簡體   English   中英

嘗試模擬grep命令

[英]Trying to simulate grep command

編譯代碼時出現分段錯誤。

如果有人可以幫助我,我不會明白我的代碼有什么問題會很高興。

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

int main(int argc,char *argv[])
{
FILE *fp;
char fline[100];
char *newline;
int i,count=0,occ=0;

fp=fopen(argv[1],"r");

while(fgets(fline,100,fp)!=NULL)
{
count++;
    if(newline=strchr(fline,'\n'))
        *newline='\0';
    if(strstr(fline,argv[2])!=NULL)
    {
        printf("%s %d %s",argv[1],count,fline);
    occ++;  
    }


}

printf("\n Occurence= %d",occ);

return 1;
}

看到男人打開男人打開

FILE *fp;
...
fp=open(argv[1],"r");

open返回一個整數,而不是文件指針。 只需將該行更改為

fp=fopen(argv[1],"r");

注意:對於那些想知道這是關於什么的人,OP從問題代碼中編輯了此錯誤。

這導致我們(也解決了其他一些小問題-請參閱評論):

+ EDIT:指向應該進行錯誤檢查的地方:

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

int main(int argc, char *argv[]) {
    FILE *fp;
    char fline[100];
    char *newline;
    int i, count = 0, occ = 0;

    // for starters, ensure that enough arguments were passed:
    if (argc < 3) {
      printf("Not enough command line parameters given!\n");
      return 3;
    } 

    fp = fopen(argv[1], "r");
    // fopen will return if something goes wrong.  In that case errno will
    // contain the error code describing the problem (could be used with
    // strerror to produce a user friendly error message
    if (fp == NULL) {
      printf("File could not be opened, found or whatever, errno is %d\n",errno);
      return 3;
    } 


    while (fgets(fline, 100, fp) != NULL) {
        count++;
        if (newline = strchr(fline, '\n'))
            *newline = '\0';
        if (strstr(fline, argv[2]) != NULL) {
            // you probably want each found line on a separate line,
            // so I added \n
            printf("%s %d %s\n", argv[1], count, fline);
            occ++;
        }
    }
    // it's good practice to end your last print in \n
    // that way at least your command prompt stars in the left column
    printf("\n Occurence= %d", occ);

    return 1;
}

ps:所以錯誤發生在運行 時而不是編譯期間 -這種區別非常關鍵,因為尋找編譯器故障並解決庫使用錯誤需要非常不同的技術...

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM