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

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xa,0x4(%rsp)
  mov    $0xa,%esi
  lea    D0(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%edi
  call   <cubeByValue@plt>
  mov    0x4(%rsp),%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rdi
  call   <cubeByReference@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D0
type_infered: string

tool: D0: val of n is %d


assistant: ```c
int main() {
  int n = 10;
  printf("val of n is %d\n", n);
  cubeByValue(n);
  printf("val of n is %d\n", n);
  cubeByReference(&n);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rbp
  mov    $0xa,%esi
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xa,0x4(%rsp)
  call   <printf@plt>
  mov    0x4(%rsp),%edi
  call   <cubeByValue@plt>
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rdi
  call   <cubeByReference@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: string

tool: D3: val of n is %d


assistant: ```c
int main() {
  int n = 10;
  printf("val of n is %d\n", n);
  cubeByValue(n);
  printf("val of n is %d\n", n);
  cubeByReference(&n);
}
```
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    D6(%rip),%rbp
  mov    $0xa,%esi
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xa,0x4(%rsp)
  call   <printf@plt>
  mov    0x4(%rsp),%edi
  call   <cubeByValue@plt>
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rdi
  call   <cubeByReference@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D6
type_guessed: string

tool: D6: val of n is %d


assistant: ```c
int main() {
  int n = 10;
  printf("val of n is %d\n", n);
  cubeByValue(n);
  printf("val of n is %d\n", n);
  cubeByReference(&n);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<invert_image>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L7:
  movl   $0x0,-0x8(%rbp)
  jmp    L49
L40:
  mov    -0xc(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  cvtss2sd %xmm0,%xmm0
  movsd  D28(%rip),%xmm2
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  pxor   %xmm0,%xmm0
  comisd %xmm1,%xmm0
  jbe    L1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  cvtss2sd %xmm0,%xmm0
  movsd  D28(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movq   D1(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  jmp    L8
L1:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  D5(%rip),%xmm1
  subss  %xmm1,%xmm0
L8:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L49:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L40
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L7
  nop
  nop
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D28
type_infered: f64
name: unpack_data
arguments: data_label: D1
type_infered: byte[8]
name: unpack_data
arguments: data_label: D5
type_infered: byte[4]

tool: D28: 1

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

tool: {"D5": [0x00,0x00,0x80,0x3f]}

assistant: ```c
void invert_image(image, nx, ny) float *image;
int nx, ny;
{
  int index, i, j;
  for (i = 0; i < ny; i++) {
    for (j = 0; j < nx; j++) {
      index = i * nx + j;
      image[index] = ((image[index] - 1.0) < 0 ? -(image[index] - 1.0) : (image[index] - 1.0));
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<invert_image>:
  endbr64
  test   %edx,%edx
  jle    L30
  push   %rbx
  mov    %rdi,%r10
  mov    %esi,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%edi
  lea    -0x1(%rsi),%ebx
  lea    0x4(%r10),%r11
  movsd  D19(%rip),%xmm3
  pxor   %xmm2,%xmm2
  movss  D16(%rip),%xmm4
  movq   D5(%rip),%xmm5
  jmp    L25
L14:
  xorpd  %xmm5,%xmm0
  pxor   %xmm1,%xmm1
  cvtsd2ss %xmm0,%xmm1
L32:
  movss  %xmm1,(%rcx)
  add    $0x4,%rax
  cmp    %rsi,%rax
  je     L70
L39:
  mov    %rax,%rcx
  movss  (%rax),%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm1,%xmm0
  subsd  %xmm3,%xmm0
  comisd %xmm0,%xmm2
  ja     L14
  subss  %xmm4,%xmm1
  jmp    L32
L70:
  add    $0x1,%edi
  add    %r9d,%r8d
  cmp    %edi,%edx
  je     L66
L25:
  test   %r9d,%r9d
  jle    L70
  movslq %r8d,%rcx
  lea    (%r10,%rcx,4),%rax
  add    %rbx,%rcx
  lea    (%r11,%rcx,4),%rsi
  jmp    L39
L66:
  pop    %rbx
  ret
L30:
  ret
```

