In this tutorial, you will learn and get code about deleting a word from a string. Here the program will ask from user to enter any string, and then a word to delete the given word from given string. The program is created with and without using two dimensional array.
To delete particular word from the given string in C programming, you have to ask from user to enter the string and then ask to enter the word that has to be delete. Then check for the presence of given word in the string. If found, then delete it as shown in the program given below:
// ----codescracker.com----
#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
char str[100], word[20];
int i, j, ls, lw, temp, chk=0;
printf("Enter the String: ");
gets(str);
printf("Enter a Word: ");
gets(word);
ls = strlen(str);
lw = strlen(word);
for(i=0; i<ls; i++)
{
temp = i;
for(j=0; j<lw; j++)
{
if(str[i]==word[j])
i++;
}
chk = i-temp;
if(chk==lw)
{
i = temp;
for(j=i; j<(ls-lw); j++)
str[j] = str[j+lw];
ls = ls-lw;
str[j]='\0';
}
}
printf("\nNew String = %s", str);
getch();
return 0;
}
This program was build and run under Code::Blocks IDE. Here is its sample run:
Now supply any string say this is codescracker and then enter a word say codescracker. Press
ENTER key to see following output:
In the above program, the statement:
for(j=0; j<lw; j++)
{
if(str[i]==word[j])
i++;
}
is used to check for the presence of word (to be delete) with character-by-character matching. But before statement, the value of i gets initialized to temp. So that after the statement, we can check whether the if statement gets evaluated for the number of times that equals to the length of the word or not. If it is, then the word is found. Therefore, the statement:
for(j=i; j<(ls-lw); j++)
str[j] = str[j+lw];
is used to move all the characters (after last character's index of word) to the place from word's first character's index in the string. That is, if word is found in given string at index number 5. And the length of word is 3, therefore we have to move character at 8th index to 5th, and character at 9th index to 6th index, and so on upto the last character of the string.
Never forgot to initialize the new length of the string to ls variable. Every time the word is deleted (if more than one time, word present), just minus the length of string by the length of word. Now place a null terminated character (\0) to the last index of new string.
This program has a limitation. For example, if the string is this is codescracker, and user wants to delete is. Then is from this also gets deleted along with lone is. Therefore, the string becomes th codescracker after deleting is. To overcome this problem, we have another program as given below:
Here is the complete version of the previous program. This program also deletes duplicate spaces from the final string.
// ----codescracker.com----
#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
char str[100], word[20];
int i, j, ls, lw, temp, chk=0, doIncrement, isSpace;
printf("Enter the String: ");
gets(str);
printf("Enter a Word: ");
gets(word);
ls = strlen(str);
lw = strlen(word);
for(i=0; i<ls; i++)
{
temp = i;
doIncrement = 0;
for(j=0; j<lw; j++)
{
if(str[i]==word[j])
{
if(temp>0 && (temp+lw)<ls)
{
if(str[temp-1]== ' ' && str[temp+lw]==' ')
doIncrement=1;
}
else if(temp==0 && (temp+lw)<ls)
{
if(str[temp+lw]==' ')
doIncrement=1;
}
else if(temp>0 && (temp+lw)==ls)
{
if(str[temp-1]== ' ')
doIncrement=1;
}
if(doIncrement==1)
i++;
else
break;
}
}
chk = i-temp;
if(chk==lw)
{
i = temp;
for(j=i; j<(ls-lw); j++)
str[j] = str[j+lw];
ls = ls-lw;
i = temp;
str[j]='\0';
}
}
ls = strlen(str);
i=0;
while(str[i]!='\0')
{
isSpace = 0;
if(str[i]==' ' && str[i+1]==' ')
{
for(j=i; j<(ls-1); j++)
{
str[j] = str[j+1];
isSpace = 1;
}
}
if(i==0 && str[i]==' ')
{
for(j=i; j<(ls-1); j++)
{
str[j] = str[j+1];
isSpace = 1;
}
}
if(isSpace==0)
i++;
else
{
str[j]='\0';
ls--;
}
}
printf("\nNew String = %s", str);
getch();
return 0;
}
Now let's provide the input string as is this is is is isth codescracker and then is as the word to be delete. Here is its sample run:
Here the word this and isth included to check whether the word is (to be delete) gets matched from any of the side or not. As you can see from both side, is does match with both the words this and isth, and also deletes all the extra spaces.
Now let's create the same purpose program using two-dimensional array. Here all the word from given string gets placed in a two dimensional array named stringIn2DArray[] one by one. And then checked for the word (to be delete). If found just perform the task of deleting it as shown here:
// ----codescracker.com----
#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
char str[400], word[20], stringIn2DArray[20][20];
int i, j=0, k=0;
printf("Enter the String: ");
gets(str);
printf("Enter the Word (to be Delete): ");
gets(word);
for(i=0; str[i]!='\0'; i++)
{
if(str[i]==' ')
{
stringIn2DArray[k][j]='\0';
k++;
j=0;
}
else
{
stringIn2DArray[k][j]=str[i];
j++;
}
}
stringIn2DArray[k][j] = '\0';
j=0;
for(i=0; i<(k+1); i++)
{
if(!strcmp(stringIn2DArray[i], word))
stringIn2DArray[i][j]='\0';
}
printf("\nThe New String is: ");
j=0;
for(i=0; i<(k+1); i++)
{
if(stringIn2DArray[i][j] == '\0')
continue;
else
printf("%s ", stringIn2DArray[i]);
}
getch();
return 0;
}
This program produces the same output as of previous one.
© Copyright 2021. All Rights Reserved.