mov    %ecx,%edx
  mov    %rbp,%rsi
  call   <readblock@plt>
  cmp    %eax,%r12d
  je     L0
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <r_write@plt>
```

assistant: ```c
int readwriteblock(int fromfd, int tofd, char *buf, int size) {
  int bytesread;
  bytesread = readblock(fromfd, buf, size);
  if (bytesread != size)
    return bytesread;
  return r_write(tofd, buf, size);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x70(%rbp)
  movl   $0x1,-0x6c(%rbp)
  movl   $0x2,-0x68(%rbp)
  movl   $0x2,-0x64(%rbp)
  mov    $0x2,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D28
data_type: string

tool: D28: %ld


assistant: ```c
int main() {
  int a[2][3][4];
  int b[][2] = {{1, 1}, {2, 2}};
  printf("%ld\n", (sizeof(a) / sizeof(a[0])));
  printf("%ld\n", (sizeof(a[0]) / sizeof(a[0][0])));
  printf("%ld\n", (sizeof(a[0][0]) / sizeof(a[0][0][0])));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2,%esi
  lea    D4(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D4
type: string

tool: D4: %ld


assistant: ```c
int main() {
  int a[2][3][4];
  int b[][2] = {{1, 1}, {2, 2}};
  printf("%ld\n", (sizeof(a) / sizeof(a[0])));
  printf("%ld\n", (sizeof(a[0]) / sizeof(a[0][0])));
  printf("%ld\n", (sizeof(a[0][0]) / sizeof(a[0][0][0])));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rbp
  mov    $0x2,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
datatype: string

tool: D9: %ld


assistant: ```c
int main() {
  int a[2][3][4];
  int b[][2] = {{1, 1}, {2, 2}};
  printf("%ld\n", (sizeof(a) / sizeof(a[0])));
  printf("%ld\n", (sizeof(a[0]) / sizeof(a[0][0])));
  printf("%ld\n", (sizeof(a[0][0]) / sizeof(a[0][0][0])));
  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
  push   %rbp
  lea    D9(%rip),%rbp
  mov    $0x2,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: label: D9
type_guessed: string

tool: D9: %ld


assistant: ```c
int main() {
  int a[2][3][4];
  int b[][2] = {{1, 1}, {2, 2}};
  printf("%ld\n", (sizeof(a) / sizeof(a[0])));
  printf("%ld\n", (sizeof(a[0]) / sizeof(a[0][0])));
  printf("%ld\n", (sizeof(a[0][0]) / sizeof(a[0][0][0])));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movss  D2(%rip),%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movd   %eax,%xmm0
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
type: byte[4]
name: disassemble_data
arguments: tag: D17
type: string

tool: {"D2": [0xde,0xb3,0x7e,0x30]}

tool: D17: %e


assistant: ```c
int main(void) {
  float num = (3.31e-8 * 2.01e-7) / (7.16e-6 + 2.01e-8);
  printf("%e\n", num);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movss  D17(%rip),%xmm0
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D17
type_infered: byte[4]
name: extract_symbolic_data
arguments: data_label: D6
type_infered: string

tool: {"D17": [0xde,0xb3,0x7e,0x30]}

tool: D6: %e


assistant: ```c
int main(void) {
  float num = (3.31e-8 * 2.01e-7) / (7.16e-6 + 2.01e-8);
  printf("%e\n", num);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movss  D1(%rip),%xmm0
  lea    D12(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_infered: byte[4]
name: parse_data
arguments: data_label: D12
type_infered: string

tool: {"D1": [0xde,0xb3,0x7e,0x30]}

tool: D12: %e


assistant: ```c
int main(void) {
  float num = (3.31e-8 * 2.01e-7) / (7.16e-6 + 2.01e-8);