tool_calls: name: read_data
arguments: data_label: D19
type_guessed: f64
name: read_data
arguments: data_label: D16
type_guessed: byte[4]
name: read_data
arguments: data_label: D5
type_guessed: byte[8]

tool: D19: 1

tool: {"D16": [0x00,0x00,0x80,0x3f]}

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void invert_image(image, nx, ny) float *image;
int nx, ny;
{
  int index, i, j;
  for (i = 0; i < ny; i++) {
    for (j = 0; j < nx; j++) {
      index = i * nx + j;
      image[index] = ((image[index] - 1.0) < 0 ? -(image[index] - 1.0) : (image[index] - 1.0));
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<invert_image>:
  endbr64
  mov    %rdi,%r9
  mov    %esi,%r8d
  test   %edx,%edx
  jle    L25
  movsd  D12(%rip),%xmm3
  xor    %edi,%edi
  xor    %esi,%esi
  lea    -0x1(%r8),%r11d
  movss  D13(%rip),%xmm4
  movq   D4(%rip),%xmm5
  lea    0x4(%r9),%r10
  pxor   %xmm2,%xmm2
  nopw   0x0(%rax,%rax,1)
L34:
  test   %r8d,%r8d
  jle    L19
  movslq %edi,%rcx
  lea    (%r9,%rcx,4),%rax
  add    %r11,%rcx
  lea    (%r10,%rcx,4),%rcx
  jmp    L41
  nopl   (%rax)
L67:
  subss  %xmm4,%xmm1
  add    $0x4,%rax
  movss  %xmm1,-0x4(%rax)
  cmp    %rcx,%rax
  je     L19
L41:
  movss  (%rax),%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm1,%xmm0
  subsd  %xmm3,%xmm0
  comisd %xmm0,%xmm2
  jbe    L67
  xorpd  %xmm5,%xmm0
  pxor   %xmm1,%xmm1
  add    $0x4,%rax
  cvtsd2ss %xmm0,%xmm1
  movss  %xmm1,-0x4(%rax)
  cmp    %rcx,%rax
  jne    L41
L19:
  add    $0x1,%esi
  add    %r8d,%edi
  cmp    %esi,%edx
  jne    L34
L25:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D12
type_guessed: f64
name: disassemble_data
arguments: data_label: D13
type_guessed: byte[4]
name: disassemble_data
arguments: data_label: D4
type_guessed: byte[8]

