ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<strindexright>:
  endbr64
  lea    -0x1(%rdx),%eax
  cmp    %ecx,%eax
  jle    L32
  push   %rbp
  push   %rbx
  mov    %rsi,%r10
  mov    %ecx,%r9d
  lea    -0x1(%rcx),%ebx
  movslq %ebx,%rdx
  movzbl (%rsi,%rdx,1),%r11d
  movslq %ecx,%rbp
  jmp    L14
L39:
  mov    %r9d,%eax
  sub    %edx,%eax
  lea    -0x1(%rcx,%rax,1),%eax
L36:
  sub    $0x1,%eax
  cmp    %eax,%r9d
  jge    L37
L14:
  movslq %eax,%rcx
  cmp    %r11b,(%rdi,%rcx,1)
  jne    L36
  sub    $0x1,%eax
  cltq
  mov    %ebx,%edx
  mov    %rbp,%r8
  sub    %rcx,%r8
  add    %r10,%r8
L52:
  lea    0x1(%rax),%ecx
  test   %edx,%edx
  jle    L70
  mov    %eax,%ecx
  movzbl (%rdi,%rax,1),%esi
  sub    $0x1,%edx
  sub    $0x1,%rax
  cmp    (%r8,%rax,1),%sil
  je     L52
L70:
  test   %edx,%edx
  jne    L39
  movslq %ecx,%rax
  movzbl (%r10),%esi
  cmp    %sil,(%rdi,%rax,1)
  jne    L39
  jmp    L4
L37:
  mov    $0xffffffff,%ecx
L4:
  mov    %ecx,%eax
  pop    %rbx
  pop    %rbp
  ret
L32:
  mov    $0xffffffff,%ecx
  mov    %ecx,%eax
  ret
```

assistant: ```c
int strindexright(char source[], char target[], int slen, int tlen) {
  int is, it;
  for (is = slen - 1, it = tlen - 1; is > tlen; --is) {
    if (source[is] == target[it]) {
      while (it > 0)
        if (source[--is] != target[--it])
          break;
      if (it == 0 && source[is] == target[it])
        return is;
      is += tlen - it - 1;
      it = tlen - 1;
    }
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strindexright>:
  endbr64
  lea    -0x1(%rdx),%eax
  cmp    %ecx,%eax
  jle    L92
  push   %rbp
  mov    %rsi,%r10
  mov    %ecx,%r9d
  movslq %ecx,%rbp
  push   %rbx
  lea    -0x1(%rcx),%ebx
  movslq %ebx,%rdx
  movzbl (%rsi,%rdx,1),%r11d
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L99:
  sub    $0x1,%eax
  cmp    %eax,%r9d
  jge    L72
L39:
  movslq %eax,%rcx
  cmp    %r11b,(%rdi,%rcx,1)
  jne    L99
  mov    %rbp,%rsi
  sub    $0x1,%eax
  mov    %ebx,%edx
  sub    %rcx,%rsi
  cltq
  add    %r10,%rsi
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L1:
  movzbl (%rdi,%rax,1),%ecx
  mov    %eax,%r8d
  sub    $0x1,%rax
  sub    $0x1,%edx
  cmp    (%rsi,%rax,1),%cl
  jne    L31
L25:
  lea    0x1(%rax),%r8d
  test   %edx,%edx
  jg     L1
L31:
  test   %edx,%edx
  jne    L27
  movslq %r8d,%rax
  movzbl (%r10),%esi
  cmp    %sil,(%rdi,%rax,1)
  je     L81
L27:
  mov    %r9d,%eax
  sub    %edx,%eax
  lea    -0x1(%r8,%rax,1),%eax
  sub    $0x1,%eax
  cmp    %eax,%r9d
  jl     L39
  cs nopw 0x0(%rax,%rax,1)
L72:
  mov    $0xffffffff,%r8d
L81:
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
L92:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int strindexright(char source[], char target[], int slen, int tlen) {
  int is, it;
  for (is = slen - 1, it = tlen - 1; is > tlen; --is) {
    if (source[is] == target[it]) {
      while (it > 0)
        if (source[--is] != target[--it])
          break;
      if (it == 0 && source[is] == target[it])
        return is;
      is += tlen - it - 1;
      it = tlen - 1;
    }
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strindexright>:
  endbr64
  lea    -0x1(%rdx),%eax
  cmp    %ecx,%eax
  jle    L53
  push   %rbp
  mov    %rsi,%r10
  mov    %ecx,%r9d
  movslq %ecx,%rbp
  push   %rbx
  lea    -0x1(%rcx),%ebx
  movslq %ebx,%rdx
  movzbl (%rsi,%rdx,1),%r11d
  jmp    L59
  nopw   0x0(%rax,%rax,1)
L14:
  sub    $0x1,%eax
  cmp    %eax,%r9d
  jge    L56
L59:
  movslq %eax,%rsi
  movzbl (%rdi,%rsi,1),%edx
  cmp    %r11b,%dl
  jne    L14
  mov    %rbp,%r8
  sub    $0x1,%eax
  mov    %ebx,%ecx
  sub    %rsi,%r8
  cltq
  lea    (%r10,%r8,1),%rsi
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L88:
  movzbl (%rdi,%rax,1),%edx
  mov    %eax,%r8d
  sub    $0x1,%rax
  sub    $0x1,%ecx
  cmp    (%rsi,%rax,1),%dl
  jne    L49
L28:
  lea    0x1(%rax),%r8d
  test   %ecx,%ecx
  jg     L88
L49:
  test   %ecx,%ecx
  jne    L18
  cmp    %dl,(%r10)
  je     L2
L18:
  mov    %r9d,%eax
  sub    %ecx,%eax
  lea    -0x1(%r8,%rax,1),%eax
  sub    $0x1,%eax
  cmp    %eax,%r9d
  jl     L59
  cs nopw 0x0(%rax,%rax,1)
L56:
  mov    $0xffffffff,%r8d
L2:
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
L53:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int strindexright(char source[], char target[], int slen, int tlen) {
  int is, it;
  for (is = slen - 1, it = tlen - 1; is > tlen; --is) {
    if (source[is] == target[it]) {
      while (it > 0)
        if (source[--is] != target[--it])
          break;
      if (it == 0 && source[is] == target[it])
        return is;
      is += tlen - it - 1;
      it = tlen - 1;
    }
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<test0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  lea    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void test0(void) {
  void **x = NULL;
  *x = (void *)&x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test0>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,0x0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test0(void) {
  void **x = NULL;
  *x = (void *)&x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test0>:
  endbr64
  mov    %fs:0x28,%rax
  mov    %rax,-0x10(%rsp)
  xor    %eax,%eax
  lea    -0x18(%rsp),%rax
  mov    %rax,0x0
  ud2
```

