我运行程序时终端挂起

时间:2014-02-17 01:04:11

标签: c linux terminal

在unix中编写了一些代码,通过函数计算argv [1]中的单词数。返回结果并显示在stdout上。

当我运行它时,该过程一直持续到我杀死它为止。没有错误显示或任何事情?有人会介意看看。

感谢

#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>

//function declaration
int countWords(char []);

int main(int argc, char* argv [])
{
  int words;

  //check 3 entered values
  if (argc != 3)
  {
    write(2,"Please enter 2 values. Seperated by Space \n", 44);
    exit(0);
  }

  words = countWords(argv[1]);
  printf("Words are %i \n", words);
  return 0;
}

//function to count words
int countWords(char a [])
{
  int counter, openStream, oTest;
  char letter;

  openStream = open(a,O_RDONLY);
  if (openStream < 0)
  {
    write(2, "Error opening specified file. \n", 32); 
    exit(1);
  }

  oTest = read(openStream, &letter, 1);
  while (oTest != 0)
  {
    if (oTest == -1)
    {
      write(2, "Error reading file \n",21);
      exit(2);
    } 
    if (oTest == '\n' || oTest == ' ')
    {
      counter++;
    }
  }
  close(openStream);
  return counter;
}

1 个答案:

答案 0 :(得分:2)

当你的输入流中还有剩余的东西时,你正在循环,但你从来没有从循环内的输入流中实际读取,这意味着你的输入流是永远的,并且永远只是超过它的第一个字符,你的oTest(第一个字符)永远不会改变。