C 低级别IO读（）和写（）

我正在编写一个程序，该程序分为多个部分，相互构建。第一个程序必须从文件中读取内容，并将内容按空格分割写入新文件。程序二应该把这些单词加上pig拉丁语，这是基于它是以元音还是辅音开头的规则，并根据它以哪一个开头在末尾加上一些字符串。我可以打开文件并读取内容，但我在查找和附加适当的规则以打印到新文件时遇到困难

          int processingWord = 1; //0 is not in the middle of a word and 1 is in the middle
  char c;
  int bytes; //Should be holding the position of the pointer in the file
  while((bytes = read(fileRead, &c, sizeof(c))) > 0) {
    //printf("%c", c);
    if(processingWord == 0) {
      processingWord = 1;
    }
    if(processingWord == 1) {
      //Figure out if a vowel or not
      if(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U') {
        //Increment the first counter
        shard1Count++;
      }
      //Get to the end of the string
      if(c == '\n') {
        c = 'r';
        write(fileWrite, &c, sizeof(c));
        c = 'a';
        write(fileWrite, &c, sizeof(c));
        c = 'y';
        write(fileWrite, &c, sizeof(c));
        c = '\n';
        write(fileWrite, &c, sizeof(c));
        processingWord = 0;
      }
    }
    write(fileWrite, &c, sizeof(c));
  }

这就是我试图查找并在单词末尾附加新的ray字符串（如果它以元音开头）的地方。文本文件如下所示

It
is
the
Zucca
Gigantopithecus,
or
Great
Pumpkin,
Charlie
Brown.

在一个新文件中，输出应该是这样的

Itray 
    isray 
    hetay 
    uccaZay 
    igantopithecusGay, 
    orray
    reatGay
    umpkinPay, 
    harlieCay 
    rownBay.

编辑：processsingWord是一个想法，我必须在检查元音之前检查我是否在这行的末尾。但这种逻辑并没有起作用，结果完全是胡说八道。 我当前的输出文件如下所示：

Itray

isray

theray

Zuccaray

Gigantopithecus,ray

orray

Greatray

Pumpkin,ray

Charlieray

Brown.ray

ray

---

以下是一个应该有效的实现：

void doStuff(void);
int startsWithVowel(char c);
int isALetter(char c);


void doStuff(){
    int processingWord = 0; 
    int already_latin = 0;
    char c = 0;
    char first_letter = 0;
    while(read(fileRead, &c, sizeof(c)) > 0) {
        if(processingWord == 0) {
            processingWord = 1;
            if(!startsWithVowel(c)){ //append constants to the end of the word in pig latin *EDIT*
                first_letter = c;
                continue;//Here we do not fall through and write
            }
        }
        else{
            if(isALetter(c)){ //This is the general case of just writing the read character
                write(fileWrite, &c, sizeof(c));
            }
            else if(c != '\n'){ //Here is handling for , and . special characters
                if(isALetter(first_letter)){ //we hit a .  or , with a vower word, need to add first letter then "ray"
                    write(fileWrite, &first_letter, sizeof(first_letter));
                }
                write(fileWrite, "ray", sizeof("ray"));
                write(fileWrite, &c, sizeof(c));
                already_latin = 1;
            }
            else if(c == '\n') { //here is the end of the string
                if(isALetter(first_letter)){
                    write(fileWrite, &first_letter, sizeof(first_letter));
                }
                if(!already_latin){
                    write(fileWrite, "ray", sizeof("ray"));
                }
                write(fileWrite, &c, sizeof(c));

                processingWord = 0; //reset all the flags for the next word.
                first_letter = 0;
                already_latin = 0;
            }//end of '\n'
        }//end of if/else block
    }//end of while loop
}//end of function


/* =========================================================
return true (1) if the character is a vowel and 0 otherwise
============================================================ */

int startsWithVowel(char c){
    if(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U') {
        return 1;
    }
    return 0;
}

/* =========================================================
return true (1) if the character is a letter and 0 otherwise
============================================================ */
int isALetter(char c){
    if (((c >= 'A') && (c <= 'Z')) || ((c >= 'a') && (c <= 'z'))){
        return 1;
    }
    return 0;
}

还有一堆未使用的东西，比如bytes变量，事情肯定会更干净，但这应该按照您需要的方式工作。试着运行它，让我知道它是如何运行的，如果我还在这里，我今晚会更新任何错误

编辑 看起来我是倒着做的，只是交换元音而不是常量。我的猪拉丁语生锈了

好的，我制作了一个本地字符串，用codechef.com/ide在线测试解析，您可以复制并粘贴到那里进行验证。将printfs更改为writes，它模仿printfs，我认为您可以这样做：