assistant: ```c
void test0(void) {
  void **x = NULL;
  *x = (void *)&x;
}
```
user: What is the c source code of the assembly code below:

```asm
<test0>:
  endbr64
  mov    %fs:0x28,%rax
  mov    %rax,-0x10(%rsp)
  xor    %eax,%eax
  lea    -0x18(%rsp),%rax
  mov    %rax,0x0
  ud2
```

assistant: ```c
void test0(void) {
  void **x = NULL;
  *x = (void *)&x;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<FakeSilenceSample>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <RandomValue@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x4(%rbp),%xmm0
  comisd D9(%rip),%xmm0
  jbe    L12
  mov    $0x1,%eax
  jmp    L26
L12:
  pxor   %xmm1,%xmm1
  cvtss2sd -0x4(%rbp),%xmm1
  movsd  D14(%rip),%xmm0
  comisd %xmm1,%xmm0
  jbe    L35
  mov    $0xffffffff,%eax
  jmp    L26
L35:
  mov    $0x0,%eax
L26:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
datatype: f64[2]
name: extract_symbolic_data
arguments: data_label: D14
datatype: f64

tool: {"D9": [0.8,0.2]}

tool: D14: 0.2

assistant: ```c
int FakeSilenceSample() {
  float x = RandomValue();
  if (x > 0.8)
    return 1;
  if (x < 0.2)
    return -1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<FakeSilenceSample>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <RandomValue@plt>
  cvtss2sd %xmm0,%xmm0
  mov    $0x1,%eax
  comisd D18(%rip),%xmm0
  ja     L1
  movsd  D8(%rip),%xmm1
  comisd %xmm0,%xmm1
  seta   %al
  movzbl %al,%eax
  neg    %eax
L1:
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D18
data_type: f64[2]
name: read_data
arguments: data_tag: D8
data_type: f64

tool: {"D18": [0.8,0.2]}

tool: D8: 0.2

assistant: ```c
int FakeSilenceSample() {
  float x = RandomValue();
  if (x > 0.8)
    return 1;
  if (x < 0.2)
    return -1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<FakeSilenceSample>:
  endbr64
  xor    %eax,%eax
  sub    $0x8,%rsp
  call   <RandomValue@plt>
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  comisd D16(%rip),%xmm0
  ja     L2
  movsd  D9(%rip),%xmm1
  xor    %eax,%eax
  comisd %xmm0,%xmm1
  seta   %al
  neg    %eax
L2:
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D16
type_guessed: f64[2]
name: unpack_data
arguments: data_label: D9
type_guessed: f64

tool: {"D16": [0.8,0.2]}

tool: D9: 0.2

assistant: ```c
int FakeSilenceSample() {
  float x = RandomValue();
  if (x > 0.8)
    return 1;
  if (x < 0.2)
    return -1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<FakeSilenceSample>:
  endbr64
  xor    %eax,%eax
  sub    $0x8,%rsp
  call   <RandomValue@plt>
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  comisd D18(%rip),%xmm0
  ja     L3
  movsd  D14(%rip),%xmm1
  xor    %eax,%eax
  comisd %xmm0,%xmm1
  seta   %al
  neg    %eax
L3:
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D18
type: f64[2]
name: read_data
arguments: label: D14
type: f64

tool: {"D18": [0.8,0.2]}

tool: D14: 0.2

assistant: ```c
int FakeSilenceSample() {
  float x = RandomValue();
  if (x > 0.8)
    return 1;
  if (x < 0.2)
    return -1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<getlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    D10(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <fseek@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %eax,-0x4(%rbp)
