c++ - fread() 调用后的段错误

Question

我有以下代码:

char*
Sender::PrepareData(char* filename, unsigned long long int bytesToTransfer)
{
    FILE* dataFile = fopen(filename, "rb"); 
    if (dataFile==NULL) {fputs ("File error",stderr); exit (1);}

    cout << "File Open: " << filename << endl;

    char* theData;
    size_t bytesRead = fread(&theData, 1, bytesToTransfer, dataFile);
    if (bytesRead != bytesToTransfer) {fputs ("Reading error",stderr); exit (3);}

    cout << "Data Read -- Num Bytes: " << bytesRead << endl;
    cout << "Data to Send: " << *theData << endl;

    return theData;
}

当这个方法被命中时，我的输出是:

File Open: t.bin

Data Read -- Num Bytes: 10

Segmentation fault (core dumped)

我的 t.bin 文件包含以下内容:

This is a test.

98172398172837129837

alsjdf89u32ijofiou2

TEST TEST...

!!## TESTING TEST!! ###(DLKAJ)

当我运行 gdb 时，段错误输出是:

File Open: t.bin Data Read -- Num Bytes: 10

Program received signal SIGSEGV, Segmentation fault. 0x00000000004015e2 in Sender::PrepareData (this=0x603010, filename=0x7fffffffe363 "t.bin", bytesToTransfer=10)
    at sender.cpp:98  98        cout << "Data to Send: " << *theData << endl;

谁能告诉我我做错了什么？

Answer 1

你需要一个缓冲区，theData 只是一个指针。

有点像

char theData[1000];
size_t bytesRead = fread(theData, 1, bytesToTransfer, dataFile);

可能取决于 bytesToTransfer 的最大值

如果您确定正在读取一个字符串，您可能还需要在写入 cout 之前终止 theData，

theData[bytesRead] = '\0';

如果要返回缓冲区，则需要在堆上分配缓冲区。

做这样的事情会容易得多

std::vector<char>
PrepareData(char* filename, unsigned long long int bytesToTransfer)
{
  std::ifstream file(filename, file.binary);

  if (!file) {
    std::cerr << "File error";
    exit(1);
  }

  std::cout << "File Open: " << filename << '\n';

  std::vector<char> data(bytesToTransfer + 1);
  file.read(data.data(), bytesToTransfer);
  data.back() = '\0';

  if (!file) {
    std::cerr << "Reading error";
    exit(3);
  }

  std::cout << "Data Read -- Num Bytes: " << bytesToTransfer << '\n';
  std::cout << "Data to Send: " << data.data() << '\n';
  return data;
}

如果您所做的只是从文件中读取 char，您应该考虑使用 string。