C delete chars from string - c

Question

I have program that asks to enter a string (mystring) and a char (ch). Then it deletes all entered chars (ch) from the string (mystring). For example "abcabc" and char 'a' then the result shoud be "bcbc".
-When I use scanf the program works nicely if the string does not have spaces. If I enter "abc abc abc" It reads and processes only the first 3 letters (until space).
Then I was advised to use gets(mystr); because it can read all the stirng. But when I use gets the result is the same as the input string and nothing happens.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 100
int main(int argc, char *argv[])
{
char mystr[N] ,result[N];
char ch;
int i,k;
k=0;
printf("enter string \n");
//gets(mystr);///////////////////////////
//scanf("%s",&mystr);///////////////////
printf("enter char \n");
scanf("%c",&ch);
scanf("%c",&ch);
for ( i = 0; i <= strlen(mystr); i++ )
{
if (mystr[i] != ch)
{
result[k]=mystr[i];
k++;
}
}
puts(result);
system("pause");
return 0;
}

Answer 1

scanf("%c",&ch);
scanf("%c",&ch);
That second scanf is your problem. It's picking up the new-line character that you enter after the letter you want to remove (and overwrites the previous value of ch).
Get rid of it.
Please note, as the man page says:
Never use gets(). Because it is impossible to tell without knowing the data in advance how many
characters gets() will read, and because gets() will continue to store characters past the end of
the buffer, it is extremely dangerous to use. It has been used to break computer security. Use
fgets() instead.

Answer 2

hmm - not sure what the problem is - use getstr, but not scanf for the string, and it works for me in visual studio
int main(int argc, char *argv[])
{
char mystr[N] ,result[N];
char ch;
int i,k;
k=0;
printf("enter string \n");
gets(mystr);///////////////////////////
//scanf("%s",&mystr);///////////////////
printf("enter char \n");
scanf("%c",&ch);
// scanf("%c",&ch);
for ( i = 0; i <= strlen(mystr); i++ )
{
if (mystr[i] != ch)
{
result[k]=mystr[i];
k++;
}
}
puts(result);
system("pause");
return 0;
}

Answer 3

Use this one:
char temp[2];
scanf("%1s",temp);
ch = temp[0];
and use gets
scanf when used with chars has some problems (it gets the "old" new line). Here we "cheat" a little and we use scanf to get a string that can have up to one character. A string of 1 character clearly needs a second character for the terminator, so an array of 2 characters.
Be aware that using a scanf for the character to search, you won't be able to insert the space character.
Note that gets is an "evil" function. You can easily do buffer overruns using it (it doesn't check that the buffer is big enough). The "right" way to do it is normally: fgets(mystr, N, stdin); (the "file" variant of gets has a maximum number of characters that can be read and will append a \0 at the end). Note that if you insert 150 characters in a fgets, 99 will go to your string (because you gave 100 of max size), 1x \0 will be appended and the other characters will remain in the buffer "ready" for the next scanf/gets/fgets... (to test it, reduce the buffer to a smaller value, like 5 characters, and do some tests)

Answer 4

You can use fgets() as suggested by xanatos with a small hack, so you can reliably handle return characters. Just change the '\n' to '\0' in the string obtained using fgets.
And in your program, you forgot to terminate the new string with a '\0'.
So here's the code you're looking for.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define N 100
int main(int argc,char **argv){
char string[N],str1[N];
char ch;
int i,k = 0;
fgets(string,N,stdin);
string[strlen(string)-1] = '\0';
scanf("%c",&ch);
printf("\n%s , %c",string,ch);
for (i=0;i<=strlen(string);i++)
if(string[i] != ch)
str1[k++] = string[i];
str1[k] = '\0';
printf("\n%s , %s\n",string,str1);
return 0;
}