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <fseek@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned int getlen(FILE *f) {
  unsigned int len;
  fseek(f, 0, SEEK_END);
  len = ftell(f);
  fseek(f, 0, SEEK_SET);
  return len;
}
```
user: What is the c source code of the assembly code below:

```asm
<getlen>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    D18(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  call   <fseek@plt>
  mov    %rbx,%rdi
  call   <ftell@plt>
  mov    %eax,%ebp
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <fseek@plt>
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned int getlen(FILE *f) {
  unsigned int len;
  fseek(f, 0, SEEK_END);
  len = ftell(f);
  fseek(f, 0, SEEK_SET);
  return len;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getlen>:
  endbr64
  push   %r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  call   <fseek@plt>
  mov    %rbp,%rdi
  call   <ftell@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  mov    %eax,%r12d
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  call   <fseek@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned int getlen(FILE *f) {
  unsigned int len;
  fseek(f, 0, SEEK_END);
  len = ftell(f);
  fseek(f, 0, SEEK_SET);
  return len;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getlen>:
  endbr64
  push   %r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  call   <fseek@plt>
  mov    %rbp,%rdi
  call   <ftell@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  mov    %eax,%r12d
  mov    D13(%rip),%rax
  mov    (%rax),%edx
  call   <fseek@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned int getlen(FILE *f) {
  unsigned int len;
  fseek(f, 0, SEEK_END);
  len = ftell(f);
  fseek(f, 0, SEEK_SET);
  return len;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Mrhor>:
  endbr64
  mov    %esi,%r11d
  movsd  (%r8),%xmm0
  movsd  %xmm0,(%r9)
  cmp    $0x1,%esi
  jle    L43
  push   %rbp
  push   %rbx
  mov    %rdx,%rsi
  mov    %r8,%rbx
  mov    %r11d,%ebp
  mov    $0x1,%r8d
L55:
  mov    -0x4(%rcx,%r8,4),%eax
  add    $0x1,%eax
  mov    (%rcx,%r8,4),%r10d
  cmp    %r10d,%eax
  jge    L40
  cltq
  pxor   %xmm1,%xmm1
L28:
  movslq (%rsi,%rax,4),%rdx
  movsd  (%r9,%rdx,8),%xmm0
  mulsd  (%rdi,%rax,8),%xmm0
  addsd  %xmm0,%xmm1
  add    $0x1,%rax
  cmp    %eax,%r10d
  jg     L28
L64:
  movsd  (%rbx,%r8,8),%xmm0
  subsd  %xmm1,%xmm0
  movslq %r10d,%r10
  divsd  (%rdi,%r10,8),%xmm0
  movsd  %xmm0,(%r9,%r8,8)
  add    $0x1,%r8
  cmp    %rbp,%r8
  jne    L55
  sub    $0x1,%r11d
  movslq %r11d,%r11
  lea    (%r9,%r11,8),%rbx
L24:
  mov    %rbx,%r10
  movslq (%rcx,%r11,4),%rax
  movsd  (%rbx),%xmm0
  divsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rbx)
  mov    -0x4(%rcx,%r11,4),%eax
  add    $0x1,%eax
  mov    (%rcx,%r11,4),%r8d
  cmp    %eax,%r8d
  jle    L13
  cltq
L72:
  movslq (%rsi,%rax,4),%rdx
  lea    (%r9,%rdx,8),%rdx
  movsd  (%rdi,%rax,8),%xmm1
  mulsd  (%r10),%xmm1
  movsd  (%rdx),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jg     L72
L13:
  sub    $0x1,%r11
  sub    $0x8,%rbx
  test   %r11d,%r11d
  jg     L24
  pop    %rbx
  pop    %rbp
  ret
L40:
  pxor   %xmm1,%xmm1
  jmp    L64
L43:
  ret
```

