ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<matrMult>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebx
  mov    %edi,0x1c(%rsp)
  mov    %esi,%r13d
  mov    %edx,%r15d
  mov    %rcx,%r14
  mov    %r8,%rbp
  lea    0x0(,%rdi,8),%edi
  call   <malloc@plt>
  mov    %rax,0x10(%rsp)
  test   %ebx,%ebx
  jle    L16
  mov    %rax,%rsi
  lea    0x0(,%r15,4),%eax
  mov    %eax,0x18(%rsp)
  lea    -0x1(%rbx),%eax
  mov    %r14,%rdi
  mov    %r14,%rbx
  lea    0x8(%r14,%rax,8),%rax
  mov    %rax,(%rsp)
  lea    -0x1(%r15),%eax
  lea    0x4(,%rax,4),%r14
  lea    -0x1(%r13),%eax
  lea    0x4(,%rax,4),%r12
  sub    %rdi,%rsi
  mov    %rsi,0x8(%rsp)
  jmp    L5
L36:
  mov    %rcx,%rdx
  add    (%r8),%rdx
  mov    (%rbx),%rdi
  mov    0x0(%rbp,%rax,2),%rsi
  movss  (%rdi,%rax,1),%xmm0
  mulss  (%rsi,%rcx,1),%xmm0
  addss  (%rdx),%xmm0
  movss  %xmm0,(%rdx)
  add    $0x4,%rax
  cmp    %rax,%r12
  jne    L36
L13:
  add    $0x4,%rcx
  cmp    %r14,%rcx
  je     L50
L76:
  mov    $0x0,%eax
  test   %r13d,%r13d
  jg     L36
  jmp    L13
L50:
  add    $0x8,%rbx
  cmp    (%rsp),%rbx
  je     L16
L5:
  mov    0x18(%rsp),%edi
  call   <malloc@plt>
  mov    0x8(%rsp),%rsi
  lea    (%rsi,%rbx,1),%r8
  mov    %rax,(%r8)
  mov    $0x0,%ecx
  test   %r15d,%r15d
  jg     L76
  jmp    L50
L16:
  mov    0x10(%rsp),%rdx
  mov    %r15d,%esi
  mov    0x1c(%rsp),%edi
  call   <printMatrix@plt>
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void matrMult(int m, int p, int n, float **a, float **b, float **c) {
  int i = 0;
  int j = 0;
  int k = 0;
  c = (float **)(malloc(m * sizeof(float *)));
  for (i = 0; i < m; i++) {
    *(c + i) = (float *)(malloc(n * sizeof(float)));
    for (j = 0; j < n; j++) {
      for (k = 0; k < p; k++) {
        *(*(c + i) + j) += (*(*(a + i) + k)) * (*(*(b + k) + j));
      }
    }
  }
  printMatrix(m, n, c);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<matrMult>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  mov    %edi,%r14d
  push   %r13
  mov    %r8,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x28,%rsp
  mov    %edi,0x14(%rsp)
  shl    $0x3,%edi
  mov    %edx,0x4(%rsp)
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %r14d,%r14d
  jle    L56
  mov    0x4(%rsp),%esi
  mov    0x18(%rsp),%rdi
  movslq %r15d,%r14
  lea    0x0(,%rsi,4),%eax
  sub    %rdi,%rbx
  mov    %eax,0x10(%rsp)
  mov    0x14(%rsp),%eax
  lea    (%rdi,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  lea    -0x1(%rsi),%eax
  lea    0x4(,%rax,4),%r12
  mov    %r15d,%eax
  mov    %rdi,%r15
  mov    %r12,%rbp
  mov    %eax,%r12d
  nopw   0x0(%rax,%rax,1)
L9:
  mov    0x10(%rsp),%edi
  call   <malloc@plt>
  xor    %edx,%edx
  mov    %rax,(%r15)
  mov    %rax,%r10
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L42
  nopl   0x0(%rax)
L20:
  test   %r12d,%r12d
  jle    L27
  lea    (%r10,%rdx,1),%rsi
  mov    (%rbx,%r15,1),%rdi
  xor    %eax,%eax
  movss  (%rsi),%xmm1
  nopl   0x0(%rax,%rax,1)
L34:
  mov    0x0(%r13,%rax,8),%rcx
  movss  (%rcx,%rdx,1),%xmm0
  mulss  (%rdi,%rax,4),%xmm0
  add    $0x1,%rax
  addss  %xmm0,%xmm1
  movss  %xmm1,(%rsi)
  cmp    %r14,%rax
  jne    L34
L27:
  add    $0x4,%rdx
  cmp    %rdx,%rbp
  jne    L20
L42:
  add    $0x8,%r15
  cmp    0x8(%rsp),%r15
  jne    L9
L56:
  mov    0x18(%rsp),%rdx
  mov    0x4(%rsp),%esi
  mov    0x14(%rsp),%edi
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <printMatrix@plt>
```

assistant: ```c
void matrMult(int m, int p, int n, float **a, float **b, float **c) {
  int i = 0;
  int j = 0;
  int k = 0;
  c = (float **)(malloc(m * sizeof(float *)));
  for (i = 0; i < m; i++) {
    *(c + i) = (float *)(malloc(n * sizeof(float)));
    for (j = 0; j < n; j++) {
      for (k = 0; k < p; k++) {
        *(*(c + i) + j) += (*(*(a + i) + k)) * (*(*(b + k) + j));
      }
    }
  }
