在循环中使用 strtof 解析字符缓冲区中的数字

Using strtof in a loop to parse numbers from a char buffer

我有一个关于分配和释放内存的问题。

我想在循环中读取一个char缓冲区并将float值保存到一个vector中。 我通过读取 fstream 获取缓冲区。

但是我的方法在最后删除缓冲区时总是崩溃。

我在循环期间更改缓冲区是否有问题? 有人知道如何解决这个问题吗?

感谢每一个提示!

char* buffer1 = new char[size]; // the size is given
char* buffer2 = NULL;

fileStream.read(buffer1,size);

while(true)
{
  // read double from buffer
  // and get pointer to the new buffer -> buffer2
  double tempDouble = strtod(buffer1, &buffer2);

  // If nothing has been read (buffer stays the same) -> break
  if (buffer1 == buffer2)   
      break;
  else // change the buffer to read the next double during the next interation step 
      buffer1= buffer2;

  // collect the doubles
  collectedDoubles.push_back(tempDouble);

  // if all numbers are read -> break
  if (++doubleCount == numDoubles) // termination condition
    break;
}

// delete the allocated buffer
delete[] buffer1;

// do I have th delete the 2nd buffer too?
// delete[] buffer2;
  1. 根据strtod的文档:

    The functions sets the pointer pointed to by str_end to point to the character past the last character interpreted. If str_end is NULL, it is ignored.

    所以你的指针 buffer2 仍然是 NULL 并且在你执行 buffer1= buffer2; - buffer1 现在也是 NULL (顺便说一句,这里是内存泄漏因为数据丢失了)。

  2. do I have th delete the 2nd buffer too?

    在这种情况下 - 不,因为删除 NULL 指针是 no-operation.

解法:

看一下文档中为strtod函数提供的示例,根据您的代码,这里是类似的:

char* buffer1 = new char[size];
char* buffer2;                           // note no NULL here !
char* p = buffer1;                       // we'll modify this in loop

for (double tempDouble = std::strtod(p, &buffer2); p != buffer2; tempDouble = std::strtod(p, &buffer2))
{
    p = buffer2;
    if (errno == ERANGE){                // check if some error occured during attempt to convertion
        std::cout << "range error\n";
        errno = 0;
    }

    collectedDoubles.push_back(tempDouble);

    if (++doubleCount == numDoubles)     // termination condition
        break;
}

delete[] buffer1;

编辑 1: 看看@JerryCoffin 在对您的问题的评论中建议的优雅且非常 'C++ like' 的解决方案。