ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<modp_itoa10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L0:
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  mov    %eax,%edx
  shr    $0x3,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L0
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strreverse@plt>
  nop
  leave
  ret
```

assistant: ```c
void modp_itoa10(int32_t value, char *str) {
  char *wstr = str;
  unsigned int uvalue = (value < 0) ? -value : value;
  do
    *wstr++ = (char)(48 + (uvalue % 10));
  while (uvalue /= 10);
  if (value < 0)
    *wstr++ = '-';
  *wstr = '\0';
  strreverse(str, wstr - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<modp_itoa10>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%edx
  mov    %rsi,%rdi
  mov    %rsi,%rcx
  mov    $0xcccccccd,%r10d
L5:
  mov    %rcx,%rsi
  add    $0x1,%rcx
  mov    %edx,%eax
  imul   %r10,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%r9d
  add    %r9d,%r9d
  mov    %edx,%r8d
  sub    %r9d,%r8d
  add    $0x30,%r8d
  mov    %r8b,-0x1(%rcx)
  mov    %edx,%r8d
  mov    %eax,%edx
  cmp    $0x9,%r8d
  ja     L5
  movb   $0x0,(%rcx)
  call   <strreverse@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void modp_itoa10(int32_t value, char *str) {
  char *wstr = str;
  unsigned int uvalue = (value < 0) ? -value : value;
  do
    *wstr++ = (char)(48 + (uvalue % 10));
  while (uvalue /= 10);
  if (value < 0)
    *wstr++ = '-';
  *wstr = '\0';
  strreverse(str, wstr - 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<modp_itoa10>:
  endbr64
  mov    %edi,%edx
  mov    %rsi,%rcx
  mov    %rsi,%rdi
  mov    $0xcccccccd,%r10d
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %edx,%eax
  mov    %edx,%r8d
  mov    %rcx,%rsi
  add    $0x1,%rcx
  imul   %r10,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%r9d
  add    %r9d,%r9d
  sub    %r9d,%r8d
  add    $0x30,%r8d
  mov    %r8b,-0x1(%rcx)
  mov    %edx,%r8d
  mov    %eax,%edx
  cmp    $0x9,%r8d
  ja     L8
  movb   $0x0,(%rcx)
  jmp    1050 <strreverse@plt>
```

