ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<CDiv>:
  endbr64
  movq   D0(%rip),%xmm5
  movapd %xmm2,%xmm4
  andpd  %xmm5,%xmm4
  andpd  %xmm3,%xmm5
  addsd  %xmm5,%xmm4
  divsd  %xmm4,%xmm0
  divsd  %xmm4,%xmm1
  divsd  %xmm4,%xmm2
  divsd  %xmm4,%xmm3
  movapd %xmm2,%xmm5
  mulsd  %xmm2,%xmm5
  movapd %xmm3,%xmm4
  mulsd  %xmm3,%xmm4
  addsd  %xmm4,%xmm5
  movapd %xmm0,%xmm4
  mulsd  %xmm2,%xmm4
  movapd %xmm1,%xmm6
  mulsd  %xmm3,%xmm6
  addsd  %xmm6,%xmm4
  divsd  %xmm5,%xmm4
  movsd  %xmm4,(%rdi)
  mulsd  %xmm2,%xmm1
  mulsd  %xmm3,%xmm0
  subsd  %xmm0,%xmm1
  divsd  %xmm5,%xmm1
  movsd  %xmm1,(%rsi)
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type_guessed: qword

tool: D0: 9223372036854775807

assistant: ```c
void CDiv(ar, ai, br, bi, cr, ci) double ar, ai, br, bi, *cr, *ci;
{
  double s, ais, bis, ars, brs;
  s = fabs(br) + fabs(bi);
  ars = ar / s;
  ais = ai / s;
  brs = br / s;
  bis = bi / s;
  s = brs * brs + bis * bis;
  *cr = (ars * brs + ais * bis) / s;
  *ci = (ais * brs - ars * bis) / s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<CDiv>:
  endbr64
  movq   D2(%rip),%xmm5
  movapd %xmm2,%xmm4
  andpd  %xmm5,%xmm4
  andpd  %xmm3,%xmm5
  addsd  %xmm5,%xmm4
  divsd  %xmm4,%xmm2
  divsd  %xmm4,%xmm3
  movapd %xmm2,%xmm5
  mulsd  %xmm2,%xmm5
  divsd  %xmm4,%xmm0
  divsd  %xmm4,%xmm1
  movapd %xmm3,%xmm4
  mulsd  %xmm3,%xmm4
  addsd  %xmm4,%xmm5
  movapd %xmm0,%xmm4
  mulsd  %xmm2,%xmm4
  mulsd  %xmm3,%xmm0
  movapd %xmm1,%xmm6
  mulsd  %xmm3,%xmm6
  mulsd  %xmm2,%xmm1
  addsd  %xmm6,%xmm4
  subsd  %xmm0,%xmm1
  divsd  %xmm5,%xmm4
  divsd  %xmm5,%xmm1
  movsd  %xmm4,(%rdi)
  movsd  %xmm1,(%rsi)
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_infered: qword

tool: D2: 9223372036854775807

assistant: ```c
void CDiv(ar, ai, br, bi, cr, ci) double ar, ai, br, bi, *cr, *ci;
{
  double s, ais, bis, ars, brs;
  s = fabs(br) + fabs(bi);
  ars = ar / s;
  ais = ai / s;
  brs = br / s;
  bis = bi / s;
  s = brs * brs + bis * bis;
  *cr = (ars * brs + ais * bis) / s;
  *ci = (ais * brs - ars * bis) / s;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<CDiv>:
  endbr64
  movq   D6(%rip),%xmm5
  movapd %xmm2,%xmm4
  andpd  %xmm5,%xmm4
  andpd  %xmm3,%xmm5
  addsd  %xmm5,%xmm4
  divsd  %xmm4,%xmm2
  divsd  %xmm4,%xmm3
  movapd %xmm2,%xmm5
  mulsd  %xmm2,%xmm5
  divsd  %xmm4,%xmm0
  divsd  %xmm4,%xmm1
  movapd %xmm3,%xmm4
  mulsd  %xmm3,%xmm4
  addsd  %xmm4,%xmm5
  movapd %xmm0,%xmm4
  mulsd  %xmm2,%xmm4
  mulsd  %xmm3,%xmm0
  movapd %xmm1,%xmm6
  mulsd  %xmm3,%xmm6
  mulsd  %xmm2,%xmm1
  addsd  %xmm6,%xmm4
  subsd  %xmm0,%xmm1
  divsd  %xmm5,%xmm4
  divsd  %xmm5,%xmm1
  movsd  %xmm4,(%rdi)
  movsd  %xmm1,(%rsi)
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
data_type: qword