  printMatrix(m, n, c);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<matrMult>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  mov    %edi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x28,%rsp
  mov    %edi,0x14(%rsp)
  shl    $0x3,%edi
  mov    %rcx,0x8(%rsp)
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %r14d,%r14d
  jle    L3
  mov    %rax,%rsi
  mov    0x14(%rsp),%eax
  lea    0x0(,%rbp,4),%r14d
  mov    %rsi,%r13
  lea    (%rsi,%rax,8),%rax
  mov    %rax,(%rsp)
  lea    -0x1(%rbp),%eax
  lea    0x4(,%rax,4),%r12
  mov    0x8(%rsp),%rax
  sub    %rsi,%rax
  mov    %rax,0x8(%rsp)
  mov    %ebp,%eax
  mov    %r12,%rbp
  mov    %eax,%r12d
  nopl   (%rax)
L28:
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  mov    %rax,%r10
  test   %r12d,%r12d
  jle    L56
  test   %ebx,%ebx
  jle    L56
  mov    0x8(%rsp),%rax
  xor    %edx,%edx
  movslq %ebx,%rsi
  mov    (%rax,%r13,1),%rdi
  xchg   %ax,%ax
L17:
  lea    (%r10,%rdx,1),%r8
  xor    %eax,%eax
  movss  (%r8),%xmm1
  nopl   0x0(%rax,%rax,1)
L29:
  mov    (%r15,%rax,8),%rcx
  movss  (%rcx,%rdx,1),%xmm0
  mulss  (%rdi,%rax,4),%xmm0
  add    $0x1,%rax
  addss  %xmm0,%xmm1
  movss  %xmm1,(%r8)
  cmp    %rax,%rsi
  jne    L29
  add    $0x4,%rdx
  cmp    %rdx,%rbp
  jne    L17
L56:
  add    $0x8,%r13
  cmp    %r13,(%rsp)
  jne    L28
  mov    %r12d,%ebp
L3:
  mov    0x18(%rsp),%rdx
  mov    0x14(%rsp),%edi
  add    $0x28,%rsp
  mov    %ebp,%esi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <printMatrix@plt>
```

assistant: ```c
void matrMult(int m, int p, int n, float **a, float **b, float **c) {
  int i = 0;
  int j = 0;
  int k = 0;
  c = (float **)(malloc(m * sizeof(float *)));
  for (i = 0; i < m; i++) {
    *(c + i) = (float *)(malloc(n * sizeof(float)));
    for (j = 0; j < n; j++) {
      for (k = 0; k < p; k++) {
        *(*(c + i) + j) += (*(*(a + i) + k)) * (*(*(b + k) + j));
      }
    }
  }
  printMatrix(m, n, c);
}
```
user: Translate the below assembly code into C programming language:

```asm
<byteswap16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x0,%eax
  call   <littleEndian@plt>
  test   %rax,%rax
  je     L4
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  movzbl %al,%edx
  mov    -0x4(%rbp),%eax
  shl    $0x8,%eax
  movzwl %ax,%eax
  or     %edx,%eax
  jmp    L6
L4:
  mov    -0x4(%rbp),%eax
L6:
  leave
  ret
```

assistant: ```c
uint16_t byteswap16(uint16_t num16) {
  if (littleEndian()) {
    return ((num16 & 0xff00) >> 8) | ((num16 & 0x00ff) << 8);
  } else {
    return num16;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<byteswap16>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <littleEndian@plt>
  test   %rax,%rax
  je     L4
  movzbl %bh,%eax
  shl    $0x8,%ebx
  movzwl %bx,%ebx
  or     %eax,%ebx
L4:
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
uint16_t byteswap16(uint16_t num16) {
  if (littleEndian()) {
    return ((num16 & 0xff00) >> 8) | ((num16 & 0x00ff) << 8);
  } else {
    return num16;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<byteswap16>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %edi,%r12d
  call   <littleEndian@plt>
  test   %rax,%rax
  je     L5
  mov    %r12d,%eax
  shl    $0x8,%r12d
  movzbl %ah,%eax
  movzwl %r12w,%r12d
  or     %eax,%r12d
L5:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint16_t byteswap16(uint16_t num16) {
  if (littleEndian()) {
    return ((num16 & 0xff00) >> 8) | ((num16 & 0x00ff) << 8);
  } else {
    return num16;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<byteswap16>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %edi,%r12d
  call   <littleEndian@plt>
  test   %rax,%rax
  je     L5
  mov    %r12d,%eax
  shl    $0x8,%r12d
  movzbl %ah,%eax
  movzwl %r12w,%r12d
  or     %eax,%r12d
L5:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint16_t byteswap16(uint16_t num16) {