assistant: ```c
void modp_itoa10(int32_t value, char *str) {
  char *wstr = str;
  unsigned int uvalue = (value < 0) ? -value : value;
  do
    *wstr++ = (char)(48 + (uvalue % 10));
  while (uvalue /= 10);
  if (value < 0)
    *wstr++ = '-';
  *wstr = '\0';
  strreverse(str, wstr - 1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<modp_itoa10>:
  endbr64
  mov    %edi,%edx
  mov    %rsi,%rcx
  mov    %rsi,%rdi
  mov    $0xcccccccd,%r10d
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %edx,%eax
  mov    %edx,%r8d
  mov    %rcx,%rsi
  add    $0x1,%rcx
  imul   %r10,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%r9d
  add    %r9d,%r9d
  sub    %r9d,%r8d
  add    $0x30,%r8d
  mov    %r8b,-0x1(%rcx)
  mov    %edx,%r8d
  mov    %eax,%edx
  cmp    $0x9,%r8d
  ja     L8
  movb   $0x0,(%rcx)
  jmp    1050 <strreverse@plt>
```

assistant: ```c
void modp_itoa10(int32_t value, char *str) {
  char *wstr = str;
  unsigned int uvalue = (value < 0) ? -value : value;
  do
    *wstr++ = (char)(48 + (uvalue % 10));
  while (uvalue /= 10);
  if (value < 0)
    *wstr++ = '-';
  *wstr = '\0';
  strreverse(str, wstr - 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sys_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L9
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L9:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <vfprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void sys_error(char *format, ...) {
  va_list args;
  va_start(args, format);
  vfprintf(stderr, format, args);
  va_end(args);
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sys_error>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <vfprintf@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void sys_error(char *format, ...) {
  va_list args;
  va_start(args, format);
  vfprintf(stderr, format, args);
  va_end(args);
  exit(1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sys_error>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    D8(%rip),%rax
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <vfprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void sys_error(char *format, ...) {
  va_list args;
  va_start(args, format);
  vfprintf(stderr, format, args);
  va_end(args);
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sys_error>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    D7(%rip),%rax
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <vfprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void sys_error(char *format, ...) {
  va_list args;
  va_start(args, format);
  vfprintf(stderr, format, args);
  va_end(args);
  exit(1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<SHA512_Init_iv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,%rdx
  mov    (%rcx),%rax
  mov    %rax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x7,-0x4(%rbp)
  jle    L10
  mov    -0x18(%rbp),%rax
  movq   $0x0,0x20(%rax)
  mov    -0x18(%rbp),%rax
  movq   $0x0,0x18(%rax)
  mov    -0x18(%rbp),%rax
  movq   $0x0,0x10(%rax)
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,0x8(%rax)
  mov    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int SHA512_Init_iv(SHA512_CTX *c, unsigned long long *h) {
  int i;
  for (i = 0; i < 8; i++)
    c->h[i] = *(h + i);
  c->Nl = 0;
  c->Nh = 0;
  c->num = 0;
  c->md_len = SHA512_DIGEST_LENGTH;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<SHA512_Init_iv>:
  endbr64
  mov    $0x0,%eax
L4:
  mov    (%rsi,%rax,1),%rcx
  mov    (%rdi),%rdx
  mov    %rcx,(%rdx,%rax,1)
  add    $0x8,%rax
  cmp    $0x40,%rax
  jne    L4
  movq   $0x0,0x20(%rdi)
  movq   $0x0,0x18(%rdi)
  movq   $0x0,0x10(%rdi)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x8(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int SHA512_Init_iv(SHA512_CTX *c, unsigned long long *h) {
  int i;
  for (i = 0; i < 8; i++)
    c->h[i] = *(h + i);
  c->Nl = 0;
  c->Nh = 0;
  c->num = 0;
  c->md_len = SHA512_DIGEST_LENGTH;
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<SHA512_Init_iv>:
  endbr64
  mov    (%rdi),%rcx
  xor    %eax,%eax
  nopl   0x0(%rax)
L7:
  mov    (%rsi,%rax,1),%rdx
  mov    %rdx,(%rcx,%rax,1)
  add    $0x8,%rax
  cmp    $0x40,%rax
  jne    L7
  mov    D1(%rip),%rax
  movq   $0x0,0x20(%rdi)
  movq   $0x0,0x18(%rdi)
  mov    (%rax),%eax
  movq   $0x0,0x10(%rdi)
  mov    %eax,0x8(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int SHA512_Init_iv(SHA512_CTX *c, unsigned long long *h) {
  int i;
  for (i = 0; i < 8; i++)
    c->h[i] = *(h + i);
  c->Nl = 0;
  c->Nh = 0;
  c->num = 0;
  c->md_len = SHA512_DIGEST_LENGTH;
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<SHA512_Init_iv>:
  endbr64
  mov    (%rdi),%rdx
  lea    0x8(%rsi),%rcx
  mov    %rdi,%rax
  cmp    %rcx,%rdx
  je     L0
  movdqu (%rsi),%xmm1
  movups %xmm1,(%rdx)
  movdqu 0x10(%rsi),%xmm2
  movups %xmm2,0x10(%rdx)
  movdqu 0x20(%rsi),%xmm3
  movups %xmm3,0x20(%rdx)
  movdqu 0x30(%rsi),%xmm4
  movups %xmm4,0x30(%rdx)
L7:
  mov    D4(%rip),%rdx
  pxor   %xmm0,%xmm0
  movq   $0x0,0x20(%rax)
  movups %xmm0,0x10(%rax)
  mov    (%rdx),%edx
  mov    %edx,0x8(%rax)
  mov    $0x1,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    (%rsi),%rcx
  mov    %rcx,(%rdx)
  mov    0x8(%rsi),%rcx
  mov    %rcx,0x8(%rdx)
  mov    0x10(%rsi),%rcx
  mov    %rcx,0x10(%rdx)
  mov    0x18(%rsi),%rcx
  mov    %rcx,0x18(%rdx)
  mov    0x20(%rsi),%rcx
  mov    %rcx,0x20(%rdx)
  mov    0x28(%rsi),%rcx
  mov    %rcx,0x28(%rdx)
  mov    0x30(%rsi),%rcx
  mov    %rcx,0x30(%rdx)
  mov    0x38(%rsi),%rcx
  mov    %rcx,0x38(%rdx)
  jmp    L7
```