tool: D6: 9223372036854775807

assistant: ```c
void CDiv(ar, ai, br, bi, cr, ci) double ar, ai, br, bi, *cr, *ci;
{
  double s, ais, bis, ars, brs;
  s = fabs(br) + fabs(bi);
  ars = ar / s;
  ais = ai / s;
  brs = br / s;
  bis = bi / s;
  s = brs * brs + bis * bis;
  *cr = (ars * brs + ais * bis) / s;
  *ci = (ais * brs - ars * bis) / s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<pushStack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,%eax
  mov    %al,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x8(%rbp),%rdx
  add    %rax,%rdx
  movzbl -0xc(%rbp),%eax
  mov    %al,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void pushStack(char *stack, char c) {
  stack[strlen(stack) + 1] = '\0';
  stack[strlen(stack)] = c;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pushStack>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  call   <strlen@plt>
  movb   $0x0,0x1(%rbx,%rax,1)
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %bpl,(%rbx,%rax,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void pushStack(char *stack, char c) {
  stack[strlen(stack) + 1] = '\0';
  stack[strlen(stack)] = c;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<pushStack>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  movb   $0x0,0x1(%rbx,%rax,1)
  call   <strlen@plt>
  mov    %bpl,(%rbx,%rax,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void pushStack(char *stack, char c) {
  stack[strlen(stack) + 1] = '\0';
  stack[strlen(stack)] = c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pushStack>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  movb   $0x0,0x1(%rbx,%rax,1)
  call   <strlen@plt>
  mov    %bpl,(%rbx,%rax,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void pushStack(char *stack, char c) {
  stack[strlen(stack) + 1] = '\0';
  stack[strlen(stack)] = c;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<multiply235>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L31
L11:
  movl   $0x0,-0x3c(%rbp)
  jmp    L20
L26:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L20:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L26
  addl   $0x1,-0x40(%rbp)
L31:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L11
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply235(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<multiply235>:
  endbr64
  test   %edi,%edi
  jle    L7
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L17:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L6:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L6
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L17
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  ret
```

assistant: ```c
void multiply235(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<multiply235>:
  endbr64
  test   %edi,%edi
  jle    L22
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L14:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L4:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L4
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L14
  pop    %rbx
  ret
L22:
  ret
```

assistant: ```c
void multiply235(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<multiply235>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L83
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L91:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L99
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L99
  nopl   0x0(%rax)
L95:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L95
  cmp    -0x24(%rsp),%r11d
  je     L93
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L93
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L93
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L93:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L91
L83:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L99:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L15:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L15
  mov    -0x18(%rsp),%r8
  jmp    L93
```

assistant: ```c
void multiply235(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D31
data_type: string
name: disassemble_data
arguments: label: D12
data_type: string
name: disassemble_data
arguments: label: D28
data_type: string
name: disassemble_data
arguments: label: D39
data_type: string

tool: D31: Usage  : >samtrial <from # trials> <to # trials> <rng> {<seed>}

tool: D12: Example: >samtrial 24 34 4 "my seed"

tool: D28: (perform 11 trials from 2**24 to 2**34 using MOTE64 on seed "my seed")

tool: D39: To change the function (MOD,SAM,LIM) uncomment its #define and re-compile.

assistant: ```c
void usage(void) {
  puts("Usage  : >samtrial <from # trials> <to # trials> <rng> {<seed>}");
  puts("Example: >samtrial 24 34 4 \"my seed\"");
  puts("(perform 11 trials from 2**24 to 2**34 using MOTE64 on seed \"my seed\")");
  puts("To change the function (MOD,SAM,LIM) uncomment its #define and re-compile.");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D35(%rip),%rdi
  call   <puts@plt>
  lea    D27(%rip),%rdi
  call   <puts@plt>
  lea    D5(%rip),%rdi
  call   <puts@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D35
type_infered: string
name: parse_data
arguments: label: D27
type_infered: string
name: parse_data
arguments: label: D5
type_infered: string
name: parse_data
arguments: label: D13
type_infered: string

tool: D35: Usage  : >samtrial <from # trials> <to # trials> <rng> {<seed>}

tool: D27: Example: >samtrial 24 34 4 "my seed"

tool: D5: (perform 11 trials from 2**24 to 2**34 using MOTE64 on seed "my seed")

tool: D13: To change the function (MOD,SAM,LIM) uncomment its #define and re-compile.

assistant: ```c
void usage(void) {
  puts("Usage  : >samtrial <from # trials> <to # trials> <rng> {<seed>}");
  puts("Example: >samtrial 24 34 4 \"my seed\"");
  puts("(perform 11 trials from 2**24 to 2**34 using MOTE64 on seed \"my seed\")");
  puts("To change the function (MOD,SAM,LIM) uncomment its #define and re-compile.");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D33(%rip),%rdi
  call   <puts@plt>
  lea    D25(%rip),%rdi
  call   <puts@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  lea    D23(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <puts@plt>
```

tool_calls: name: read_data
arguments: data_label: D33
datatype: string
name: read_data
arguments: data_label: D25
datatype: string
name: read_data
arguments: data_label: D10
datatype: string
name: read_data
arguments: data_label: D23
datatype: string

tool: D33: Usage  : >samtrial <from # trials> <to # trials> <rng> {<seed>}

tool: D25: Example: >samtrial 24 34 4 "my seed"

tool: D10: (perform 11 trials from 2**24 to 2**34 using MOTE64 on seed "my seed")

tool: D23: To change the function (MOD,SAM,LIM) uncomment its #define and re-compile.