  if (littleEndian()) {
    return ((num16 & 0xff00) >> 8) | ((num16 & 0x00ff) << 8);
  } else {
    return num16;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cart2_3d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  addsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double cart2_3d(double *x1, double *x2) {
  double out;
  double a;
  a = *(x2++) - *(x1++);
  out = a * a;
  a = *(x2++) - *(x1++);
  out += a * a;
  a = *(x2++) - *(x1++);
  return (out + a * a);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cart2_3d>:
  endbr64
  movsd  (%rsi),%xmm0
  subsd  (%rdi),%xmm0
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  movsd  0x8(%rsi),%xmm0
  subsd  0x8(%rdi),%xmm0
  movsd  0x10(%rsi),%xmm1
  subsd  0x10(%rdi),%xmm1
  mulsd  %xmm0,%xmm0
  addsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double cart2_3d(double *x1, double *x2) {
  double out;
  double a;
  a = *(x2++) - *(x1++);
  out = a * a;
  a = *(x2++) - *(x1++);
  out += a * a;
  a = *(x2++) - *(x1++);
  return (out + a * a);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cart2_3d>:
  endbr64
  movsd  (%rsi),%xmm0
  subsd  (%rdi),%xmm0
  movsd  0x10(%rsi),%xmm1
  subsd  0x10(%rdi),%xmm1
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  movsd  0x8(%rsi),%xmm0
  subsd  0x8(%rdi),%xmm0
  mulsd  %xmm1,%xmm1
  mulsd  %xmm0,%xmm0
  addsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double cart2_3d(double *x1, double *x2) {
  double out;
  double a;
  a = *(x2++) - *(x1++);
  out = a * a;
  a = *(x2++) - *(x1++);
  out += a * a;
  a = *(x2++) - *(x1++);
  return (out + a * a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cart2_3d>:
  endbr64
  movsd  (%rsi),%xmm0
  subsd  (%rdi),%xmm0
  movsd  0x10(%rsi),%xmm1
  subsd  0x10(%rdi),%xmm1
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  movsd  0x8(%rsi),%xmm0
  subsd  0x8(%rdi),%xmm0
  mulsd  %xmm1,%xmm1
  mulsd  %xmm0,%xmm0
  addsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double cart2_3d(double *x1, double *x2) {
  double out;
  double a;
  a = *(x2++) - *(x1++);
  out = a * a;
  a = *(x2++) - *(x1++);
  out += a * a;
  a = *(x2++) - *(x1++);
  return (out + a * a);
}
```
user: Translate the below assembly code into C programming language:

```asm
<str_replace>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %rsi,%rsi
  je     L23
  mov    %rsi,%r13
  mov    %rdx,%r14
  test   %rdx,%rdx
  je     L23
  call   <strdup@plt>
  mov    %rax,%rbp
L52:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strstr@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L13
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r13,%rdi
  call   <strlen@plt>
  sub    %eax,%ebx
  mov    %r14,%rdi
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L57
  mov    %r12,%r15
  sub    %rbp,%r15
  mov    %r15d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    (%rbx,%r15,1),%rdi
  mov    %r14,%rsi
  call   <memcpy@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,0x8(%rsp)
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,0xc(%rsp)
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  add    %rax,%r12
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    0x8(%rsp),%edx
  sub    0xc(%rsp),%edx
  sub    %r15d,%edx
  movslq %eax,%rdi
  add    %r15,%rdi
  add    %rbx,%rdi
  mov    %r12,%rsi
  call   <memcpy@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%r15d
  mov    %r14,%rdi
  call   <strlen@plt>
  movslq %r12d,%r12
  cltq
  add    %rax,%r12
  movslq %r15d,%r15
  sub    %r15,%r12
  lea    (%rbx,%r12,1),%rdi
  mov    $0x1,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rbp
  jmp    L52
L23:
  call   <strdup@plt>
  mov    %rax,%rbp
L13:
  mov    %rbp,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L57:
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rbp