  printf("%e\n", num);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movss  D8(%rip),%xmm0
  lea    D13(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D13
data_type: string

tool: {"D8": [0xde,0xb3,0x7e,0x30]}

tool: D13: %e


assistant: ```c
int main(void) {
  float num = (3.31e-8 * 2.01e-7) / (7.16e-6 + 2.01e-8);
  printf("%e\n", num);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ulpscale_f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movd   %eax,%xmm0
  call   <asuint@plt>
  sar    $0x17,%eax
  and    $0xff,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L4
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  sub    $0x96,%eax
  leave
  ret
```

assistant: ```c
int ulpscale_f(float x) {
  int e = asuint(x) >> 23 & 0xff;
  if (!e)
    e++;
  return e - 0x7f - 23;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ulpscale_f>:
  endbr64
  sub    $0x8,%rsp
  call   <asuint@plt>
  sar    $0x17,%eax
  and    $0xff,%eax
  mov    $0x1,%edx
  cmove  %edx,%eax
  sub    $0x96,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ulpscale_f(float x) {
  int e = asuint(x) >> 23 & 0xff;
  if (!e)
    e++;
  return e - 0x7f - 23;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ulpscale_f>:
  endbr64
  sub    $0x8,%rsp
  call   <asuint@plt>
  sar    $0x17,%eax
  movzbl %al,%edx
  test   %edx,%edx
  lea    -0x96(%rdx),%eax
  mov    $0xffffff6b,%edx
  cmove  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ulpscale_f(float x) {
  int e = asuint(x) >> 23 & 0xff;
  if (!e)
    e++;
  return e - 0x7f - 23;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ulpscale_f>:
  endbr64
  sub    $0x8,%rsp
  call   <asuint@plt>
  sar    $0x17,%eax
  movzbl %al,%edx
  test   %edx,%edx
  lea    -0x96(%rdx),%eax
  mov    $0xffffff6b,%edx
  cmove  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ulpscale_f(float x) {
  int e = asuint(x) >> 23 & 0xff;
  if (!e)
    e++;
  return e - 0x7f - 23;
}
```
user: What is the c source code of the assembly code below:

```asm
<SKP_Silk_fit_LTP>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L56
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sar    %eax
  add    $0x1,%eax
  sar    %eax
  cmp    $0x7fff,%eax
  jg     L57
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sar    %eax
  add    $0x1,%eax
  sar    %eax
  cmp    $0xffff8000,%eax
  jl     L44
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sar    %eax
  add    $0x1,%eax
  sar    %eax
  jmp    L25
L44:
  mov    $0xffff8000,%eax
  jmp    L25
L57:
  mov    $0x7fff,%eax
L25:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    (%rdx,%rdx,1),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    %ax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L56:
  cmpl   $0x4,-0x4(%rbp)
  jle    L17
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void SKP_Silk_fit_LTP(int LTP_coefs_Q16[5], short LTP_coefs_Q14[5]) {
  int i;
  for (i = 0; i < 5; i++) {
    LTP_coefs_Q14[i] = (short)((((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1)) > 0x7FFF ? 0x7FFF : ((((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1)) < ((short)0x8000) ? ((short)0x8000) : (((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1))));
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<SKP_Silk_fit_LTP>:
  endbr64
  mov    $0x0,%edx
  mov    $0x7fff,%r9d
  mov    $0xffff8000,%r10d
  jmp    L16
L4:
  mov    %cx,(%rsi,%rdx,2)