assistant: ```c
void usage(void) {
  puts("Usage  : >samtrial <from # trials> <to # trials> <rng> {<seed>}");
  puts("Example: >samtrial 24 34 4 \"my seed\"");
  puts("(perform 11 trials from 2**24 to 2**34 using MOTE64 on seed \"my seed\")");
  puts("To change the function (MOD,SAM,LIM) uncomment its #define and re-compile.");
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <puts@plt>
  lea    D34(%rip),%rdi
  call   <puts@plt>
  lea    D22(%rip),%rdi
  call   <puts@plt>
  lea    D37(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <puts@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_guessed: string
name: parse_data
arguments: data_tag: D34
type_guessed: string
name: parse_data
arguments: data_tag: D22
type_guessed: string
name: parse_data
arguments: data_tag: D37
type_guessed: string

tool: D3: Usage  : >samtrial <from # trials> <to # trials> <rng> {<seed>}

tool: D34: Example: >samtrial 24 34 4 "my seed"

tool: D22: (perform 11 trials from 2**24 to 2**34 using MOTE64 on seed "my seed")

tool: D37: To change the function (MOD,SAM,LIM) uncomment its #define and re-compile.

assistant: ```c
void usage(void) {
  puts("Usage  : >samtrial <from # trials> <to # trials> <rng> {<seed>}");
  puts("Example: >samtrial 24 34 4 \"my seed\"");
  puts("(perform 11 trials from 2**24 to 2**34 using MOTE64 on seed \"my seed\")");
  puts("To change the function (MOD,SAM,LIM) uncomment its #define and re-compile.");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<subfactor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L9
L7:
  mov    -0x18(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0xc(%rbp)
L9:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L7
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  pop    %rbp
  ret
```

assistant: ```c
int subfactor(int k, int n) {
  int i, f, l;
  f = n;
  l = 1;
  for (i = 1; i < k; i++) {
    f *= n - i;
    l *= i + 1;
  }
  return f / l;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<subfactor>:
  endbr64
  cmp    $0x1,%edi
  jle    L28
  mov    %esi,%eax
  mov    $0x1,%ecx
  mov    $0x1,%edx
L26:
  mov    %esi,%r8d
  sub    %edx,%r8d
  imul   %r8d,%eax
  add    $0x1,%edx
  imul   %edx,%ecx
  cmp    %edi,%edx
  jne    L26
L10:
  cltd
  idiv   %ecx
  ret
L28:
  mov    %esi,%eax
  mov    $0x1,%ecx
  jmp    L10
```