assistant: ```c
void Mrhor(double *C, int neq, int *ia, int *iz, double *r, double *rho) {
  int i = 0, j = 0, jlo = 0, jup = 0;
  double s = 0.0;
  rho[0] = r[0];
  for (i = 1; i < neq; i++) {
    s = 0.0;
    jlo = iz[i - 1] + 1;
    jup = iz[i];
    for (j = jlo; j < jup; j++)
      s += C[j] * rho[ia[j]];
    rho[i] = (r[i] - s) / C[jup];
  }
  for (i = neq - 1; i > 0; i--) {
    rho[i] /= C[iz[i]];
    jlo = iz[i - 1] + 1;
    jup = iz[i] - 1;
    for (j = jlo; j <= jup; j++)
      rho[ia[j]] -= C[j] * rho[i];
  }
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<Mrhor>:
  endbr64
  push   %rbp
  mov    %rdx,%r10
  push   %rbx
  movsd  (%r8),%xmm0
  mov    %r8,%rbx
  lea    -0x1(%rsi),%r8d
  movsd  %xmm0,(%r9)
  cmp    $0x1,%esi
  jle    L85
  movslq %esi,%rbp
  mov    $0x1,%r11d
  pxor   %xmm2,%xmm2
L13:
  mov    -0x4(%rcx,%r11,4),%eax
  movslq (%rcx,%r11,4),%rsi
  add    $0x1,%eax
  cmp    %esi,%eax
  jge    L54
  cltq
  movapd %xmm2,%xmm1
  nopw   0x0(%rax,%rax,1)
L72:
  movslq (%r10,%rax,4),%rdx
  movsd  (%r9,%rdx,8),%xmm0
  mulsd  (%rdi,%rax,8),%xmm0
  add    $0x1,%rax
  addsd  %xmm0,%xmm1
  cmp    %eax,%esi
  jg     L72
L87:
  movsd  (%rbx,%r11,8),%xmm0
  subsd  %xmm1,%xmm0
  divsd  (%rdi,%rsi,8),%xmm0
  movsd  %xmm0,(%r9,%r11,8)
  add    $0x1,%r11
  cmp    %r11,%rbp
  jne    L13
L4:
  movslq %r8d,%r11
  lea    (%r9,%r11,8),%rsi
L86:
  movslq (%rcx,%r11,4),%rax
  movsd  (%rsi),%xmm0
  mov    %rax,%r8
  divsd  (%rdi,%rax,8),%xmm0
  mov    -0x4(%rcx,%r11,4),%eax
  movsd  %xmm0,(%rsi)
  add    $0x1,%eax
  cmp    %eax,%r8d
  jle    L14
  cltq
  jmp    L2
  nopl   (%rax)
L1:
  movsd  (%rsi),%xmm0
L2:
  mulsd  (%rdi,%rax,8),%xmm0
  movslq (%r10,%rax,4),%rdx
  add    $0x1,%rax
  lea    (%r9,%rdx,8),%rdx
  movsd  (%rdx),%xmm1
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  cmp    %eax,%r8d
  jg     L1
L14:
  sub    $0x1,%r11
  sub    $0x8,%rsi
  test   %r11d,%r11d
  jg     L86
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L54:
  movapd %xmm2,%xmm1
  jmp    L87
L85:
  test   %r8d,%r8d
  jg     L4
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void Mrhor(double *C, int neq, int *ia, int *iz, double *r, double *rho) {
  int i = 0, j = 0, jlo = 0, jup = 0;
  double s = 0.0;
  rho[0] = r[0];
  for (i = 1; i < neq; i++) {
    s = 0.0;
    jlo = iz[i - 1] + 1;
    jup = iz[i];
    for (j = jlo; j < jup; j++)
      s += C[j] * rho[ia[j]];
    rho[i] = (r[i] - s) / C[jup];
  }
  for (i = neq - 1; i > 0; i--) {
    rho[i] /= C[iz[i]];
    jlo = iz[i - 1] + 1;
    jup = iz[i] - 1;
    for (j = jlo; j <= jup; j++)
      rho[ia[j]] -= C[j] * rho[i];
  }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Mrhor>:
  endbr64
  push   %rbp
  mov    %rdx,%r10
  mov    %rcx,%r11
  push   %rbx
  movsd  (%r8),%xmm0
  mov    %r8,%rbx
  lea    -0x1(%rsi),%r8d
  movsd  %xmm0,(%r9)
  cmp    $0x1,%esi
  jle    L19
  movslq %esi,%rbp
  pxor   %xmm2,%xmm2
  mov    $0x1,%esi
  nopw   0x0(%rax,%rax,1)
L55:
  mov    -0x4(%r11,%rsi,4),%eax
  movslq (%r11,%rsi,4),%rcx
  add    $0x1,%eax
  cmp    %ecx,%eax
  jge    L2
  cltq