  add    $0x1,%rdx
  cmp    $0x5,%rdx
  je     L19
L16:
  mov    (%rdi,%rdx,4),%eax
  sar    %eax
  add    $0x1,%eax
  mov    %eax,%r8d
  sar    %r8d
  mov    %r9d,%ecx
  cmp    $0xffff,%eax
  jg     L4
  cmp    $0xffff8000,%r8d
  mov    %r8d,%ecx
  cmovl  %r10d,%ecx
  jmp    L4
L19:
  ret
```

assistant: ```c
void SKP_Silk_fit_LTP(int LTP_coefs_Q16[5], short LTP_coefs_Q14[5]) {
  int i;
  for (i = 0; i < 5; i++) {
    LTP_coefs_Q14[i] = (short)((((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1)) > 0x7FFF ? 0x7FFF : ((((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1)) < ((short)0x8000) ? ((short)0x8000) : (((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1))));
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<SKP_Silk_fit_LTP>:
  endbr64
  xor    %edx,%edx
  mov    $0xffff8000,%r9d
L9:
  mov    (%rdi,%rdx,4),%eax
  mov    $0x7fff,%ecx
  sar    %eax
  add    $0x1,%eax
  mov    %eax,%r8d
  sar    %r8d
  cmp    $0xffff,%eax
  jg     L19
  cmp    $0xffff8000,%r8d
  mov    %r8d,%ecx
  cmovl  %r9d,%ecx
L19:
  mov    %cx,(%rsi,%rdx,2)
  add    $0x1,%rdx
  cmp    $0x5,%rdx
  jne    L9
  ret
```

assistant: ```c
void SKP_Silk_fit_LTP(int LTP_coefs_Q16[5], short LTP_coefs_Q14[5]) {
  int i;
  for (i = 0; i < 5; i++) {
    LTP_coefs_Q14[i] = (short)((((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1)) > 0x7FFF ? 0x7FFF : ((((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1)) < ((short)0x8000) ? ((short)0x8000) : (((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1))));
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<SKP_Silk_fit_LTP>:
  endbr64
  mov    (%rdi),%ecx
  mov    %rdi,%rdx
  mov    %rsi,%rax
  mov    $0x7fff,%esi
  sar    %ecx
  add    $0x1,%ecx
  mov    %ecx,%edi
  sar    %edi
  cmp    $0xffff,%ecx
  jg     L32
  cmp    $0xffff8000,%edi
  mov    $0xffff8000,%esi
  cmovge %edi,%esi
L32:
  mov    0x4(%rdx),%ecx
  mov    %si,(%rax)
  mov    $0x7fff,%esi
  sar    %ecx
  add    $0x1,%ecx
  mov    %ecx,%edi
  sar    %edi
  cmp    $0xffff,%ecx
  jg     L47
  cmp    $0xffff8000,%edi
  mov    $0xffff8000,%esi
  cmovge %edi,%esi
L47:
  mov    0x8(%rdx),%ecx
  mov    %si,0x2(%rax)
  mov    $0x7fff,%esi
  sar    %ecx
  add    $0x1,%ecx
  mov    %ecx,%edi
  sar    %edi
  cmp    $0xffff,%ecx
  jg     L42
  cmp    $0xffff8000,%edi
  mov    $0xffff8000,%esi
  cmovge %edi,%esi
L42:
  mov    0xc(%rdx),%ecx
  mov    %si,0x4(%rax)
  mov    $0x7fff,%esi
  sar    %ecx
  add    $0x1,%ecx
  mov    %ecx,%edi
  sar    %edi
  cmp    $0xffff,%ecx
  jg     L37
  cmp    $0xffff8000,%edi
  mov    $0xffff8000,%esi
  cmovge %edi,%esi
L37:
  mov    0x10(%rdx),%edx
  mov    %si,0x6(%rax)
  mov    $0x7fff,%ecx
  sar    %edx
  add    $0x1,%edx
  mov    %edx,%esi
  sar    %esi
  cmp    $0xffff,%edx
  jg     L20
  cmp    $0xffff8000,%esi
  mov    $0xffff8000,%ecx
  cmovge %esi,%ecx
L20:
  mov    %cx,0x8(%rax)
  ret
```

assistant: ```c
void SKP_Silk_fit_LTP(int LTP_coefs_Q16[5], short LTP_coefs_Q14[5]) {
  int i;
  for (i = 0; i < 5; i++) {
    LTP_coefs_Q14[i] = (short)((((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1)) > 0x7FFF ? 0x7FFF : ((((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1)) < ((short)0x8000) ? ((short)0x8000) : (((2) == 1 ? ((LTP_coefs_Q16[i]) >> 1) + ((LTP_coefs_Q16[i]) & 1) : (((LTP_coefs_Q16[i]) >> ((2) - 1)) + 1) >> 1))));
  }
}
```