assistant: ```c
int subfactor(int k, int n) {
  int i, f, l;
  f = n;
  l = 1;
  for (i = 1; i < k; i++) {
    f *= n - i;
    l *= i + 1;
  }
  return f / l;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<subfactor>:
  endbr64
  mov    %esi,%eax
  cmp    $0x1,%edi
  jle    L9
  mov    $0x1,%ecx
  mov    $0x1,%edx
  nopl   (%rax)
L12:
  mov    %esi,%r8d
  sub    %edx,%r8d
  add    $0x1,%edx
  imul   %r8d,%eax
  imul   %edx,%ecx
  cmp    %edi,%edx
  jne    L12
  cltd
  idiv   %ecx
  ret
L9:
  ret
```

assistant: ```c
int subfactor(int k, int n) {
  int i, f, l;
  f = n;
  l = 1;
  for (i = 1; i < k; i++) {
    f *= n - i;
    l *= i + 1;
  }
  return f / l;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<subfactor>:
  endbr64
  cmp    $0x1,%edi
  jle    L50
  lea    -0x2(%rdi),%eax
  lea    -0x1(%rdi),%r8d
  cmp    $0x91,%eax
  jbe    L28
  movdqa D8(%rip),%xmm5
  movd   %esi,%xmm6
  mov    %r8d,%edx
  xor    %eax,%eax
  pshufd $0x0,%xmm6,%xmm7
  movdqa D0(%rip),%xmm4
  movdqa D11(%rip),%xmm6
  shr    $0x2,%edx
  movdqa %xmm5,%xmm3
  movdqa %xmm5,%xmm2
L32:
  movdqa %xmm4,%xmm0
  movdqa %xmm7,%xmm1
  paddd  %xmm6,%xmm4
  add    $0x1,%eax
  psubd  %xmm0,%xmm1
  paddd  %xmm5,%xmm0
  movdqa %xmm1,%xmm8
  psrlq  $0x20,%xmm1
  pmuludq %xmm2,%xmm8
  psrlq  $0x20,%xmm2
  pmuludq %xmm2,%xmm1
  pshufd $0x8,%xmm8,%xmm2
  pshufd $0x8,%xmm1,%xmm1
  punpckldq %xmm1,%xmm2
  movdqa %xmm0,%xmm1
  pmuludq %xmm3,%xmm1
  psrlq  $0x20,%xmm0
  psrlq  $0x20,%xmm3
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  movdqa %xmm1,%xmm3
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm3
  cmp    %edx,%eax
  jne    L32
  movdqa %xmm3,%xmm0
  mov    %r8d,%r9d
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%r9d
  movdqa %xmm0,%xmm1
  psrlq  $0x20,%xmm0
  lea    0x1(%r9),%edx
  pmuludq %xmm3,%xmm1
  psrlq  $0x20,%xmm3
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqa %xmm1,%xmm0
  psrldq $0x4,%xmm0
  pmuludq %xmm1,%xmm0
  movd   %xmm0,%ecx
  movdqa %xmm2,%xmm0
  psrldq $0x8,%xmm0
  movdqa %xmm0,%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  psrlq  $0x20,%xmm2
  pmuludq %xmm2,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqa %xmm1,%xmm0
  psrldq $0x4,%xmm0
  pmuludq %xmm1,%xmm0
  movd   %xmm0,%eax
  imul   %esi,%eax
  cmp    %r9d,%r8d
  je     L19
  nopl   0x0(%rax)
L0:
  mov    %esi,%r8d
  sub    %edx,%r8d
  add    $0x1,%edx
  imul   %r8d,%eax
  imul   %edx,%ecx
  cmp    %edi,%edx
  jl     L0
L19:
  cltd
  idiv   %ecx
  ret
  nopl   0x0(%rax,%rax,1)
L50:
  mov    %esi,%eax
  ret
L28:
  mov    %esi,%eax
  mov    $0x1,%ecx
  mov    $0x1,%edx
  jmp    L0
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type_guessed: byte[16]
name: unpack_data
arguments: data_tag: D0
type_guessed: byte[16]
name: unpack_data
arguments: data_tag: D11
type_guessed: byte[16]

tool: {"D8": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D0": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D11": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int subfactor(int k, int n) {
  int i, f, l;
  f = n;
  l = 1;
  for (i = 1; i < k; i++) {
    f *= n - i;
    l *= i + 1;
  }
  return f / l;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<remove_cmdline_parser_ext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    $0x0,%r8d
  mov    %eax,%edi
  call   <remove_cmdline_parser_internal@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int remove_cmdline_parser_ext(int argc, char **argv, struct remove_args_info *args_info, struct remove_cmdline_parser_params *params) {
  int result;
  result = remove_cmdline_parser_internal(argc, argv, args_info, params, 0);
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<remove_cmdline_parser_ext>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%r8d
  call   <remove_cmdline_parser_internal@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int remove_cmdline_parser_ext(int argc, char **argv, struct remove_args_info *args_info, struct remove_cmdline_parser_params *params) {
  int result;
  result = remove_cmdline_parser_internal(argc, argv, args_info, params, 0);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<remove_cmdline_parser_ext>:
  endbr64
  xor    %r8d,%r8d
  jmp    1050 <remove_cmdline_parser_internal@plt>
```

assistant: ```c
int remove_cmdline_parser_ext(int argc, char **argv, struct remove_args_info *args_info, struct remove_cmdline_parser_params *params) {
  int result;
  result = remove_cmdline_parser_internal(argc, argv, args_info, params, 0);
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<remove_cmdline_parser_ext>:
  endbr64
  xor    %r8d,%r8d
  jmp    1050 <remove_cmdline_parser_internal@plt>
```

assistant: ```c
int remove_cmdline_parser_ext(int argc, char **argv, struct remove_args_info *args_info, struct remove_cmdline_parser_params *params) {
  int result;
  result = remove_cmdline_parser_internal(argc, argv, args_info, params, 0);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x8(%rbp)
  movl   $0x5,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  mov    -0x4(%rbp),%edx
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
data_type: string

tool: D1: a++=%d,++b=%d

assistant: ```c
int main() {
  int a = 1, b = 5;
  printf("a++=%d,++b=%d", a++, ++b);
  return 0;
}
```