tool: D12: 1

tool: {"D13": [0x00,0x00,0x80,0x3f]}

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void invert_image(image, nx, ny) float *image;
int nx, ny;
{
  int index, i, j;
  for (i = 0; i < ny; i++) {
    for (j = 0; j < nx; j++) {
      index = i * nx + j;
      image[index] = ((image[index] - 1.0) < 0 ? -(image[index] - 1.0) : (image[index] - 1.0));
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<invert_image>:
  endbr64
  mov    %edx,%r8d
  test   %edx,%edx
  jle    L70
  test   %esi,%esi
  jle    L70
  lea    -0x1(%rsi),%eax
  movslq %esi,%r9
  movsd  D15(%rip),%xmm3
  xor    %ecx,%ecx
  shl    $0x2,%rax
  shl    $0x2,%r9
  pxor   %xmm2,%xmm2
  mov    $0xfffffffffffffffc,%rsi
  movss  D20(%rip),%xmm4
  movq   D12(%rip),%xmm5
  lea    0x4(%rdi,%rax,1),%rdx
  sub    %rax,%rsi
  xchg   %ax,%ax
L60:
  lea    (%rsi,%rdx,1),%rax
  nopl   0x0(%rax)
L54:
  movss  (%rax),%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm1,%xmm0
  subsd  %xmm3,%xmm0
  comisd %xmm0,%xmm2
  ja     L76
  subss  %xmm4,%xmm1
  add    $0x4,%rax
  movss  %xmm1,-0x4(%rax)
  cmp    %rax,%rdx
  jne    L54
  add    $0x1,%ecx
  add    %r9,%rdx
  cmp    %ecx,%r8d
  jne    L60
L70:
  ret
  nopl   0x0(%rax)
L76:
  xorpd  %xmm5,%xmm0
  add    $0x4,%rax
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x4(%rax)
  cmp    %rax,%rdx
  jne    L54
  add    $0x1,%ecx
  add    %r9,%rdx
  cmp    %ecx,%r8d
  jne    L60
  jmp    L70
```

tool_calls: name: retrieve_data
arguments: data_tag: D15
data_type: f64
name: retrieve_data
arguments: data_tag: D20
data_type: byte[4]
name: retrieve_data
arguments: data_tag: D12
data_type: byte[8]

tool: D15: 1

tool: {"D20": [0x00,0x00,0x80,0x3f]}

tool: {"D12": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void invert_image(image, nx, ny) float *image;
int nx, ny;
{
  int index, i, j;
  for (i = 0; i < ny; i++) {
    for (j = 0; j < nx; j++) {
      index = i * nx + j;
      image[index] = ((image[index] - 1.0) < 0 ? -(image[index] - 1.0) : (image[index] - 1.0));
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<reverseEndianness>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L13
L1:
  mov    -0x14(%rbp),%eax
  shl    $0x3,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x28(%rbp),%eax
  sub    -0x14(%rbp),%eax
  sub    $0x1,%eax
  shl    $0x3,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    $0xff,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  and    -0x8(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
L13:
  mov    -0x14(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L1
  mov    -0x18(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int reverseEndianness(int value, int numbytes) {
  int rev = 0;
  int i;
  for (i = 0; i < numbytes; i++) {
    int srcshift = i * 8;
    int dstshift = (numbytes - i - 1) * 8;
    int srcmask = 0xFF << srcshift;
    int b = (value & srcmask) >> srcshift;
    rev |= (b << dstshift);
  }
  return rev;
}
```
user: What is the c source code of the assembly code below:

```asm
<reverseEndianness>:
  endbr64
  test   %esi,%esi
  jle    L8
  lea    -0x8(,%rsi,8),%r8d
  mov    %r8d,%edx
  mov    $0x0,%esi
  mov    $0xff,%r9d
L21:
  mov    %r8d,%ecx
  sub    %edx,%ecx
  mov    %r9d,%eax
  shl    %cl,%eax
  and    %edi,%eax
  sar    %cl,%eax
  mov    %edx,%ecx
  shl    %cl,%eax
  or     %eax,%esi
  sub    $0x8,%edx
  cmp    $0xfffffff8,%edx
  jne    L21
L1:
  mov    %esi,%eax
  ret
L8:
  mov    $0x0,%esi
  jmp    L1
```

assistant: ```c
int reverseEndianness(int value, int numbytes) {
  int rev = 0;
  int i;
  for (i = 0; i < numbytes; i++) {
    int srcshift = i * 8;
    int dstshift = (numbytes - i - 1) * 8;
    int srcmask = 0xFF << srcshift;
    int b = (value & srcmask) >> srcshift;
    rev |= (b << dstshift);
  }
  return rev;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverseEndianness>:
  endbr64
  test   %esi,%esi
  jle    L13
  lea    -0x8(,%rsi,8),%esi
  xor    %r8d,%r8d
  mov    $0xff,%r9d
  mov    %esi,%edx
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %esi,%ecx
  mov    %r9d,%eax
  sub    %edx,%ecx
  shl    %cl,%eax
  and    %edi,%eax
  sar    %cl,%eax
  mov    %edx,%ecx
  sub    $0x8,%edx
  shl    %cl,%eax
  or     %eax,%r8d
  cmp    $0xfffffff8,%edx
  jne    L11
  mov    %r8d,%eax
  ret
L13:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int reverseEndianness(int value, int numbytes) {
  int rev = 0;
  int i;
  for (i = 0; i < numbytes; i++) {
    int srcshift = i * 8;
    int dstshift = (numbytes - i - 1) * 8;
    int srcmask = 0xFF << srcshift;
    int b = (value & srcmask) >> srcshift;
    rev |= (b << dstshift);
  }
  return rev;
}
```
user: What is the c source code of the assembly code below?