assistant: ```c
int SHA512_Init_iv(SHA512_CTX *c, unsigned long long *h) {
  int i;
  for (i = 0; i < 8; i++)
    c->h[i] = *(h + i);
  c->Nl = 0;
  c->Nh = 0;
  c->num = 0;
  c->md_len = SHA512_DIGEST_LENGTH;
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<adgold_pseudo_stereo_iir>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  movss  %xmm0,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movss  D158(%rip),%xmm0
  movss  %xmm0,-0x20(%rbp)
  movss  D168(%rip),%xmm0
  movss  %xmm0,-0x1c(%rbp)
  movss  D158(%rip),%xmm0
  movss  %xmm0,-0x18(%rbp)
  movss  D41(%rip),%xmm0
  movss  %xmm0,-0x14(%rbp)
  movss  D105(%rip),%xmm0
  movss  %xmm0,-0x10(%rbp)
  movss  D127(%rip),%xmm0
  movss  %xmm0,-0xc(%rbp)
  movl   $0x2,-0x24(%rbp)
  jmp    L22
L16:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    D88(%rip),%rax
  movss  (%rdx,%rax,1),%xmm0
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    D88(%rip),%rax
  movss  %xmm0,(%rdx,%rax,1)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    D72(%rip),%rax
  movss  (%rdx,%rax,1),%xmm0
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    D72(%rip),%rax
  movss  %xmm0,(%rdx,%rax,1)
  subl   $0x1,-0x24(%rbp)
L22:
  cmpl   $0x0,-0x24(%rbp)
  jg     L16
  movss  -0x34(%rbp),%xmm0
  movss  %xmm0,D98(%rip)
  movss  -0x20(%rbp),%xmm1
  movss  D98(%rip),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,D98(%rip)
  movl   $0x1,-0x24(%rbp)
  jmp    L43
L33:
  movss  D98(%rip),%xmm1
  mov    -0x24(%rbp),%eax
  cltq
  movss  -0x20(%rbp,%rax,4),%xmm2
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    D98(%rip),%rax
  movss  (%rdx,%rax,1),%xmm0
  mulss  %xmm2,%xmm0
  mov    -0x24(%rbp),%eax
  cltq
  movss  -0x14(%rbp,%rax,4),%xmm3
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    D98(%rip),%rax
  movss  (%rdx,%rax,1),%xmm2
  mulss  %xmm3,%xmm2
  subss  %xmm2,%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,D98(%rip)
  addl   $0x1,-0x24(%rbp)
L43:
  cmpl   $0x2,-0x24(%rbp)
  jle    L33
  movss  D98(%rip),%xmm0
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D158
datatype: f32
name: retrieve_data
arguments: data_label: D168
datatype: f32
name: retrieve_data
arguments: data_label: D41
datatype: f32
name: retrieve_data
arguments: data_label: D105
datatype: byte[4]
name: retrieve_data
arguments: data_label: D127
datatype: f32