  jmp    L13
```

assistant: ```c
char *str_replace(const char *string, const char *substr, const char *replacement) {
  char *tok = NULL;
  char *newstr = NULL;
  char *oldstr = NULL;
  if (substr == NULL || replacement == NULL)
    return strdup(string);
  newstr = strdup(string);
  while ((tok = strstr(newstr, substr))) {
    oldstr = newstr;
    newstr = malloc(strlen(oldstr) - strlen(substr) + strlen(replacement) + 1);
    if (newstr == NULL) {
      free(oldstr);
      return NULL;
    }
    memcpy(newstr, oldstr, tok - oldstr);
    memcpy(newstr + (tok - oldstr), replacement, strlen(replacement));
    memcpy(newstr + (tok - oldstr) + strlen(replacement), tok + strlen(substr), strlen(oldstr) - strlen(substr) - (tok - oldstr));
    memset(newstr + strlen(oldstr) - strlen(substr) + strlen(replacement), 0, 1);
    free(oldstr);
  }
  return newstr;
}
```
user: What is the c source code of the assembly code below:

```asm
<str_replace>:
  endbr64
  test   %rsi,%rsi
  je     L40
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %rdx,%rdx
  je     L51
  mov    %rsi,%r12
  call   <strdup@plt>
  mov    %rax,%r15
  jmp    L45
  nopw   0x0(%rax,%rax,1)
L33:
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %r13,%rdi
  sub    %eax,%ebx
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L20
  mov    %r14,%rbx
  mov    %r15,%rsi
  mov    %rax,%rdi
  sub    %r15,%rbx
  mov    %ebx,%edx
  call   <memcpy@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rsi
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %eax,%edx
  call   <memcpy@plt>
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,0x8(%rsp)
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  mov    %r13,%rdi
  cltq
  add    %rax,%r14
  call   <strlen@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edx
  movslq %eax,%rdi
  sub    %esi,%edx
  add    %rbx,%rdi
  mov    %r14,%rsi
  sub    %ebx,%edx
  add    %rbp,%rdi
  call   <memcpy@plt>
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  movslq %eax,%rbx
  call   <strlen@plt>
  mov    %r13,%rdi
  movslq %eax,%r14
  call   <strlen@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  cltq
  add    %rax,%rbx
  sub    %r14,%rbx
  lea    0x0(%rbp,%rbx,1),%rdi
  call   <memset@plt>
  mov    %r15,%rdi
  mov    %rbp,%r15
  call   <free@plt>
L45:
  mov    %r12,%rsi
  mov    %r15,%rdi
  call   <strstr@plt>
  mov    %rax,%r14
  test   %rax,%rax
  jne    L33
L43:
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  mov    %r15,%rdi
  xor    %r15d,%r15d
  call   <free@plt>
  jmp    L43
  nopl   (%rax)
L51:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1100 <strdup@plt>
  nopl   0x0(%rax,%rax,1)
L40:
  jmp    1100 <strdup@plt>
```

assistant: ```c
char *str_replace(const char *string, const char *substr, const char *replacement) {
  char *tok = NULL;
  char *newstr = NULL;
  char *oldstr = NULL;
  if (substr == NULL || replacement == NULL)
    return strdup(string);
  newstr = strdup(string);
  while ((tok = strstr(newstr, substr))) {
    oldstr = newstr;
    newstr = malloc(strlen(oldstr) - strlen(substr) + strlen(replacement) + 1);
    if (newstr == NULL) {
      free(oldstr);
      return NULL;
    }
    memcpy(newstr, oldstr, tok - oldstr);
    memcpy(newstr + (tok - oldstr), replacement, strlen(replacement));
    memcpy(newstr + (tok - oldstr) + strlen(replacement), tok + strlen(substr), strlen(oldstr) - strlen(substr) - (tok - oldstr));
    memset(newstr + strlen(oldstr) - strlen(substr) + strlen(replacement), 0, 1);
    free(oldstr);
  }
  return newstr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<str_replace>:
  endbr64
  test   %rsi,%rsi
  je     L13
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %rdx,%rdx
  je     L35
  mov    %rsi,%r12
  call   <strdup@plt>
  mov    %rax,%r15
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L59:
