Using self made strstr() function to check an embedded string in some other string and then replacing it by some other string (input by user)

Question

I am writing a program that uses an array of pointers to strings str[].It receives two strings str1 and str2 and check if str1 is embedded in any of the strings in str[].If str1 is found, then replace it with str2. for example: if str1 contains "mountain" and str2 contains "car", then the second string in str should get changed to "Move a car" .When i try to execute this program , in gcc it is crashing,and when i tried to compile it online it is giving segmentation fault ,i am not able to understand where this program went wrong and what correction is needed in this.

#include<stdio.h>
#include<string.h>
char* xstrstr(char* ,char* );
int main()
{
char *str[]={
                "We will teach you how to...",
                "Move a mountain",
                "Level a building",
                "Erase the past",
                "Make a million",
                "...all through C!"
            };
char* p, *temp, *cpy;
int len,i,j;
char str1[28];
char str2[10];
printf("Enter string1\n");
scanf("%s",str1);
printf("Enter string 2\n");
scanf("%s",str2);
printf("After modifying the saved strings\n\n");
for(i=0;i<6;i++)
{
    p=xstrstr(str[i],str1);
    if(p)
    {
       /*Copy the remaining string*/
       temp=p+strlen(str1);
       strcpy(cpy,temp);
       /*Replace the old string*/
       strcpy(p,str2);
      /*Finally append the remaining part*/
       strcat(p,cpy);
       break;
    }
}
if(p==NULL)
{
    printf("No string match found\n");
    return 1;
}
for(i=0;i<6;i++)
{
    printf("%s\n",str[i]);
}
return 0;
}
char* xstrstr(char *string1,char* string2)
{
    while(*string1)
    {
        char* begin=string1;
        char* find=string2;

           while((*string1)&&(*find)&&(*string1==*find))
           {
               string1++;
               find++;
           }
           if(!(*find))
           {
               return begin;
           }
           string1=begin+1;
    }
return NULL;
}

I am lazy to dig into this code, but I suspect you are trying to write a string literal. `str` is array of pointers to literals. — Eugene Sh., Aug 07 '18 at 14:37
I will not even take a look on it. You did not even bother to format it properly. — 0___________, Aug 07 '18 at 14:40
in place of xstrstr use library function strstr to analyze my code,it will be very much easy for you to go through my code — LocalHost, Aug 07 '18 at 14:41
Take the time to consistently indent and format your code. Failure to do so makes your code almost unreadable - even to you. — Andrew Henle, Aug 07 '18 at 14:43
I suggest you read and follow https://ericlippert.com/2014/03/05/how-to-debug-small-programs/ — SergeyA, Aug 07 '18 at 14:43
sorry to all for my bad code , i will properly indent it and update it so it becomes easy to read. — LocalHost, Aug 07 '18 at 14:48

score 1 · Accepted Answer · answered Aug 07 '18 at 14:45

1

Two main problems cpy is not initalised

 strcpy(cpy,temp);

p points to the read only string literal and you write to it.

 strcpy(p,str2);

answered Aug 07 '18 at 14:45

0___________

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Andre Motta · Answer 2 · 2018-08-07T14:45:23.410

Namely you are using strcpy to copy strings but your destiny string is just a pointer with no size. You need to do a malloc first. That's why it is giving you a segmentation fault error. Beware of pointers without memory allocation and it should work. Please update your code after this and if there is any other error I will edit this answer

char *cpy;

These variable is never initialized. But still you strcpy to it bellow.

EDIT: just go with

cpy = malloc(strlen(sourcestring)+1);

Using self made strstr() function to check an embedded string in some other string and then replacing it by some other string (input by user)

2 Answers2