```asm
<reverseEndianness>:
  endbr64
  test   %esi,%esi
  jle    L16
  lea    -0x8(,%rsi,8),%esi
  xor    %r8d,%r8d
  mov    $0xff,%r9d
  mov    %esi,%edx
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %esi,%ecx
  mov    %r9d,%eax
  sub    %edx,%ecx
  shl    %cl,%eax
  and    %edi,%eax
  sar    %cl,%eax
  mov    %edx,%ecx
  sub    $0x8,%edx
  shl    %cl,%eax
  or     %eax,%r8d
  cmp    $0xfffffff8,%edx
  jne    L9
  mov    %r8d,%eax
  ret
L16:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int reverseEndianness(int value, int numbytes) {
  int rev = 0;
  int i;
  for (i = 0; i < numbytes; i++) {
    int srcshift = i * 8;
    int dstshift = (numbytes - i - 1) * 8;
    int srcmask = 0xFF << srcshift;
    int b = (value & srcmask) >> srcshift;
    rev |= (b << dstshift);
  }
  return rev;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D2
datatype: string

tool: D2: hello from mylib.c


assistant: ```c
void print(void) {
  printf("hello from mylib.c\n");
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<print>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_guessed: string

tool: D1: hello from mylib.c


assistant: ```c
void print(void) {
  printf("hello from mylib.c\n");
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print>:
  endbr64
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D8
type_infered: string

tool: D8: hello from mylib.c


assistant: ```c
void print(void) {
  printf("hello from mylib.c\n");
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print>:
  endbr64
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type: string

tool: D8: hello from mylib.c


assistant: ```c
void print(void) {
  printf("hello from mylib.c\n");
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<memcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L13
L1:
  subq   $0x1,-0x28(%rbp)
  addq   $0x8,-0x10(%rbp)
  addq   $0x8,-0x8(%rbp)
L13:
  cmpq   $0x0,-0x28(%rbp)
  je     L31
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L1
L31:
  cmpq   $0x0,-0x28(%rbp)
  je     L35
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  jmp    L21
L35:
  mov    $0x0,%eax
L21:
  pop    %rbp
  ret
```

assistant: ```c
int memcmp(const void *vl, const void *vr, size_t n) {
  const uint8_t *l = vl, *r = vr;
  for (; n && *l == *r; n--, l++, r++)
    ;
  return n ? *l - *r : 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<memcmp>:
  endbr64
  test   %rdx,%rdx
  je     L3
L18:
  mov    (%rsi),%rax
  cmp    %rax,(%rdi)
  jne    L32
  add    $0x8,%rdi
  add    $0x8,%rsi
  sub    $0x1,%rdx
  jne    L18
  mov    $0x0,%eax
  ret
L3:
  mov    $0x0,%eax
  ret
L32:
  test   %rdx,%rdx
  je     L21
  mov    (%rdi),%rax
  sub    (%rsi),%eax
  ret
L21:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int memcmp(const void *vl, const void *vr, size_t n) {
  const uint8_t *l = vl, *r = vr;
  for (; n && *l == *r; n--, l++, r++)
    ;
  return n ? *l - *r : 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<memcmp>:
  endbr64
  xor    %ecx,%ecx
  test   %rdx,%rdx
  jne    L30
  jmp    L1
  nopl   (%rax)
L27:
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L1
L30:
  mov    (%rdi,%rcx,8),%rax
  mov    (%rsi,%rcx,8),%r8
  cmp    %r8,%rax
  je     L27
  sub    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  ret
```