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %r13,%rdi
  sub    %eax,%ebx
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L40
  mov    %r14,%rbx
  mov    %r15,%rsi
  mov    %rax,%rdi
  sub    %r15,%rbx
  mov    %ebx,%edx
  call   <memcpy@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rsi
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %eax,%edx
  call   <memcpy@plt>
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,0x8(%rsp)
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  mov    %r13,%rdi
  cltq
  add    %rax,%r14
  call   <strlen@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edx
  movslq %eax,%rdi
  sub    %esi,%edx
  add    %rbx,%rdi
  mov    %r14,%rsi
  sub    %ebx,%edx
  add    %rbp,%rdi
  call   <memcpy@plt>
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  movslq %eax,%rbx
  call   <strlen@plt>
  mov    %r13,%rdi
  movslq %eax,%r14
  call   <strlen@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  cltq
  add    %rax,%rbx
  sub    %r14,%rbx
  lea    0x0(%rbp,%rbx,1),%rdi
  call   <memset@plt>
  mov    %r15,%rdi
  mov    %rbp,%r15
  call   <free@plt>
L0:
  mov    %r12,%rsi
  mov    %r15,%rdi
  call   <strstr@plt>
  mov    %rax,%r14
  test   %rax,%rax
  jne    L59
L23:
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L40:
  mov    %r15,%rdi
  xor    %r15d,%r15d
  call   <free@plt>
  jmp    L23
  nopl   (%rax)
L35:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1100 <strdup@plt>
  nopl   0x0(%rax,%rax,1)
L13:
  jmp    1100 <strdup@plt>
```

assistant: ```c
char *str_replace(const char *string, const char *substr, const char *replacement) {
  char *tok = NULL;
  char *newstr = NULL;
  char *oldstr = NULL;
  if (substr == NULL || replacement == NULL)
    return strdup(string);
  newstr = strdup(string);
  while ((tok = strstr(newstr, substr))) {
    oldstr = newstr;
    newstr = malloc(strlen(oldstr) - strlen(substr) + strlen(replacement) + 1);
    if (newstr == NULL) {
      free(oldstr);
      return NULL;
    }
    memcpy(newstr, oldstr, tok - oldstr);
    memcpy(newstr + (tok - oldstr), replacement, strlen(replacement));
    memcpy(newstr + (tok - oldstr) + strlen(replacement), tok + strlen(substr), strlen(oldstr) - strlen(substr) - (tok - oldstr));
    memset(newstr + strlen(oldstr) - strlen(substr) + strlen(replacement), 0, 1);
    free(oldstr);
  }
  return newstr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<aloneBubbleSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L62
L69:
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <printarray@plt>
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L28
  cmpl   $0x0,-0x10(%rbp)
  jne    L18
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L18:
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  cmpl   $0x0,-0xc(%rbp)
  jle    L70
  subl   $0x2,-0xc(%rbp)
  jmp    L6
L70:
  movl   $0x0,-0xc(%rbp)
  jmp    L6
L28:
  cmpl   $0x1,-0x10(%rbp)
  jne    L6
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
L6:
  addl   $0x1,-0xc(%rbp)
L62:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L69
  nop
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D12
datatype: string
name: read_data
arguments: tag: D5
datatype: string

tool: D12: first reverse place at %d	

tool: D5: exchange(%d,%d)


assistant: ```c
void aloneBubbleSort(int a[], int n) {
  int f = 0;
  int i, d = 0;
  for (i = 0; i < n - 1; i++) {
    printf("first reverse place at %d\t", d);
    printarray(a, n);
    if (a[i] > a[i + 1]) {
      if (f == 0) {
        d = i;
      }
      printf("exchange(%d,%d)\n", a[i], a[i + 1]);
      f = 1;
      int t = a[i];
      a[i] = a[i + 1];
      a[i + 1] = t;
      if (i > 0) {
        i -= 2;
      } else {
        i = 0;
      }
    } else {
      if (f == 1) {
        i = d;
        f = 0;
      }
    }
  }
}
```