tool: D158: 1.409030866231767e-05

tool: D168: 2.818061732463533e-05

tool: D41: 1

tool: {"D105": [0xd6,0x60,0xfe,0xbf]}

tool: D127: 0.9873836100406357

assistant: ```c
float adgold_pseudo_stereo_iir(float NewSample) {
  float ACoef[2 + 1] = {0.00001409030866231767, 0.00002818061732463533, 0.00001409030866231767};
  float BCoef[2 + 1] = {1.00000000000000000000, -1.98733021473466760000, 0.98738361004063568000};
  static float y[2 + 1];
  static float x[2 + 1];
  int n;
  for (n = 2; n > 0; n--) {
    x[n] = x[n - 1];
    y[n] = y[n - 1];
  }
  x[0] = NewSample;
  y[0] = ACoef[0] * x[0];
  for (n = 1; n <= 2; n++)
    y[0] += ACoef[n] * x[n] - BCoef[n] * y[n];
  return y[0];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xrealloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  mov    $0x0,%eax
  call   <abort@plt>
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *xrealloc(void *ptr, size_t size) {
  ptr = realloc(ptr, size);
  if (!ptr)
    abort();
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<xrealloc>:
  endbr64
  push   %rbx
  call   <realloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L5
L7:
  mov    %rbx,%rax
  pop    %rbx
  ret
L5:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L7
```

assistant: ```c
void *xrealloc(void *ptr, size_t size) {
  ptr = realloc(ptr, size);
  if (!ptr)
    abort();
  return ptr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<xrealloc>:
  endbr64
  push   %r12
  call   <realloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L10
L2:
  mov    %r12,%rax
  pop    %r12
  ret
L10:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L2
```

assistant: ```c
void *xrealloc(void *ptr, size_t size) {
  ptr = realloc(ptr, size);
  if (!ptr)
    abort();
  return ptr;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<xrealloc>:
  endbr64
  push   %r12
  call   <realloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L10
L16:
  mov    %r12,%rax
  pop    %r12
  ret
L10:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L16
```

assistant: ```c
void *xrealloc(void *ptr, size_t size) {
  ptr = realloc(ptr, size);
  if (!ptr)
    abort();
  return ptr;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<SAC_strntrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  jmp    L9
L32:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L14
  addq   $0x1,-0x8(%rbp)
  subq   $0x1,-0x10(%rbp)
L9:
  cmpq   $0x0,-0x10(%rbp)
  jne    L32
  jmp    L74
L14:
  nop
  jmp    L74
L55:
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L43
  subq   $0x1,-0x10(%rbp)
L74:
  cmpq   $0x0,-0x10(%rbp)
  jne    L55
  jmp    L18
L43:
  nop
L18:
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void SAC_strntrim(const char *s, size_t n, const char **t, size_t *tn, const char *match) {
  for (; n != 0; ++s, --n) {
    if (strchr(match, *s) == NULL)
      break;
  }
  for (; n != 0; --n) {
    if (strchr(match, s[n - 1]) == NULL)
      break;
  }
  *t = s;
  *tn = n;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<SAC_strntrim>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r14
  mov    %rcx,%r15
  mov    %r8,%r12
  test   %rsi,%rsi
  je     L42
L25:
  movsbl 0x0(%rbp),%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L53
  add    $0x1,%rbp
  sub    $0x1,%rbx
  jne    L25
L42:
  mov    $0x0,%ebx
L52:
  mov    %rbp,(%r14)
  mov    %rbx,(%r15)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  test   %rbx,%rbx
  je     L52
  lea    -0x1(%rbx),%r13
  movsbl -0x1(%rbp,%rbx,1),%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L52
  mov    %r13,%rbx
  jmp    L53
```