assistant: ```c
int memcmp(const void *vl, const void *vr, size_t n) {
  const uint8_t *l = vl, *r = vr;
  for (; n && *l == *r; n--, l++, r++)
    ;
  return n ? *l - *r : 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<memcmp>:
  endbr64
  xor    %ecx,%ecx
  test   %rdx,%rdx
  jne    L16
  jmp    L37
  nopl   (%rax)
L9:
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L37
L16:
  mov    (%rdi,%rcx,8),%rax
  mov    (%rsi,%rcx,8),%r8
  cmp    %r8,%rax
  je     L9
  sub    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L37:
  xor    %eax,%eax
  ret
```

assistant: ```c
int memcmp(const void *vl, const void *vr, size_t n) {
  const uint8_t *l = vl, *r = vr;
  for (; n && *l == *r; n--, l++, r++)
    ;
  return n ? *l - *r : 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ide_padstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L38
L27:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L28
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L22
L28:
  movl   $0x20,-0x4(%rbp)
L22:
  mov    -0x8(%rbp),%eax
  xor    $0x1,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x8(%rbp)
L38:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L27
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ide_padstr(char *str, const char *src, int len) {
  int i, v;
  for (i = 0; i < len; i++) {
    if (*src != '\0')
      v = *src++;
    else
      v = ' ';
    str[i ^ 1] = v;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ide_padstr>:
  endbr64
  test   %edx,%edx
  jle    L7
  mov    $0x0,%eax
  mov    $0x20,%r9d
  jmp    L9
L10:
  mov    %eax,%ecx
  xor    $0x1,%ecx
  movslq %ecx,%rcx
  mov    %r8b,(%rdi,%rcx,1)
  add    $0x1,%eax
  cmp    %eax,%edx
  je     L7
L9:
  movzbl (%rsi),%ecx
  mov    %r9d,%r8d
  test   %cl,%cl
  je     L10
  add    $0x1,%rsi
  movsbl %cl,%r8d
  jmp    L10
L7:
  ret
```

assistant: ```c
void ide_padstr(char *str, const char *src, int len) {
  int i, v;
  for (i = 0; i < len; i++) {
    if (*src != '\0')
      v = *src++;
    else
      v = ' ';
    str[i ^ 1] = v;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<ide_padstr>:
  endbr64
  xor    %eax,%eax
  test   %edx,%edx
  jg     L40
  jmp    L43
  nopl   0x0(%rax)
L7:
  add    $0x1,%rsi
L30:
  mov    %eax,%ecx
  add    $0x1,%eax
  xor    $0x1,%ecx
  movslq %ecx,%rcx
  mov    %r8b,(%rdi,%rcx,1)
  cmp    %eax,%edx
  je     L43
L40:
  movzbl (%rsi),%r8d
  test   %r8b,%r8b
  jne    L7
  mov    $0x20,%r8d
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L43:
  ret
```

assistant: ```c
void ide_padstr(char *str, const char *src, int len) {
  int i, v;
  for (i = 0; i < len; i++) {
    if (*src != '\0')
      v = *src++;
    else
      v = ' ';
    str[i ^ 1] = v;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<ide_padstr>:
  endbr64
  xor    %eax,%eax
  test   %edx,%edx
  jg     L36
  jmp    L46
  nopl   0x0(%rax)
L8:
  add    $0x1,%rsi
L32:
  mov    %eax,%ecx
  add    $0x1,%eax
  xor    $0x1,%ecx
  movslq %ecx,%rcx
  mov    %r8b,(%rdi,%rcx,1)
  cmp    %eax,%edx
  je     L46
L36:
  movzbl (%rsi),%r8d
  test   %r8b,%r8b
  jne    L8
  mov    $0x20,%r8d
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L46:
  ret
```