#include <stdio.h>
#include <string.h>

void doStuff(void);
int startsWithVowel(char c);
int isALetter(char c);

char * str = "It\nis\nthe\nZucca\nGigantopithecus,\nor\nGreat\nPumpkin,\nCharlie\nBrown.";

int main(void) {

    doStuff();

    return 0;
}

void doStuff(){
    int processingWord = 0; 
    char c = 0;
    char first_letter = 0;
    int already_latin = 0;
    //while(read(fileRead, &c, sizeof(c)) > 0) {
    while(strlen(str) > 0){        //Made local for testing, no file io here
        c = str[0];
        str++;                    //end of local nonsense you wont have to use
        if(processingWord == 0) {
            processingWord = 1;
            if(!startsWithVowel(c)){
                first_letter = c;
                continue;//Here we don not fall through and write
            }
        }
        if(processingWord == 1) {
            if(isALetter(c)){ //This is the general case of just writing the read character
                //write(fileWrite, &c, sizeof(c));
                printf("%c",c);
                //printf(" SHOULD PRINT FIRST LETTER VOWEL HERE ");
            }
            else if(c != '\n'){ //Here is handling for , and . special characters
                if(isALetter(first_letter)){ //we hit a .  or , with a vower word, need to add first letter then "ray"
                    //write(fileWrite, &first_letter, sizeof(first_letter));
                    printf("%cay%c",first_letter,c);
                }
                else{
                    //write(fileWrite, "ray", sizeof("ray"));
                    //write(fileWrite, &c, sizeof(c));
                    printf("ray%c", c);   
                }
                already_latin = 1;
            }
            else if(c == '\n') { //here is the end of the string
                if(!already_latin){
                    if(isALetter(first_letter)){
                        //write(fileWrite, &first_letter, sizeof(first_letter));
                        printf("%cay",first_letter);
                        //printf(" SHOULD PRINT FIRST LETTER CONSTANT HERE  ");
                    }
                    else{
                        //write(fileWrite, "ray", sizeof("ray"));
                        printf("ray");
                    }
                }
                //write(fileWrite, &c, sizeof(c));
                printf("%c", c);
                processingWord = 0;
                first_letter = 0;
                already_latin = 0;
            }//end of '\n'
        }//end of if/else block
    }//end of while loop
}//end of function


/* =========================================================
return true (1) if the character is a vowel and 0 otherwise
============================================================ */

int startsWithVowel(char c){
    if(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U') {
        return 1;
    }
    return 0;
}

/* =========================================================
return true (1) if the character is a letter and 0 otherwise
============================================================ */
int isALetter(char c){
    if (((c >= 'A') && (c <= 'Z')) || ((c >= 'a') && (c <= 'z'))){
        return 1;
    }
    return 0;
}

输出： 伊特雷 伊斯雷 赫泰 加州大学 伊甘托古猿盖， 奥雷 雷特盖 umpkinPay， 哈利埃

---

rownBay。

您的评论说processingWord应该是0或1，但在您的程序中始终是1。你是想在某个时候把它设回零吗？是的，这就是我最初的逻辑。在检查元音等之前，我有一个if语句首先检查我是否在这行的末尾。当我检查temp2.data时，输出的都是jibberish和一堆？标志。我出于测试目的无意中删除了它。当前的输出是什么，看起来strcat没有做你认为它应该做的事情。使用调试器或sprintf，尽管如果使用strcat，它应该可以正常工作。同样，将缓冲区初始化为0，如下所示：char BUFFER[250]={0}；当我使用缓冲区时，我的输出只是作为输出的随机符号。我没有得到像IIt、iis、E等那样的输出。。所以我越来越近了，但是我不能正确地附加光线。我已经更新了上面的代码。Printf strlen是什么，我怀疑您的字符串不是以null结尾的，因此strcat没有正确追加。初始化为零将有助于第一次写入iTray感谢您的响应。事实上，我正在使它与我的逻辑一起工作，尽管它是混乱和漫长的。虽然我能相对容易地理解你的代码。非常感谢您的回答，我将在尝试我的实现后尝试。一些清理：如果您将continue移到If之外，您可以消除后面的最外层，即更少的缩进。同样，你也可以在同一级别添加continue to if。我在工作时使用了2个空格缩进，codechef上的这4个空格缩进在我看来非常大。只要它起作用，我就让OP按照他想要的方式来做，我想这是他的家庭作业。@SaiPeri如果你最终使用了这个工作示例代码，你能接受这个评论作为你问题的答案吗。我最终得到了我的实现，因为我害怕因为学术上的不诚实而被抓住，但在我们的第二部分，我们必须利用线程，所以我一定会看看你的实现，让我的生活更轻松当我们接到下一个任务时。非常感谢你