  movapd %xmm2,%xmm1
  nopw   0x0(%rax,%rax,1)
L79:
  movslq (%r10,%rax,4),%rdx
  movsd  (%r9,%rdx,8),%xmm0
  mulsd  (%rdi,%rax,8),%xmm0
  add    $0x1,%rax
  addsd  %xmm0,%xmm1
  cmp    %eax,%ecx
  jg     L79
L89:
  movsd  (%rbx,%rsi,8),%xmm0
  subsd  %xmm1,%xmm0
  divsd  (%rdi,%rcx,8),%xmm0
  movsd  %xmm0,(%r9,%rsi,8)
  add    $0x1,%rsi
  cmp    %rsi,%rbp
  jne    L55
L12:
  movslq %r8d,%r8
  lea    (%r9,%r8,8),%rcx
  nopw   0x0(%rax,%rax,1)
L32:
  movslq (%r11,%r8,4),%rax
  movsd  (%rcx),%xmm0
  mov    %rax,%rsi
  divsd  (%rdi,%rax,8),%xmm0
  mov    -0x4(%r11,%r8,4),%eax
  movsd  %xmm0,(%rcx)
  lea    0x1(%rax),%edx
  cmp    %edx,%esi
  jle    L85
  movslq %edx,%rdx
  mulsd  (%rdi,%rdx,8),%xmm0
  movslq (%r10,%rdx,4),%rbx
  lea    0x2(%rax),%edx
  movslq %edx,%rax
  lea    (%r9,%rbx,8),%rbx
  movsd  (%rbx),%xmm1
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%rbx)
  cmp    %edx,%esi
  jle    L85
  cs nopw 0x0(%rax,%rax,1)
L54:
  movslq (%r10,%rax,4),%rdx
  movsd  (%rdi,%rax,8),%xmm1
  add    $0x1,%rax
  mulsd  (%rcx),%xmm1
  lea    (%r9,%rdx,8),%rdx
  movsd  (%rdx),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx)
  cmp    %eax,%esi
  jg     L54
L85:
  sub    $0x1,%r8
  sub    $0x8,%rcx
  test   %r8d,%r8d
  jg     L32
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L2:
  movapd %xmm2,%xmm1
  jmp    L89
L19:
  test   %r8d,%r8d
  jg     L12
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void Mrhor(double *C, int neq, int *ia, int *iz, double *r, double *rho) {
  int i = 0, j = 0, jlo = 0, jup = 0;
  double s = 0.0;
  rho[0] = r[0];
  for (i = 1; i < neq; i++) {
    s = 0.0;
    jlo = iz[i - 1] + 1;
    jup = iz[i];
    for (j = jlo; j < jup; j++)
      s += C[j] * rho[ia[j]];
    rho[i] = (r[i] - s) / C[jup];
  }
  for (i = neq - 1; i > 0; i--) {
    rho[i] /= C[iz[i]];
    jlo = iz[i - 1] + 1;
    jup = iz[i] - 1;
    for (j = jlo; j <= jup; j++)
      rho[ia[j]] -= C[j] * rho[i];
  }
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<adjust>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void adjust(char **s) {
  char *tmp = *s;
  char *result = (char *)calloc(strlen(tmp) + 1, sizeof(char));
  memcpy(result, tmp, strlen(tmp));
  free(tmp);
  *s = result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<adjust>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%rbp
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void adjust(char **s) {
  char *tmp = *s;
  char *result = (char *)calloc(strlen(tmp) + 1, sizeof(char));
  memcpy(result, tmp, strlen(tmp));
  free(tmp);
  *s = result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<adjust>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%r12
  mov    %rdi,%rbx
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %rax,%rdx
  call   <memcpy@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void adjust(char **s) {
  char *tmp = *s;
  char *result = (char *)calloc(strlen(tmp) + 1, sizeof(char));
  memcpy(result, tmp, strlen(tmp));
  free(tmp);
  *s = result;
}
```
user: What is the c source code of the assembly code below?