assistant: ```c
void ide_padstr(char *str, const char *src, int len) {
  int i, v;
  for (i = 0; i < len; i++) {
    if (*src != '\0')
      v = *src++;
    else
      v = ' ';
    str[i ^ 1] = v;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<gmt2local>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  mov    -0x48(%rbp),%rax
  test   %rax,%rax
  jne    L8
  mov    $0x0,%edi
  call   <time@plt>
  mov    %rax,-0x48(%rbp)
L8:
  lea    -0x40(%rbp),%rbx
  lea    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <gmtime@plt>
  mov    0x8(%rax),%rdx
  mov    (%rax),%rax
  mov    %rax,(%rbx)
  mov    %rdx,0x8(%rbx)
  lea    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <localtime@plt>
  mov    %rax,%r12
  mov    (%r12),%eax
  mov    (%rbx),%edx
  sub    %edx,%eax
  imul   $0xe10,%eax,%edx
  mov    0x4(%r12),%eax
  mov    0x4(%rbx),%ecx
  sub    %ecx,%eax
  imul   $0x3c,%eax,%eax
  lea    (%rdx,%rax,1),%r14d
  mov    0x8(%r12),%eax
  mov    0x8(%rbx),%edx
  sub    %edx,%eax
  mov    %eax,%r13d
  test   %r13d,%r13d
  jne    L19
  mov    0xc(%r12),%eax
  mov    0xc(%rbx),%edx
  sub    %edx,%eax
  mov    %eax,%r13d
L19:
  imul   $0x15180,%r13d,%eax
  add    %eax,%r14d
  mov    %r14d,%eax
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  add    $0x30,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
```

assistant: ```c
int32_t gmt2local(time_t t) {
  register int dt, dir;
  register struct tm *gmt, *loc;
  struct tm sgmt;
  if (t == 0)
    t = time(NULL);
  gmt = &sgmt;
  *gmt = *gmtime(&t);
  loc = localtime(&t);
  dt = (loc->tm_hour - gmt->tm_hour) * 60 * 60 + (loc->tm_min - gmt->tm_min) * 60;
  dir = loc->tm_year - gmt->tm_year;
  if (dir == 0)
    dir = loc->tm_yday - gmt->tm_yday;
  dt += dir * 24 * 60 * 60;
  return (dt);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<gmt2local>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,0x8(%rsp)
  test   %rdi,%rdi
  je     L3
L2:
  lea    0x8(%rsp),%r14
  mov    %r14,%rdi
  call   <gmtime@plt>
  mov    (%rax),%r13d
  mov    0x4(%rax),%r12d
  mov    0x8(%rax),%ebp
  mov    0xc(%rax),%ebx
  mov    %r14,%rdi
  call   <localtime@plt>
  mov    (%rax),%ecx
  sub    %r13d,%ecx
  imul   $0xe10,%ecx,%ecx
  mov    0x4(%rax),%edx
  sub    %r12d,%edx
  imul   $0x3c,%edx,%edx
  add    %edx,%ecx
  mov    0x8(%rax),%edx
  sub    %ebp,%edx
  jne    L28
  mov    0xc(%rax),%edx
  sub    %ebx,%edx
L28:
  imul   $0x15180,%edx,%eax
  add    %ecx,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  mov    $0x0,%edi
  call   <time@plt>
  mov    %rax,0x8(%rsp)
  jmp    L2
```

assistant: ```c
int32_t gmt2local(time_t t) {
  register int dt, dir;
  register struct tm *gmt, *loc;
  struct tm sgmt;
  if (t == 0)
    t = time(NULL);
  gmt = &sgmt;
  *gmt = *gmtime(&t);
  loc = localtime(&t);
  dt = (loc->tm_hour - gmt->tm_hour) * 60 * 60 + (loc->tm_min - gmt->tm_min) * 60;
  dir = loc->tm_year - gmt->tm_year;
  if (dir == 0)
    dir = loc->tm_yday - gmt->tm_yday;
  dt += dir * 24 * 60 * 60;
  return (dt);
}
```