assistant: ```c
void SAC_strntrim(const char *s, size_t n, const char **t, size_t *tn, const char *match) {
  for (; n != 0; ++s, --n) {
    if (strchr(match, *s) == NULL)
      break;
  }
  for (; n != 0; --n) {
    if (strchr(match, s[n - 1]) == NULL)
      break;
  }
  *t = s;
  *tn = n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<SAC_strntrim>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L70
  mov    %rsi,%rbx
  mov    %r8,%r13
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L22:
  add    $0x1,%rbp
  sub    $0x1,%rbx
  je     L70
L1:
  movsbl 0x0(%rbp),%esi
  mov    %r13,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L22
  jmp    L36
  nopl   (%rax)
L33:
  test   %rbx,%rbx
  je     L70
L36:
  mov    %rbx,%r12
  mov    %r13,%rdi
  lea    -0x1(%rbx),%rbx
  movsbl -0x1(%rbp,%r12,1),%esi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L33
L64:
  mov    %rbp,(%r15)
  mov    %r12,(%r14)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L70:
  xor    %r12d,%r12d
  jmp    L64
```

assistant: ```c
void SAC_strntrim(const char *s, size_t n, const char **t, size_t *tn, const char *match) {
  for (; n != 0; ++s, --n) {
    if (strchr(match, *s) == NULL)
      break;
  }
  for (; n != 0; --n) {
    if (strchr(match, s[n - 1]) == NULL)
      break;
  }
  *t = s;
  *tn = n;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<SAC_strntrim>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L75
  mov    %rsi,%rbx
  mov    %r8,%r13
  jmp    L55
  nopl   0x0(%rax,%rax,1)
L42:
  add    $0x1,%rbp
  sub    $0x1,%rbx
  je     L75
L55:
  movsbl 0x0(%rbp),%esi
  mov    %r13,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L42
  jmp    L73
  nopl   (%rax)
L59:
  test   %rbx,%rbx
  je     L75
L73:
  mov    %rbx,%r12
  mov    %r13,%rdi
  lea    -0x1(%rbx),%rbx
  movsbl -0x1(%rbp,%r12,1),%esi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L59
L79:
  mov    %rbp,(%r15)
  mov    %r12,(%r14)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L75:
  xor    %r12d,%r12d
  jmp    L79
```

assistant: ```c
void SAC_strntrim(const char *s, size_t n, const char **t, size_t *tn, const char *match) {
  for (; n != 0; ++s, --n) {
    if (strchr(match, *s) == NULL)
      break;
  }
  for (; n != 0; --n) {
    if (strchr(match, s[n - 1]) == NULL)
      break;
  }
  *t = s;
  *tn = n;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<genpush>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <gentext@plt>
  mov    $0x0,%eax
  call   <cgpush@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void genpush(void) {
  gentext();
  cgpush();
}
```
user: 请将以下的汇编代码转换成C语言代码:

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

assistant: ```c
void genpush(void) {
  gentext();
  cgpush();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<genpush>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <gentext@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <cgpush@plt>
```

assistant: ```c
void genpush(void) {
  gentext();
  cgpush();
}
```
user: What is the c source code of the assembly code below?

```asm
<genpush>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <gentext@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <cgpush@plt>
```

assistant: ```c
void genpush(void) {
  gentext();
  cgpush();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L13:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %eax,-0x4(%rbp)
  subl   $0x1,-0x1c(%rbp)
L8:
  cmpl   $0x0,-0x1c(%rbp)
  jg     L13
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int sum(unsigned char *data, int num) {
  unsigned int sum = 0;
  while (num > 0) {
    sum += *data++;
    num--;
  }
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L7
  mov    %esi,%esi
  lea    (%rdi,%rsi,1),%rcx
  mov    $0x0,%eax
L5:
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%edx
  add    %edx,%eax
  cmp    %rcx,%rdi
  jne    L5
  ret
L7:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned int sum(unsigned char *data, int num) {
  unsigned int sum = 0;
  while (num > 0) {
    sum += *data++;
    num--;
  }
  return sum;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L8
  movslq %esi,%rsi
  xor    %eax,%eax
  lea    (%rdi,%rsi,1),%rcx
  nopl   0x0(%rax)
L5:
  movzbl (%rdi),%edx
  add    $0x1,%rdi
  add    %edx,%eax
  cmp    %rcx,%rdi
  jne    L5
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned int sum(unsigned char *data, int num) {
  unsigned int sum = 0;
  while (num > 0) {
    sum += *data++;
    num--;
  }
  return sum;
}
```