```asm
<adjust>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%r12
  mov    %rdi,%rbx
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %rax,%rdx
  call   <memcpy@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void adjust(char **s) {
  char *tmp = *s;
  char *result = (char *)calloc(strlen(tmp) + 1, sizeof(char));
  memcpy(result, tmp, strlen(tmp));
  free(tmp);
  *s = result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test_multprec_Newton_step>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D22(%rip),%rdi
  call   <puts@plt>
  lea    D54(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D64(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%edi
  call   <syscon_read_multprec_system@plt>
  lea    0xc(%rsp),%rbx
  mov    %rbx,%rdi
  call   <syscon_number_of_multprec_polynomials@plt>
  mov    0xc(%rsp),%esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <solcon_read_multprec_solutions@plt>
  lea    0x10(%rsp),%rdi
  call   <solcon_number_of_multprec_solutions@plt>
  mov    0x10(%rsp),%esi
  lea    D47(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <solcon_dimension_of_multprec_solutions@plt>
  mov    0xc(%rsp),%esi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%edi
  call   <multprec_Newton_step@plt>
  lea    D65(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <solcon_write_multprec_solutions@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x20,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D22
datatype: string
name: unpack_data
arguments: label: D54
datatype: string
name: unpack_data
arguments: label: D64
datatype: string
name: unpack_data
arguments: label: D23
datatype: string
name: unpack_data
arguments: label: D47
datatype: string
name: unpack_data
arguments: label: D30
datatype: string
name: unpack_data
arguments: label: D65
datatype: string

tool: D22: 
Running Newton step with multiprecision arithmetic ...

tool: D54: 
give the number of decimal places in the working precision : 

tool: D64: %d

tool: D23: The system container has %d polynomials.


tool: D47: The solution container has size %d.


tool: D30: The solutions in the container have dimension %d.


tool: D65: The solutions after the Newton step :

assistant: ```c
void test_multprec_Newton_step(void) {
  int fail, dim, len, deci;
  printf("\nRunning Newton step with multiprecision arithmetic ...\n");
  printf("\ngive the number of decimal places in the working precision : ");
  scanf("%d", &deci);
  fail = syscon_read_multprec_system(deci);
  fail = syscon_number_of_multprec_polynomials(&dim);
  printf("The system container has %d polynomials.\n", dim);
  fail = solcon_read_multprec_solutions();
  fail = solcon_number_of_multprec_solutions(&len);
  printf("The solution container has size %d.\n", len);
  fail = solcon_dimension_of_multprec_solutions(&dim);
  printf("The solutions in the container have dimension %d.\n", dim);
  fail = multprec_Newton_step(deci);
  printf("The solutions after the Newton step :\n");
  fail = solcon_write_multprec_solutions();
}
```